Multivariable & Vector Calculus

1. The boundary consists of all points where the inequalities become equalities.
2. The side is the surface $x^2 + y^2 = 1$ for $0 \le z \le 1$.
3. The top is the disk $x^2 + y^2 \le 1$ at $z = 1$.
4. The bottom is the disk $x^2 + y^2 \le 1$ at $z = 0$.

Answer: The boundary is the union of the lateral cylinder wall and the two circular end-caps.

1. The boundary interior principle states that the total expansion inside equals the net flow through the boundary.
2. If the boundary flow is zero, the total integral of the expansion rate inside must be zero.
3. Expansion represents sources (positive) and sinks (negative).

Answer: The total strength of sources inside the region exactly cancels the total strength of sinks.

1. The boundary interior principle for surfaces states that the total twist through a surface equals the flow along its boundary.
2. The boundary of the bowl $S$ is the circle $C$.
3. The flow along $C$ is given as $5$.

Answer: The total twist through the bowl is $5$.

1. Calculate the dot product: $(3)(1) + (-2)(2) + (5)(1)$
2. Simplify: $3 - 4 + 5 = 4$
3. Since the dot product is $4$, which is not $0$, the vectors are not perpendicular.

Answer: 4, No

1. Set up the determinant: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & -1 \\ 0 & 3 & 2 \end{vmatrix}$
2. i component: $(0)(2) - (-1)(3) = 3$
3. j component: $-((2)(2) - (-1)(0)) = -4$
4. k component: $(2)(3) - (0)(0) = 6$

Answer: \langle 3, -4, 6 \rangle

1. Find the cross product $\vec{n} = \vec{u} \times \vec{v} = \langle (0)(1)-(1)(1), -((1)(1)-(1)(0)), (1)(1)-(0)(0) \rangle = \langle -1, -1, 1 \rangle$
2. Calculate the magnitude: $|\vec{n}| = \sqrt{(-1)^2 + (-1)^2 + 1^2} = \sqrt{3}$
3. Normalize the vector: $\hat{n} = \frac{1}{\sqrt{3}} \langle -1, -1, 1 \rangle$

Answer: \langle -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle

1. Differentiate the x-component: the derivative of $t^3$ is $3t^2$.
2. Differentiate the y-component: the derivative of $\sin(t)$ is $\cos(t)$.
3. Differentiate the z-component: the derivative of $\sqrt{t}$ is $\frac{1}{2\sqrt{t}}$.
4. Assemble the velocity vector: $\vec{v}(t) = \langle 3t^2, \cos(t), \frac{1}{2\sqrt{t}} \rangle$.
5. Differentiate the velocity components for acceleration: $3t^2$ becomes $6t$, $\cos(t)$ becomes $-\sin(t)$, and $\frac{1}{2}t^{-1/2}$ becomes $-\frac{1}{4}t^{-3/2}$.
6. Assemble the acceleration vector: $\vec{a}(t) = \langle 6t, -\sin(t), -\frac{1}{4t\sqrt{t}} \rangle$.

Answer: \vec{v}(t) = \langle 3t^2, \cos(t), \frac{1}{2\sqrt{t}} \rangle, \vec{a}(t) = \langle 6t, -\sin(t), -\frac{1}{4t\sqrt{t}} \rangle

1. Find the velocity vector $\vec{v}(t)$ by differentiating each component: $\vec{v}(t) = \langle 2e^{2t}, 2t, 5 \rangle$.
2. Evaluate the velocity at $t = 0$: $\vec{v}(0) = \langle 2e^0, 2(0), 5 \rangle = \langle 2, 0, 5 \rangle$.
3. Calculate the magnitude of the velocity vector: $|\vec{v}(0)| = \sqrt{2^2 + 0^2 + 5^2}$.
4. Simplify the result: $\sqrt{4 + 0 + 25} = \sqrt{29}$.

Answer: \sqrt{29}

1. Find the velocity vector using the chain rule: $\vec{v}(t) = \langle -3\sin(3t), 3\cos(3t), 2 \rangle$.
2. Differentiate the velocity vector to find acceleration.
3. The derivative of $-3\sin(3t)$ is $-3 \cdot 3\cos(3t) = -9\cos(3t)$.
4. The derivative of $3\cos(3t)$ is $3 \cdot (-3\sin(3t)) = -9\sin(3t)$.
5. The derivative of the constant $2$ is $0$.
6. Assemble the acceleration vector: $\vec{a}(t) = \langle -9\cos(3t), -9\sin(3t), 0 \rangle$.

Answer: \vec{a}(t) = \langle -9\cos(3t), -9\sin(3t), 0 \rangle

1. Compute the velocity vector: $\vec{v}(t) = \langle 1, t^{1/2}, 0 \rangle$.
2. Compute the speed: $|\vec{v}(t)| = \sqrt{1^2 + (t^{1/2})^2 + 0^2} = \sqrt{1+t}$.
3. Set up the integral: $L = \int_0^3 \sqrt{1+t} \, dt$.
4. Use u-substitution with $u=1+t$, $du=dt$: $\int_1^4 u^{1/2} \, du$.
5. Evaluate: $[\frac{2}{3}u^{3/2}]_1^4 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8-1) = \frac{14}{3}$.

Answer: 14/3

1. Compute velocity: $\vec{v}(t) = \langle 1, 2t, 0 \rangle$. At $t=0$, $\vec{v}(0) = \langle 1, 0, 0 \rangle$.
2. Compute acceleration: $\vec{a}(t) = \langle 0, 2, 0 \rangle$. At $t=0$, $\vec{a}(0) = \langle 0, 2, 0 \rangle$.
3. Compute cross product: $\vec{v}(0) \times \vec{a}(0) = \langle 1, 0, 0 \rangle \times \langle 0, 2, 0 \rangle = \langle 0, 0, 2 \rangle$.
4. Find magnitudes: $|\vec{v} \times \vec{a}| = 2$ and $|\vec{v}| = 1$.
5. Apply formula: $\kappa = \frac{2}{1^3} = 2$.

Answer: 2

1. Compute velocity: $\vec{v}(t) = \langle e^t, e^t, 1 \rangle$.
2. Compute acceleration: $\vec{a}(t) = \langle e^t, e^t, 0 \rangle$.
3. Compute cross product: $\vec{v} \times \vec{a} = \langle -e^t, e^t, 0 \rangle$.
4. Find magnitudes: $|\vec{v} \times \vec{a}| = \sqrt{(-e^t)^2 + (e^t)^2} = \sqrt{2e^{2t}} = e^t\sqrt{2}$.
5. Find speed: $|\vec{v}| = \sqrt{(e^t)^2 + (e^t)^2 + 1^2} = \sqrt{2e^{2t} + 1}$.
6. Apply formula: $\kappa(t) = \frac{e^t\sqrt{2}}{(2e^{2t} + 1)^{3/2}}$.

Answer: \frac{e^t\sqrt{2}}{(2e^{2t} + 1)^{3/2}}

1. Compute velocity: $\vec{v}(t) = \langle 1, 2t, 0 \rangle$. At $t=1$, $\vec{v}(1) = \langle 1, 2, 0 \rangle$.
2. Normalize velocity: $|\vec{v}(1)| = \sqrt{1^2 + 2^2 + 0^2} = \sqrt{5}$. So $\mathbf{T}(1) = \langle 1/\sqrt{5}, 2/\sqrt{5}, 0 \rangle$.
3. Find $\mathbf{T}(t) = \frac{1}{\sqrt{1+4t^2}} \langle 1, 2t, 0 \rangle$.
4. Differentiate $\mathbf{T}(t)$ using the quotient rule to find $\mathbf{T}'(t) = \frac{1}{(1+4t^2)^{3/2}} \langle -4t, 2, 0 \rangle$.
5. At $t=1$, $\mathbf{T}'(1) = \frac{1}{5\sqrt{5}} \langle -4, 2, 0 \rangle$. Normalize this to find $\mathbf{N}(1) = \langle -4/\sqrt{20}, 2/\sqrt{20}, 0 \rangle = \langle -2/\sqrt{5}, 1/\sqrt{5}, 0 \rangle$.

Answer: \mathbf{T}(1) = \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, 0 \rangle, \mathbf{N}(1) = \langle \frac{-2}{\sqrt{5}}, \frac{1}{\sqrt{5}}, 0 \rangle

1. From the lesson, $\mathbf{T}(t) = \frac{1}{\sqrt{2}} \langle -\sin t, \cos t, 1 \rangle$.
2. From the lesson, $\mathbf{N}(t) = \langle -\cos t, -\sin t, 0 \rangle$.
3. Compute the dot product: $\mathbf{T} \cdot \mathbf{N} = \frac{1}{\sqrt{2}} ((-\sin t)(-\cos t) + (\cos t)(-\sin t) + (1)(0))$.
4. Simplify: $\frac{1}{\sqrt{2}} (\sin t \cos t - \sin t \cos t + 0) = 0$.

Answer: The dot product is $0$, confirming they are orthogonal.

1. Velocity: $\vec{v}(t) = \langle 1, 2t, 3t^2 \rangle$. At $t=0$, $\vec{v}(0) = \langle 1, 0, 0 \rangle$.
2. Unit tangent: $\mathbf{T}(0) = \langle 1, 0, 0 \rangle$.
3. Find $\mathbf{T}(t) = \frac{1}{\sqrt{1+4t^2+9t^4}} \langle 1, 2t, 3t^2 \rangle$.
4. Differentiate $\mathbf{T}(t)$ and evaluate at $t=0$. $\mathbf{T}'(0) = \langle 0, 2, 0 \rangle$.
5. Normalize to get $\mathbf{N}(0) = \langle 0, 1, 0 \rangle$.
6. Compute $\mathbf{B}(0) = \mathbf{T}(0) \times \mathbf{N}(0) = \langle 1, 0, 0 \rangle \times \langle 0, 1, 0 \rangle = \langle 0, 0, 1 \rangle$.

Answer: \mathbf{B}(0) = \langle 0, 0, 1 \rangle

1. Divide the entire equation by $16$ to put it in standard form: $\frac{x^2}{16} + \frac{y^2}{4} + \frac{z^2}{16} = 1$.
2. Recognize this as the equation of an ellipsoid centered at the origin.
3. To find the intersection with the $xy$-plane, set $z = 0$.
4. The resulting equation is $x^2 + 4y^2 = 16$, which is an ellipse in the $xy$-plane.

Answer: The surface is an ellipsoid; the intersection is the ellipse $x^2 + 4y^2 = 16$.

1. For $k = 0$, $0 = 4 - x^2 - y^2 \implies x^2 + y^2 = 4$ (circle radius $2$).
2. For $k = 1$, $1 = 4 - x^2 - y^2 \implies x^2 + y^2 = 3$ (circle radius $\sqrt{3}$).
3. For $k = 3$, $3 = 4 - x^2 - y^2 \implies x^2 + y^2 = 1$ (circle radius $1$).
4. For $k = 4$, $4 = 4 - x^2 - y^2 \implies x^2 + y^2 = 0$ (the point $(0,0)$).
5. The surface is a circular paraboloid opening downwards with its peak at $(0,0,4)$.

Answer: The level curves are concentric circles with decreasing radii as $k$ increases; the surface is a downward-opening circular paraboloid.

1. A graph $z = f(x, y)$ can always be rewritten as $z - f(x, y) = 0$.
2. Substitute the given function: $z - \frac{1}{x^2 + y^2} = 0$.
3. Alternatively, multiply by $(x^2 + y^2)$ to clear the fraction: $z(x^2 + y^2) - 1 = 0$.
4. Define $F(x, y, z) = z(x^2 + y^2)$ and set the constant $c = 1$.

Answer: Yes, it can be represented as $F(x, y, z) = z(x^2 + y^2) = 1$ (or $F(x, y, z) = z - \frac{1}{x^2 + y^2} = 0$).

1. Divide the entire equation by $6$ to get the standard form: $\frac{x^2}{6} + \frac{y^2}{3} - \frac{z^2}{6} = 1$.
2. Observe the signs of the squared terms: two are positive and one is negative.
3. Check the traces: setting $z=0$ gives an ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$; setting $x=0$ gives a hyperbola $\frac{y^2}{3} - \frac{z^2}{6} = 1$.
4. A surface with one negative squared term and a constant of $1$ is a hyperboloid of one sheet.

Answer: Hyperboloid of one sheet

1. Use spherical coordinates where $u$ is the polar angle $\phi$ and $v$ is the azimuthal angle $\theta$.
2. The radius is fixed at $R=3$.
3. The coordinates are $x = 3 \sin u \cos v$, $y = 3 \sin u \sin v$, and $z = 3 \cos u$.
4. Combine these into a vector function.

Answer: \vec{r}(u, v) = \langle 3\sin u \cos v, 3\sin u \sin v, 3\cos u \rangle

1. Since the surface is given as a graph $z = f(x, y)$, use $x$ and $y$ as the parameters.
2. Let $u = x$ and $v = y$.
3. The vector components are $x(u, v) = u$, $y(u, v) = v$, and $z(u, v) = u^2 - v^2$.
4. Specify the bounds for the parameters based on the given constraints.

Answer: \vec{r}(u, v) = \langle u, v, u^2 - v^2 \rangle, \text{ for } 0 \le u \le 1, 0 \le v \le 1

1. For cylindrical: $r = \sqrt{0^2 + 2^2} = 2$. Since $x=0$ and $y=2$, the point is on the positive y-axis, so $\theta = \pi/2$. $z$ remains $2$. Result: $(2, \pi/2, 2)$.
2. For spherical: $\rho = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. $\theta$ is the same as cylindrical, $\pi/2$. For $\phi$, $\cos(\phi) = z/\rho = 2/(2\sqrt{2}) = 1/\sqrt{2}$, so $\phi = \pi/4$. Result: $(2\sqrt{2}, \pi/2, \pi/4)$.

Answer: Cylindrical: $(2, \pi/2, 2)$; Spherical: $(2\sqrt{2}, \pi/2, \pi/4)$

1. Recall the definition of $\rho$ in spherical coordinates: $\rho^2 = x^2 + y^2 + z^2$.
2. Substitute $\rho^2$ into the equation: $\rho^2 = 16$.
3. Taking the square root (since $\rho \ge 0$), we get $\rho = 4$.

Answer: \rho = 4

1. The equation $\phi = \pi/6$ represents all points at a constant angle of $30$ degrees from the positive z-axis, which is a right circular cone opening upwards.
2. In cylindrical coordinates, we know $\tan(\phi) = r/z$.
3. Substituting $\phi = \pi/6$, we get $\tan(\pi/6) = r/z$.
4. Since $\tan(\pi/6) = 1/\sqrt{3}$, the equation is $1/\sqrt{3} = r/z$, or $z = r\sqrt{3}$.

Answer: A cone; $z = r\sqrt{3}$

1. Test the path $x=0$: the function becomes $0/y^2 = 0$.
2. Test the path $y=0$: the function becomes $0/x^2 = 0$.
3. Use the inequality $|x|/\sqrt{x^2+y^2} \le 1$.
4. Rewrite the function as $|x| \cdot |y^2 / (x^2+y^2)|$.
5. Since $y^2 / (x^2+y^2) \le 1$, the absolute value of the function is $\le |x|$.
6. As $(x,y) \to (0,0)$, $|x| \to 0$, so by the Squeeze Theorem, the limit is $0$.

Answer: 0

1. Test the path $y=0$: the limit is $\lim_{x \to 0} x^2/x^2 = 1$.
2. Test the path $x=0$: the limit is $\lim_{y \to 0} -y^2/y^2 = -1$.
3. Since $1 \neq -1$, the limit depends on the path.

Answer: Does not exist

1. Check the limit as $(x,y) \to (0,0)$.
2. Convert to polar: $f(r, \theta) = (r^3 \cos^3(\theta)) / (r^2) = r \cos^3(\theta)$.
3. Since $|\cos^3(\theta)| \le 1$, we have $|f(r, \theta)| \le r$.
4. As $r \to 0$, the limit is $0$.
5. Since the limit $0$ equals the defined value $f(0,0) = 0$, the function is continuous.

Answer: Continuous

1. Compute $f_x = 9x^2 y^2 - 10xy$.
2. Compute $f_y = 6x^3 y - 5x^2$.
3. Differentiate $f_x$ wrt $x$ to get $f_{xx} = 18xy^2 - 10y$.
4. Differentiate $f_y$ wrt $y$ to get $f_{yy} = 6x^3$.
5. Differentiate $f_x$ wrt $y$ to get $f_{xy} = 18x^2 y - 10x$.
6. Differentiate $f_y$ wrt $x$ to get $f_{yx} = 18x^2 y - 10x$.

Answer: f_{xx} = 18xy^2 - 10y, f_{yy} = 6x^3, f_{xy} = f_{yx} = 18x^2 y - 10x

1. Compute $f_x = y e^{xy}$ using the chain rule.
2. Compute $f_y = x e^{xy}$ using the chain rule.
3. Compute $f_{xy}$ by differentiating $y e^{xy}$ wrt $y$ using the product rule: $1 \cdot e^{xy} + y \cdot (x e^{xy}) = (1 + xy)e^{xy}$.
4. Compute $f_{yx}$ by differentiating $x e^{xy}$ wrt $x$ using the product rule: $1 \cdot e^{xy} + x \cdot (y e^{xy}) = (1 + xy)e^{xy}$.

Answer: f_{xy} = f_{yx} = (1 + xy)e^{xy}

1. Compute $f_x = \sec(x+y)\tan(x+y)$ using the chain rule.
2. Differentiate $f_x$ wrt $y$ using the product rule: $\frac{\partial}{\partial y}[\sec(x+y)] \tan(x+y) + \sec(x+y) \frac{\partial}{\partial y}[\tan(x+y)]$.
3. The derivative of $\sec(u)$ is $\sec(u)\tan(u)u'$, and the derivative of $\tan(u)$ is $\sec^2(u)u'$.
4. Substitute $u = x+y$ and $u' = 1$: $[\sec(x+y)\tan(x+y)]\tan(x+y) + \sec(x+y)[\sec^2(x+y)]$.
5. Simplify to $\sec(x+y)\tan^2(x+y) + \sec^3(x+y)$.

Answer: f_{xy} = \sec(x+y)\tan^2(x+y) + \sec^3(x+y)

1. Step 1: Find $z_0 = 1^3 + 2^3 = 1 + 8 = 9$.
2. Step 2: Compute partial derivatives: $f_x = 3x^2$ and $f_y = 3y^2$.
3. Step 3: Evaluate partials at $(1, 2)$: $f_x(1, 2) = 3(1)^2 = 3$ and $f_y(1, 2) = 3(2)^2 = 12$.
4. Step 4: Plug into the formula: $z - 9 = 3(x - 1) + 12(y - 2)$.
5. Step 5: Simplify: $z = 3x - 3 + 12y - 24 + 9 = 3x + 12y - 18$.

Answer: $z = 3x + 12y - 18$

1. Step 1: Check partial derivatives at $(0,0)$. $f_x(0,0) = \lim_{h \to 0} \frac{|h| + 0 - 0}{h} = \lim_{h \to 0} \frac{|h|}{h}$.
2. Step 2: The limit $\lim_{h \to 0} \frac{|h|}{h}$ does not exist because it is $1$ from the right and $-1$ from the left.
3. Step 3: Since the partial derivatives do not exist at $(0,0)$, the function cannot be differentiable there.

Answer: Not differentiable

1. Step 1: Find $z_0 = e^{0 \cdot 1} = e^0 = 1$.
2. Step 2: Compute partials: $f_x = y e^{xy}$ and $f_y = x e^{xy}$.
3. Step 3: Evaluate at $(0, 1)$: $f_x(0, 1) = 1 \cdot e^0 = 1$ and $f_y(0, 1) = 0 \cdot e^0 = 0$.
4. Step 4: Plug into formula: $z - 1 = 1(x - 0) + 0(y - 1)$.
5. Step 5: Simplify: $z = x + 1$.

Answer: $z = x + 1$

1. Compute the partial derivative with respect to $x$: $f_x = 3x^2 - 6y$.
2. Compute the partial derivative with respect to $y$: $f_y = 3y^2 - 6x$.
3. Evaluate $f_x$ at $(1,1)$: $3(1)^2 - 6(1) = -3$.
4. Evaluate $f_y$ at $(1,1)$: $3(1)^2 - 6(1) = -3$.
5. Combine into a vector: $\nabla f(1,1) = \langle -3, -3 \rangle$.

Answer: \langle -3, -3 \rangle

1. Compute $f_x = 2e^{2x-y}$ using the chain rule.
2. Compute $f_y = -e^{2x-y}$ using the chain rule.
3. Evaluate at $(0,0)$: $f_x(0,0) = 2e^0 = 2$ and $f_y(0,0) = -e^0 = -1$.
4. The direction of steepest ascent is $\nabla f(0,0) = \langle 2, -1 \rangle$.
5. The maximum slope is $|\nabla f(0,0)| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$.

Answer: \text{Direction: } \langle 2, -1 \rangle, \text{ Slope: } \sqrt{5}

1. The level curve is $g(x,y) = x^2 + y^2 - 25 = 0$.
2. The tangent direction to the curve at $(3,4)$ can be found by the vector $\langle -y, x \rangle = \langle -4, 3 \rangle$.
3. Check the dot product of the gradient $\langle 6, 8 \rangle$ and the tangent vector $\langle -4, 3 \rangle$.
4. Compute: $(6)(-4) + (8)(3) = -24 + 24 = 0$.
5. Since the dot product is zero, the gradient is perpendicular to the tangent of the level curve.

Answer: \text{Verified: } \langle 6, 8 \rangle \cdot \langle -4, 3 \rangle = 0

1. Compute the gradient: $\nabla f = \langle 2x, 2y \rangle$.
2. Evaluate the gradient at $(2,3)$: $\nabla f(2,3) = \langle 4, 6 \rangle$.
3. Normalize the direction vector $v$: $|v| = \sqrt{4^2 + (-3)^2} = 5$, so $u = \langle 4/5, -3/5 \rangle$.
4. Compute the dot product: $D_u f = \langle 4, 6 \rangle \cdot \langle 4/5, -3/5 \rangle = 16/5 - 18/5 = -2/5$.

Answer: -2/5

1. Compute the gradient: $\nabla f = \langle 3, 2 \rangle$.
2. Find the unit vector for $\theta = \pi/4$: $u = \langle \cos(\pi/4), \sin(\pi/4) \rangle = \langle \sqrt{2}/2, \sqrt{2}/2 \rangle$.
3. Compute the dot product: $D_u f = \langle 3, 2 \rangle \cdot \langle \sqrt{2}/2, \sqrt{2}/2 \rangle = 3\sqrt{2}/2 + 2\sqrt{2}/2 = 5\sqrt{2}/2$.

Answer: 5\sqrt{2}/2

1. Compute the gradient at $(1,2)$: $\nabla f = \langle 2xy, x^2 \rangle \implies \nabla f(1,2) = \langle 4, 1 \rangle$.
2. The directional derivative is zero when $u$ is perpendicular to the gradient: $\langle 4, 1 \rangle \cdot \langle u_1, u_2 \rangle = 0$.
3. This implies $4u_1 + u_2 = 0$, or $u_2 = -4u_1$.
4. Since $u$ is a unit vector, $u_1^2 + (-4u_1)^2 = 1 \implies 17u_1^2 = 1 \implies u_1 = \pm 1/\sqrt{17}$.
5. Thus, $u = \langle 1/\sqrt{17}, -4/\sqrt{17} \rangle$ or $u = \langle -1/\sqrt{17}, 4/\sqrt{17} \rangle$.

Answer: \langle \pm 1/\sqrt{17}, \mp 4/\sqrt{17} \rangle

1. Identify dependencies: $z$ depends on $x$ and $y$, and both depend on $t$.
2. Apply chain rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$.
3. Compute partials: $\frac{\partial z}{\partial x} = 2xy$, $\frac{\partial z}{\partial y} = x^2$.
4. Compute inner derivatives: $\frac{dx}{dt} = e^t$, $\frac{dy}{dt} = 2t$.
5. Substitute: $\frac{dz}{dt} = (2xy)(e^t) + (x^2)(2t)$.
6. Evaluate at $t=1$: $x=e$, $y=1$. So $\frac{dz}{dt} = (2e)(e) + (e^2)(2) = 2e^2 + 2e^2 = 4e^2$.

Answer: 4e^2

1. Find the gradient of $f$: $\nabla f = \begin{pmatrix} 1 & 2 \end{pmatrix}$.
2. Find the Jacobian of the transformation: $J = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 2u & -2v \\ 2v & 2u \end{pmatrix}$.
3. Multiply $\nabla f$ by $J$: $\begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2u & -2v \\ 2v & 2u \end{pmatrix} = \begin{pmatrix} 1(2u) + 2(2v) & 1(-2v) + 2(2u) \end{pmatrix}$.
4. Simplify: $\begin{pmatrix} 2u + 4v & 4u - 2v \end{pmatrix}$.

Answer: \frac{\partial f}{\partial u} = 2u + 4v, \frac{\partial f}{\partial v} = 4u - 2v

1. Identify the paths from $w$ to $u$.
2. The paths are $w \to x \to u$, $w \to y \to u$, and $w \to z \to u$.
3. The contribution of each path is the product of the partial derivatives along that path.
4. Sum the contributions: $\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}$.
5. Express as a sum over indices $i=1, 2, 3$ where $x_1=x, x_2=y, x_3=z$.

Answer: \frac{\partial w}{\partial u} = \sum_{i=1}^3 \frac{\partial w}{\partial x_i} \frac{\partial x_i}{\partial u}

1. Define $F(x, y, z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$.
2. Compute $F_x = 3x^2 + 6yz$.
3. Compute $F_z = 3z^2 + 6xy$.
4. Apply the formula $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$.

Answer: $\frac{\partial z}{\partial x} = -\frac{3x^2 + 6yz}{3z^2 + 6xy} = -\frac{x^2 + 2yz}{z^2 + 2xy}$

1. Define $F(x, y, z) = \ln(z) + x^2 y + z - 10 = 0$.
2. Compute $F_y = x^2$.
3. Compute $F_z = \frac{1}{z} + 1$.
4. At $(1, 1, e)$, $F_y = 1^2 = 1$ and $F_z = \frac{1}{e} + 1 = \frac{1+e}{e}$.
5. Apply $\frac{\partial z}{\partial y} = -\frac{1}{(1+e)/e} = -\frac{e}{1+e}$.

Answer: $-\frac{e}{1+e}$

1. Define $F(x, y, z) = x^2 + 2y^2 + 3z^2 - 6 = 0$.
2. Compute $F_x = 2x$.
3. Compute $F_z = 6z$.
4. Apply the formula $\frac{\partial z}{\partial x} = -\frac{2x}{6z}$.
5. Simplify the fraction to get $-\frac{x}{3z}$.

Answer: $\frac{\partial z}{\partial x} = -\frac{x}{3z}$

1. Compute $f(0, 0) = \sin(0) + \cos(0) = 1$.
2. Compute first partials: $f_x = \cos(x)$, $f_y = -\sin(y)$. At $(0, 0)$, $f_x = 1$ and $f_y = 0$.
3. Compute second partials: $f_{xx} = -\sin(x)$, $f_{yy} = -\cos(y)$, $f_{xy} = 0$. At $(0, 0)$, $f_{xx} = 0$, $f_{yy} = -1$, $f_{xy} = 0$.
4. Plug into formula: $f(x, y) \approx 1 + 1(x-0) + 0(y-0) + \frac{1}{2}(0x^2 + 2(0)xy + (-1)y^2)$.

Answer: f(x, y) \approx 1 + x - \frac{1}{2}y^2

1. First partials: $f_x = 3x^2 y^2$, $f_y = 2x^3 y$.
2. Second partials: $f_{xx} = 6xy^2$, $f_{yy} = 2x^3$, $f_{xy} = 6x^2 y$.
3. Evaluate at $(1, 2)$: $f_{xx} = 6(1)(4) = 24$, $f_{yy} = 2(1)^3 = 2$, $f_{xy} = 6(1)^2(2) = 12$.

Answer: H = \begin{pmatrix} 24 & 12 \\ 12 & 2 \end{pmatrix}

1. Value: $f(0, 0) = e^0 = 1$.
2. First partials: $f_x = e^{x+y}$, $f_y = e^{x+y}$. At $(0, 0)$, both are $1$.
3. Second partials: $f_{xx} = e^{x+y}$, $f_{yy} = e^{x+y}$, $f_{xy} = e^{x+y}$. At $(0, 0)$, all are $1$.
4. Assemble: $f(x, y) \approx 1 + 1x + 1y + \frac{1}{2}(1x^2 + 2(1)xy + 1y^2)$.

Answer: f(x, y) \approx 1 + x + y + \frac{1}{2}(x^2 + 2xy + y^2)

1. Compute partials: $f_x = 2x + y$ and $f_y = 4y + x$.
2. Set to zero: $2x + y = 0$ and $x + 4y = 0$. Solving this system gives the only critical point $(0, 0)$.
3. Compute second partials: $f_{xx} = 2$, $f_{yy} = 4$, $f_{xy} = 1$.
4. Calculate determinant $D = (2)(4) - (1)^2 = 7$.
5. Since $D > 0$ and $f_{xx} = 2 > 0$, the point is a local minimum.

Answer: Local minimum at $(0, 0)$

1. Verify critical point: $f_x = 2x + 2$, $f_y = -6y$. Setting these to zero gives $(-1, 0)$.
2. Compute second partials: $f_{xx} = 2$, $f_{yy} = -6$, $f_{xy} = 0$.
3. Calculate determinant $D = (2)(-6) - 0^2 = -12$.
4. Since $D < 0$, the point is a saddle point.

Answer: Saddle point at $(-1, 0)$

1. Compute partials: $f_x = 3x^2 - 3$ and $f_y = 2y$.
2. Set to zero: $3x^2 = 3 \implies x = \pm 1$ and $2y = 0 \implies y = 0$. Critical points are $(1, 0)$ and $(-1, 0)$.
3. Compute second partials: $f_{xx} = 6x$, $f_{yy} = 2$, $f_{xy} = 0$.
4. At $(1, 0)$: $D = (6)(2) - 0^2 = 12$. Since $D > 0$ and $f_{xx} = 6 > 0$, it is a local minimum.
5. At $(-1, 0)$: $D = (-6)(2) - 0^2 = -12$. Since $D < 0$, it is a saddle point.

Answer: Local minimum at $(1, 0)$ and saddle point at $(-1, 0)$

1. Find interior critical points: $f_x = 2x = 0$ and $f_y = 2y = 0$, so $(0,0)$ is a critical point. $f(0,0) = 0$.
2. Check the boundary $x^2 + y^2 = 1$. On this boundary, $f(x, y) = 1$ for all points.
3. Compare values: the minimum is $0$ at $(0,0)$ and the maximum is $1$ at any point on the boundary circle.

Answer: Min: 0, Max: 1

1. Interior: $f_x = 2, f_y = 3$. No critical points since the gradient is never zero.
2. Boundary $x=0$: $f(0, y) = 3y$, max $3$ at $y=1$, min $0$ at $y=0$.
3. Boundary $x=2$: $f(2, y) = 4 + 3y$, max $7$ at $y=1$, min $4$ at $y=0$.
4. Boundary $y=0$: $f(x, 0) = 2x$, max $4$ at $x=2$, min $0$ at $x=0$.
5. Boundary $y=1$: $f(x, 1) = 2x + 3$, max $7$ at $x=2$, min $3$ at $x=0$.
6. Comparing all: absolute min is $0$ at $(0,0)$ and absolute max is $7$ at $(2,1)$.

Answer: Min: 0, Max: 7

1. Let $x$ be the side of the square base and $h$ be the height. Volume $V = x^2 h = 32$, so $h = 32/x^2$.
2. Surface area $S = x^2 + 4xh$. Substitute $h$: $S(x) = x^2 + 4x(32/x^2) = x^2 + 128/x$.
3. Find critical points: $S'(x) = 2x - 128/x^2 = 0$.
4. Solve $2x^3 = 128 \implies x^3 = 64 \implies x = 4$.
5. Find $h$: $h = 32/(4^2) = 32/16 = 2$.
6. Check second derivative: $S''(x) = 2 + 256/x^3$, which is positive for $x=4$, confirming a minimum.

Answer: Base side 4m, Height 2m

1. Set up the gradients: $\nabla f = (1, 2)$ and $\nabla g = (2x, 2y)$.
2. Apply Lagrange equations: $1 = \lambda(2x)$ and $2 = \lambda(2y)$.
3. Divide the equations: $\frac{1}{2} = \frac{2x}{2y} \implies y = 2x$.
4. Substitute into constraint: $x^2 + (2x)^2 = 5 \implies 5x^2 = 5 \implies x = \pm 1$.
5. Find points: $(1, 2)$ and $(-1, -2)$.
6. Evaluate $f$: $f(1, 2) = 5$ and $f(-1, -2) = -5$.

Answer: Max is 5, Min is -5

1. Minimize distance squared: $f(x, y) = x^2 + y^2$ subject to $g(x, y) = 2x - y = -3$.
2. Gradients: $\nabla f = (2x, 2y)$ and $\nabla g = (2, -1)$.
3. Equations: $2x = 2\lambda$ and $2y = -\lambda$.
4. Express in terms of $\lambda$: $x = \lambda$ and $y = -\lambda/2$.
5. Substitute into constraint: $2(\lambda) - (-\lambda/2) = -3 \implies 2.5\lambda = -3 \implies \lambda = -1.2$.
6. Find coordinates: $x = -1.2$, $y = 0.6$.

Answer: (-1.2, 0.6)

1. Gradients: $\nabla f = (yz, xz, xy)$ and $\nabla g = (1, 1, 1)$.
2. Equations: $yz = \lambda$, $xz = \lambda$, $xy = \lambda$.
3. Equating them: $yz = xz \implies y = x$ (since $z > 0$) and $xz = xy \implies z = y$.
4. Thus $x = y = z$.
5. Substitute into constraint: $x + x + x = 12 \implies 3x = 12 \implies x = 4$.
6. Point is $(4, 4, 4)$, and $f(4, 4, 4) = 64$.

Answer: 64

1. Compute partial derivatives: $f_x = y e^{xy} \cos(x+y) - e^{xy} \sin(x+y)$ and $f_y = x e^{xy} \cos(x+y) - e^{xy} \sin(x+y)$.
2. Evaluate at $(0, 0)$: $f_x(0, 0) = 0(1)(1) - (1)(0) = 0$ and $f_y(0, 0) = 0(1)(1) - (1)(0) = 0$.
3. The gradient is $\nabla f(0, 0) = \langle 0, 0 \rangle$.
4. The directional derivative is the dot product of the gradient and the unit vector $\mathbf{u} = \langle 1/\sqrt{2}, 1/\sqrt{2} \rangle$.
5. Result: $0(1/\sqrt{2}) + 0(1/\sqrt{2}) = 0$.

Answer: 0

1. Find partials: $f_x = 2x - 2y - 4$ and $f_y = 4y - 2x$.
2. Set to zero: $2x - 2y = 4$ and $2x = 4y \implies x = 2y$.
3. Substitute $x=2y$ into first eq: $2(2y) - 2y = 4 \implies 2y = 4 \implies y = 2$. Then $x = 4$.
4. Critical point is $(4, 2)$.
5. Compute Hessian: $f_{xx} = 2, f_{yy} = 4, f_{xy} = -2$.
6. Determinant $D = (2)(4) - (-2)^2 = 8 - 4 = 4$.
7. Since $D > 0$ and $f_{xx} > 0$, it is a local minimum.

Answer: Local minimum at (4, 2)

1. Set up equations: $\nabla f = \langle y, x \rangle$ and $\nabla g = \langle 2x, 2y \rangle$.
2. System: $y = 2\lambda x$ and $x = 2\lambda y$.
3. Substitute $y$: $x = 2\lambda (2\lambda x) = 4\lambda^2 x$.
4. Either $x=0$ (which implies $y=0$, but $(0,0)$ is not on the circle) or $4\lambda^2 = 1 \implies \lambda = \pm 1/2$.
5. If $\lambda = 1/2$, $y=x$. Constraint $x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x = \pm 1/\sqrt{2}$. Points: $(1/\sqrt{2}, 1/\sqrt{2})$ and $(-1/\sqrt{2}, -1/\sqrt{2})$. $f = 1/2$.
6. If $\lambda = -1/2$, $y=-x$. Constraint $x^2 + (-x)^2 = 1 \implies 2x^2 = 1 \implies x = \pm 1/\sqrt{2}$. Points: $(1/\sqrt{2}, -1/\sqrt{2})$ and $(-1/\sqrt{2}, 1/\sqrt{2})$. $f = -1/2$.

Answer: Max value 1/2, Min value -1/2

1. Set up the iterated integral: $\int_{0}^{1} \int_{0}^{1} (x^2 + y^2) \, dy \, dx$.
2. Integrate with respect to $y$: $[x^2 y + \frac{1}{3}y^3]_{0}^{1} = x^2 + \frac{1}{3}$.
3. Integrate the result with respect to $x$: $\int_{0}^{1} (x^2 + \frac{1}{3}) \, dx = [\frac{1}{3}x^3 + \frac{1}{3}x]_{0}^{1}$.
4. Evaluate: $\frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.

Answer: 2/3

1. Set up the integral with $y$ limits from $0$ to $1-x$ and $x$ limits from $0$ to $1$: $\int_{0}^{1} \int_{0}^{1-x} (4 - x - y) \, dy \, dx$.
2. Inner integral: $[4y - xy - \frac{1}{2}y^2]_{0}^{1-x} = 4(1-x) - x(1-x) - \frac{1}{2}(1-x)^2$.
3. Simplify the integrand: $4 - 4x - x + x^2 - \frac{1}{2}(1 - 2x + x^2) = 3.5 - 4x + 0.5x^2$.
4. Outer integral: $\int_{0}^{1} (3.5 - 4x + 0.5x^2) \, dx = [3.5x - 2x^2 + \frac{1}{6}x^3]_{0}^{1}$.
5. Evaluate: $3.5 - 2 + \frac{1}{6} = 1.5 + \frac{1}{6} = \frac{3}{2} + \frac{1}{6} = \frac{9+1}{6} = \frac{10}{6} = \frac{5}{3}$.

Answer: 5/3

1. The curves intersect at $x=0$ and $x=1$. The line $y=x$ is the upper boundary.
2. Set up the integral: $\int_{0}^{1} \int_{x^2}^{x} x \, dy \, dx$.
3. Inner integral: $[xy]_{x^2}^{x} = x(x) - x(x^2) = x^2 - x^3$.
4. Outer integral: $\int_{0}^{1} (x^2 - x^3) \, dx = [\frac{1}{3}x^3 - \frac{1}{4}x^4]_{0}^{1}$.
5. Evaluate: $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$.

Answer: 1/12

1. Find intersection points: $x^2 = 4$ implies $x = -2$ and $x = 2$.
2. For a vertical slice, the bottom boundary is $y = x^2$ and the top boundary is $y = 4$.
3. The outer limits are the x-range from $-2$ to $2$.

Answer: \int_{-2}^{2} \int_{x^2}^{4} f(x, y) \, dy \, dx

1. Find intersection points: $x = x^2$ implies $x = 0$ and $x = 1$. This means $y$ also ranges from $0$ to $1$.
2. Solve boundaries for $x$: $y = x$ becomes $x = y$, and $y = x^2$ becomes $x = \sqrt{y}$ (for $x \ge 0$).
3. For a horizontal slice, the left boundary is $x = y$ and the right boundary is $x = \sqrt{y}$.

Answer: \int_{0}^{1} \int_{y}^{\sqrt{y}} f(x, y) \, dx \, dy

1. The region is bounded by $y = e^x$ (top), $y = 1$ (bottom), and $x = 1$ (right).
2. The left intersection is where $e^x = 1$, which is $x = 0$.
3. Inner limits: $y$ goes from $1$ to $e^x$.
4. Outer limits: $x$ goes from $0$ to $1$.

Answer: \int_{0}^{1} \int_{1}^{e^x} f(x, y) \, dy \, dx

1. Identify the region: $x \le y \le 1$ and $0 \le x \le 1$.
2. Swap to type two: for a fixed $y$, $x$ ranges from $0$ to $y$.
3. The outer bounds for $y$ are $0$ to $1$.
4. The new integral is $\int_{0}^{1} \int_{0}^{y} \sin(y^2) \, dx \, dy$.
5. Integrate with respect to $x$: $\int_{0}^{1} [x \sin(y^2)]_{0}^{y} \, dy = \int_{0}^{1} y \sin(y^2) \, dy$.
6. Use u-substitution $u = y^2$, $du = 2y \, dy$: $\frac{1}{2} \int_{0}^{1} \sin(u) \, du = [-\frac{1}{2} \cos(u)]_{0}^{1} = \frac{1}{2}(1 - \cos(1))$.

Answer: \frac{1}{2}(1 - \cos(1))

1. Region: $y/2 \le x \le 1$ and $0 \le y \le 2$.
2. This is a triangle with vertices $(0,0), (1,0), (1,2)$.
3. Swap to type one: $x$ ranges from $0$ to $1$.
4. For a fixed $x$, $y$ ranges from $0$ to $2x$.
5. New integral: $\int_{0}^{1} \int_{0}^{2x} e^{x^2} \, dy \, dx$.
6. Inner integral: $[y e^{x^2}]_{0}^{2x} = 2x e^{x^2}$.
7. Outer integral: $\int_{0}^{1} 2x e^{x^2} \, dx = [e^{x^2}]_{0}^{1} = e - 1$.

Answer: e - 1

1. Region: $x \le y \le 1$ and $0 \le x \le 1$.
2. Swap to type two: $y$ ranges from $0$ to $1$.
3. For a fixed $y$, $x$ ranges from $0$ to $y$.
4. New integral: $\int_{0}^{1} \int_{0}^{y} e^{y^2} \, dx \, dy$.
5. Inner integral: $[x e^{y^2}]_{0}^{y} = y e^{y^2}$.
6. Outer integral: $\int_{0}^{1} y e^{y^2} \, dy = [\frac{1}{2} e^{y^2}]_{0}^{1} = \frac{1}{2}(e - 1)$.

Answer: \frac{1}{2}(e - 1)

1. Convert to polar: $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$.
2. Set bounds: $r$ goes from $0$ to $3$, and $\theta$ goes from $0$ to $\pi/2$.
3. Set up integral: $\int_{0}^{\pi/2} \int_{0}^{3} r^2 \cdot r \, dr \, d\theta$.
4. Integrate $r^3$: $[\frac{1}{4}r^4]_0^3 = \frac{81}{4}$.
5. Integrate with respect to $\theta$: $\int_{0}^{\pi/2} \frac{81}{4} \, d\theta = \frac{81\pi}{8}$.

Answer: $\frac{81\pi}{8}$

1. Convert to polar: $z = e^{-r^2}$ and $dA = r \, dr \, d\theta$.
2. Set bounds: $r$ goes from $0$ to $1$, and $\theta$ goes from $0$ to $\pi$.
3. Set up integral: $\int_{0}^{\pi} \int_{0}^{1} e^{-r^2} r \, dr \, d\theta$.
4. Use u-substitution $u = -r^2$, $du = -2r \, dr$: $\int_{0}^{1} r e^{-r^2} \, dr = [-\frac{1}{2}e^{-r^2}]_0^1 = \frac{1}{2}(1 - e^{-1})$.
5. Integrate with respect to $\theta$: $\int_{0}^{\pi} \frac{1}{2}(1 - e^{-1}) \, d\theta = \frac{\pi}{2}(1 - e^{-1})$.

Answer: $\frac{\pi}{2}(1 - e^{-1})$

1. The region $r = 2 \sin \theta$ is a circle centered at $(0, 1)$ with radius $1$.
2. Bounds: $r$ goes from $0$ to $2 \sin \theta$, and $\theta$ goes from $0$ to $\pi$.
3. Convert $y$ to polar: $y = r \sin \theta$.
4. Set up integral: $\int_{0}^{\pi} \int_{0}^{2\sin\theta} (r \sin \theta) r \, dr \, d\theta$.
5. Integrate $r^2$: $[\frac{1}{3}r^3]_0^{2\sin\theta} = \frac{8}{3} \sin^3 \theta \cdot \sin \theta = \frac{8}{3} \sin^4 \theta$.
6. Integrate $\int_{0}^{\pi} \frac{8}{3} \sin^4 \theta \, d\theta$ using power reduction: $\sin^4 \theta = \frac{3}{8} - \frac{1}{2}\cos 2\theta + \frac{1}{8}\cos 4\theta$.
7. Evaluating from $0$ to $\pi$: $\frac{8}{3} [\frac{3}{8}\theta]_0^{\pi} = \pi$.

Answer: $\pi$

1. Convert density to polar: $\rho = r^2$.
2. Set up the integral: $m = \int_{0}^{\pi} \int_{0}^{1} r^2 \cdot r \, dr \, d\theta$.
3. Integrate $r^3$ to get $r^4/4$ from $0$ to $1$, which is $1/4$.
4. Integrate $1/4$ with respect to $\theta$ from $0$ to $\pi$.

Answer: \frac{\pi}{4}

1. Compute mass: $m = \int_{0}^{1} \int_{0}^{2} x \, dy \, dx = \int_{0}^{1} 2x \, dx = 1$.
2. Compute $M_y$: $\int_{0}^{1} \int_{0}^{2} x(x) \, dy \, dx = \int_{0}^{1} 2x^2 \, dx = 2/3$.
3. Compute $M_x$: $\int_{0}^{1} \int_{0}^{2} y(x) \, dy \, dx = \int_{0}^{1} [y^2/2]_0^2 x \, dx = \int_{0}^{1} 2x \, dx = 1$.
4. Calculate coordinates: $\bar{x} = (2/3)/1 = 2/3$ and $\bar{y} = 1/1 = 1$.

Answer: (2/3, 1)

1. Use polar coordinates: $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$.
2. Set up integral: $I_z = \int_{0}^{2\pi} \int_{0}^{1} r^2 \cdot 1 \cdot r \, dr \, d\theta$.
3. Integrate $r^3$ to get $r^4/4$ from $0$ to $1$, which is $1/4$.
4. Integrate $1/4$ from $0$ to $2\pi$.

Answer: \frac{\pi}{2}

1. Identify the $z$ bounds: the solid is bounded below by $z = 0$ and above by $z = x + y$.
2. Identify the shadow region $R$ in the $xy$-plane: it is the quarter disk $x^2 + y^2 \le 1$ where $x \ge 0$ and $y \ge 0$.
3. Convert the shadow region to polar coordinates: $0 \le r \le 1$ and $0 \le \theta \le \pi/2$.
4. Substitute $x = r \cos \theta$ and $y = r \sin \theta$ into the $z$ bound: $z_{top} = r \cos \theta + r \sin \theta$.
5. Assemble the integral with the Jacobian $r$.

Answer: \int_{0}^{\pi/2} \int_{0}^{1} \int_{0}^{r(\cos\theta + \sin\theta)} r \, dz \, dr \, d\theta

1. The cube is centered at the origin with side length 2, so the bounds for $x$, $y$, and $z$ are all from $-1$ to $1$.
2. Since the bounds are constants, the order of integration does not matter.
3. The density function is $\rho(x, y, z) = |x| + |y| + |z|$.
4. The integral is $\int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (|x| + |y| + |z|) \, dz \, dy \, dx$.

Answer: \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (|x| + |y| + |z|) \, dz \, dy \, dx

1. The order $dx \, dz \, dy$ means $x$ is the innermost integral.
2. The $x$ bounds are given as $x = 0$ to $x = 2$.
3. The remaining region is the shadow in the $yz$-plane, which is the semi-disk $y^2 + z^2 \le 1$ where $z \ge 0$.
4. For a fixed $y$, $z$ goes from $0$ to $\sqrt{1 - y^2}$.
5. The outermost variable $y$ ranges from $-1$ to $1$.

Answer: \int_{-1}^{1} \int_{0}^{\sqrt{1-y^2}} \int_{0}^{2} 1 \, dx \, dz \, dy

1. Convert to cylindrical: $x = r \cos \theta$, $y = r \sin \theta$, $z = z$.
2. Bounds: $0 \le z \le r \cos \theta + 4$, $0 \le r \le 3$, $0 \le \theta \le 2\pi$.
3. Integral: $\int_0^{2\pi} \int_0^3 \int_0^{r \cos \theta + 4} r \, dz \, dr \, d\theta$.
4. Inner: $\int_0^{r \cos \theta + 4} r \, dz = r(r \cos \theta + 4) = r^2 \cos \theta + 4r$.
5. Middle: $\int_0^3 (r^2 \cos \theta + 4r) \, dr = [\frac{1}{3}r^3 \cos \theta + 2r^2]_0^3 = 9 \cos \theta + 18$.
6. Outer: $\int_0^{2\pi} (9 \cos \theta + 18) \, d\theta = [9 \sin \theta + 18\theta]_0^{2\pi} = 36\pi$.

Answer: 36\pi

1. The sphere is $z^2 = 4 - r^2$, so $z = \pm \sqrt{4 - r^2}$.
2. The cylinder gives $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.
3. Integral: $\int_0^{2\pi} \int_0^1 \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$.
4. Inner: $2r \sqrt{4-r^2}$.
5. Middle: $\int_0^1 2r \sqrt{4-r^2} \, dr$. Let $u = 4-r^2, du = -2r dr$. $\int_4^3 -\sqrt{u} \, du = [\frac{2}{3}u^{3/2}]_3^4 = \frac{2}{3}(8 - 3\sqrt{3})$.
6. Outer: $2\pi \cdot \frac{2}{3}(8 - 3\sqrt{3}) = \frac{4\pi}{3}(8 - 3\sqrt{3})$.

Answer: \frac{4\pi}{3}(8 - 3\sqrt{3})

1. Cone: $z = \sqrt{3}r$. Sphere: $z = \sqrt{4-r^2}$.
2. Intersection: $\sqrt{3}r = \sqrt{4-r^2} \implies 3r^2 = 4-r^2 \implies 4r^2 = 4 \implies r = 1$.
3. Bounds: $0 \le \theta \le 2\pi$, $0 \le r \le 1$, $\sqrt{3}r \le z \le \sqrt{4-r^2}$.
4. Integral: $\int_0^{2\pi} \int_0^1 \int_{\sqrt{3}r}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$.
5. Inner: $r(\sqrt{4-r^2} - \sqrt{3}r) = r\sqrt{4-r^2} - \sqrt{3}r^2$.
6. Middle: $\int_0^1 (r\sqrt{4-r^2} - \sqrt{3}r^2) \, dr = [-\frac{1}{3}(4-r^2)^{3/2} - \frac{\sqrt{3}}{3}r^3]_0^1 = (-\frac{1}{3}(3\sqrt{3}) - \frac{\sqrt{3}}{3}) - (-\frac{1}{3}(8) - 0) = -\sqrt{3} - \frac{\sqrt{3}}{3} + \frac{8}{3} = \frac{8 - 4\sqrt{3}}{3}$.
7. Outer: $2\pi \cdot \frac{8 - 4\sqrt{3}}{3} = \frac{8\pi(2 - \sqrt{3})}{3}$.

Answer: \frac{8\pi(2 - \sqrt{3})}{3}

1. Set up the integral in spherical coordinates: $M = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{R} \rho_0 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.
2. Integrate $\rho^2$ to get $R^3/3$.
3. Integrate $\sin \phi$ from $0$ to $\pi/2$ to get $1$.
4. Integrate $d\theta$ from $0$ to $2\pi$ to get $2\pi$.
5. Multiply results: $\rho_0 \cdot (R^3/3) \cdot 1 \cdot 2\pi$.

Answer: M = \frac{2}{3}\pi \rho_0 R^3

1. Set up the integral: $V = \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{4} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.
2. The $\rho$ integral is $4^3/3 = 64/3$.
3. The $\phi$ integral is $[-\cos \phi]_0^{\pi/6} = 1 - \sqrt{3}/2$.
4. The $\theta$ integral is $2\pi$.
5. Combine: $2\pi \cdot (1 - \sqrt{3}/2) \cdot 64/3$.

Answer: V = \frac{64\pi}{3}(2 - \sqrt{3})

1. Convert the function to spherical: $f = \rho$.
2. Include the Jacobian: $\rho \cdot \rho^2 \sin \phi = \rho^3 \sin \phi$.
3. Set bounds: $\rho \in [0, 1], \phi \in [0, \pi], \theta \in [0, 2\pi]$.
4. Integrate $\rho^3$ to get $1/4$.
5. Integrate $\sin \phi$ to get $2$.
6. Integrate $d\theta$ to get $2\pi$.
7. Multiply: $(1/4) \cdot 2 \cdot 2\pi$.

Answer: \pi

1. Let $u = x+y$ and $v = x-y$. The bounds are $1 \le u \le 3$ and $0 \le v \le 2$.
2. Solve for $x$ and $y$: $x = (u+v)/2$ and $y = (u-v)/2$.
3. Compute the Jacobian: $\det J = \det \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} = -1/2$. The absolute value is $1/2$.
4. Substitute into the integrand: $x^2 + y^2 = ((u+v)/2)^2 + ((u-v)/2)^2 = (u^2 + 2uv + v^2 + u^2 - 2uv + v^2)/4 = (2u^2 + 2v^2)/4 = (u^2 + v^2)/2$.
5. Set up the integral: $\int_0^2 \int_1^3 \frac{u^2 + v^2}{2} \cdot \frac{1}{2} \, du \, dv = \frac{1}{4} \int_0^2 [\frac{u^3}{3} + v^2 u]_1^3 \, dv$.
6. Evaluate: $\frac{1}{4} \int_0^2 (\frac{26}{3} + 2v^2) \, dv = \frac{1}{4} [\frac{26}{3}v + \frac{2v^3}{3}]_0^2 = \frac{1}{4} (\frac{52}{3} + \frac{16}{3}) = \frac{1}{4} \cdot \frac{68}{3} = \frac{17}{3}$.

Answer: 17/3

1. Compute partials of $x$: $\frac{\partial x}{\partial u} = 2u$, $\frac{\partial x}{\partial v} = -2v$.
2. Compute partials of $y$: $\frac{\partial y}{\partial u} = 2v$, $\frac{\partial y}{\partial v} = 2u$.
3. Set up the determinant: $\det J = (2u)(2u) - (-2v)(2v)$.
4. Simplify: $4u^2 + 4v^2 = 4(u^2 + v^2)$.

Answer: 4(u^2 + v^2)

1. Let $u = x-y$ and $v = x+y$. The bounds are $0 \le u \le 1$ and $0 \le v \le 2$.
2. The Jacobian for this transformation is $1/2$ (as calculated in the lesson example).
3. The integral becomes $\int_0^2 \int_0^1 e^u \cdot \frac{1}{2} \, du \, dv$.
4. Integrate $e^u$: $\frac{1}{2} \int_0^2 [e^u]_0^1 \, dv = \frac{1}{2} \int_0^2 (e - 1) \, dv$.
5. Integrate with respect to $v$: $\frac{1}{2} (e-1) [v]_0^2 = \frac{1}{2} (e-1) \cdot 2 = e-1$.

Answer: e - 1

1. Identify semi axes $a=4$ and $b=1$.
2. Use transformation $x=4r\cos\theta, y=r\sin\theta$.
3. Compute Jacobian $\det J = (4)(1)r = 4r$.
4. Set up integral $\int_0^{2\pi} \int_0^1 4r \, dr \, d\theta$.
5. Integrate $4r$ to get $2r^2$, evaluated from $0$ to $1$ is $2$.
6. Integrate $2$ from $0$ to $2\pi$ to get $4\pi$.

Answer: 4\pi

1. Let $u=x+y$ and $v=x-y$.
2. The bounds are $u \in [0, 2]$ and $v \in [0, 2]$.
3. The Jacobian absolute value is $|\det J| = 1/2$.
4. The integrand $x+y$ becomes $u$.
5. Set up integral $\int_0^2 \int_0^2 u \cdot (1/2) \, du \, dv$.
6. Inner integral $\int_0^2 (u/2) \, du = [u^2/4]_0^2 = 1$.
7. Outer integral $\int_0^2 1 \, dv = 2$.

Answer: 2

1. Switch to polar coordinates: $x^2+y^2 = r^2$ and $dA = r \, dr \, d\theta$.
2. Set up integral $\int_0^{2\pi} \int_0^{\infty} e^{-2r^2} r \, dr \, d\theta$.
3. Use substitution $w = 2r^2$, $dw = 4r \, dr$, so $r \, dr = dw/4$.
4. Inner integral $\int_0^{\infty} e^{-w} (1/4) \, dw = [ -1/4 e^{-w} ]_0^{\infty} = 1/4$.
5. Outer integral $\int_0^{2\pi} (1/4) \, d\theta = 2\pi/4 = \pi/2$.

Answer: \pi/2

1. Recognize the region is a quarter disk of radius 1, suggesting polar coordinates.
2. Substitute $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$.
3. Set bounds: $r$ from $0$ to $1$, $\theta$ from $0$ to $\pi/2$.
4. Compute $\int_0^{\pi/2} \int_0^1 r^2 \cdot r \, dr \, d\theta = \int_0^{\pi/2} [r^4/4]_0^1 \, d\theta = \int_0^{\pi/2} 1/4 \, d\theta = \pi/8$.

Answer: \pi/8

1. The region is a spherical sector, suggesting spherical coordinates.
2. The cone $z = \sqrt{x^2 + y^2}$ corresponds to $\phi = \pi/4$.
3. The sphere $x^2 + y^2 + z^2 = 1$ corresponds to $\rho = 1$.
4. First octant means $\theta$ goes from $0$ to $\pi/2$.
5. Compute $\int_0^{\pi/2} \int_0^{\pi/4} \int_0^1 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.
6. $\int_0^1 \rho^2 \, d\rho = 1/3$. $\int_0^{\pi/4} \sin \phi \, d\phi = [-\cos \phi]_0^{\pi/4} = 1 - \sqrt{2}/2$.
7. Total volume is $(\pi/2) \cdot (1/3) \cdot (1 - \sqrt{2}/2) = \frac{\pi(2-\sqrt{2})}{12}$.

Answer: \frac{\pi(2-\sqrt{2})}{12}

1. The region is a rotated square. Define $u = x+y$ and $v = x-y$.
2. The vertices in $uv$ coordinates are $(0,0), (2,0), (2,2), (0,2)$. This is a square $0 \le u \le 2, 0 \le v \le 2$.
3. Compute the Jacobian: $x = (u+v)/2, y = (u-v)/2$. $\det J = |(1/2)(1/2) - (1/2)(-1/2)| = 1/2$.
4. The integral becomes $\int_0^2 \int_0^2 u^2 e^v (1/2) \, du \, dv$.
5. $\int_0^2 u^2 \, du = 8/3$. $\int_0^2 e^v \, dv = e^2 - 1$.
6. Result is $(1/2) \cdot (8/3) \cdot (e^2 - 1) = \frac{4}{3}(e^2 - 1)$.

Answer: \frac{4}{3}(e^2 - 1)

1. Sketch the region: $y$ goes from $0$ to $1$, and $x$ goes from $y$ to $1$. This is a triangle with vertices $(0,0), (1,0), (1,1)$.
2. Swap the order: $x$ now goes from $0$ to $1$, and for a fixed $x$, $y$ goes from $0$ to $x$.
3. Set up the new integral: $\int_{0}^{1} \int_{0}^{x} \sin(x^2) \, dy \, dx$.
4. Integrate with respect to $y$: $\int_{0}^{1} [y \sin(x^2)]_{0}^{x} \, dx = \int_{0}^{1} x \sin(x^2) \, dx$.
5. Use u-substitution $u=x^2$, $du=2x dx$: $[-\frac{1}{2}\cos(x^2)]_{0}^{1} = \frac{1}{2}(1 - \cos(1))$.

Answer: $\frac{1}{2}(1 - \cos(1))$

1. Use cylindrical coordinates: $x = r \cos \theta, y = r \sin \theta, z = z$.
2. The region is $0 \le r \le 2, 0 \le \theta \le 2\pi, 0 \le z \le r \cos \theta + 3$.
3. Set up the integral: $\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{r \cos \theta + 3} r \, dz \, dr \, d\theta$.
4. Integrate $z$: $\int_{0}^{2\pi} \int_{0}^{2} r(r \cos \theta + 3) \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{2} (r^2 \cos \theta + 3r) \, dr \, d\theta$.
5. Integrate $r$: $\int_{0}^{2\pi} [\frac{1}{3}r^3 \cos \theta + \frac{3}{2}r^2]_{0}^{2} \, d\theta = \int_{0}^{2\pi} (\frac{8}{3} \cos \theta + 6) \, d\theta$.
6. Integrate $\theta$: $[\frac{8}{3} \sin \theta + 6\theta]_{0}^{2\pi} = 12\pi$.

Answer: $12\pi$

1. Define $u = x+y$ and $v = x-y$.
2. The vertices in $(u,v)$ are: $(1,0) \to (1,1), (2,1) \to (3,1), (1,2) \to (3,-1), (0,1) \to (1,-1)$.
3. The region $S$ is $1 \le u \le 3$ and $-1 \le v \le 1$.
4. Compute Jacobian: $x = (u+v)/2, y = (u-v)/2$. $\det J = 1/2$.
5. Set up integral: $\int_{-1}^{1} \int_{1}^{3} u^2 e^v \cdot \frac{1}{2} \, du \, dv$.
6. Integrate $u$: $\frac{1}{2} \int_{-1}^{1} e^v [\frac{1}{3}u^3]_{1}^{3} \, dv = \frac{1}{2} \int_{-1}^{1} e^v (9 - 1/3) \, dv = \frac{1}{2} \cdot \frac{26}{3} \int_{-1}^{1} e^v \, dv$.
7. Integrate $v$: $\frac{13}{3} [e^v]_{-1}^{1} = \frac{13}{3}(e - e^{-1})$.

Answer: $\frac{13}{3}(e - e^{-1})$

1. Identify $P(x, y) = x + y$ and $Q(x, y) = x^2y$.
2. Substitute $x = 1$ and $y = -2$ into $P$: $P(1, -2) = 1 + (-2) = -1$.
3. Substitute $x = 1$ and $y = -2$ into $Q$: $Q(1, -2) = (1)^2(-2) = -2$.
4. Combine the components into a vector.

Answer: \mathbf{F}(1, -2) = -1\mathbf{i} - 2\mathbf{j} \text{ or } \langle -1, -2 \rangle

1. Compute the partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = y e^{xy}$.
2. Compute the partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = x e^{xy}$.
3. Write the gradient field: $\nabla f = y e^{xy}\mathbf{i} + x e^{xy}\mathbf{j}$.
4. Substitute $x = 0$ and $y = 1$: $\nabla f(0, 1) = (1)e^0\mathbf{i} + (0)e^0\mathbf{j} = 1\mathbf{i} + 0\mathbf{j}$.

Answer: \nabla f(0, 1) = \mathbf{i} \text{ or } \langle 1, 0 \rangle

1. Observe that the vector is a scalar multiple of the position vector $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.
2. The scalar multiple is $1/(x^2+y^2+z^2)$, which is always positive. Thus, all vectors point radially away from the origin.
3. Substitute $x=1, y=1, z=1$ into the components.
4. The denominator is $1^2+1^2+1^2 = 3$.
5. The components are $1/3, 1/3, 1/3$.

Answer: \mathbf{F}(1, 1, 1) = \frac{1}{3}\mathbf{i} + \frac{1}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}

1. Parameterize the line: $\mathbf{r}(t) = \langle 2t, 3t \rangle$ for $0 \le t \le 1$.
2. Compute the derivative: $\mathbf{r}'(t) = \langle 2, 3 \rangle$.
3. Find the speed: $|\mathbf{r}'(t)| = \sqrt{2^2 + 3^2} = \sqrt{13}$.
4. Substitute into the density function: $f(2t, 3t) = (2t)^2 + 3t = 4t^2 + 3t$.
5. Set up the integral: $\int_0^1 (4t^2 + 3t)\sqrt{13} \, dt$.
6. Integrate: $\sqrt{13} [\frac{4}{3}t^3 + \frac{3}{2}t^2]_0^1 = \sqrt{13}(\frac{4}{3} + \frac{3}{2}) = \sqrt{13}(\frac{17}{6})$.

Answer: \frac{17\sqrt{13}}{6}

1. Parameterize the curve: $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$ for $0 \le t \le \pi/2$.
2. Compute the derivative: $\mathbf{r}'(t) = \langle -2\sin t, 2\cos t \rangle$.
3. Find the speed: $|\mathbf{r}'(t)| = \sqrt{4\sin^2 t + 4\cos^2 t} = 2$.
4. Substitute into the integrand: $x = 2\cos t$.
5. Set up the integral: $\int_0^{\pi/2} (2\cos t)(2) \, dt = 4 \int_0^{\pi/2} \cos t \, dt$.
6. Evaluate: $4 [\sin t]_0^{\pi/2} = 4(1 - 0) = 4$.

Answer: 4

1. Compute the derivative: $\mathbf{r}'(t) = \langle 1, 2t, 2t^2 \rangle$.
2. Find the speed: $|\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2 + (2t^2)^2} = \sqrt{1 + 4t^2 + 4t^4}$.
3. Recognize the perfect square: $\sqrt{(1 + 2t^2)^2} = 1 + 2t^2$.
4. Set up the integral: $\int_0^1 (1 + 2t^2) \, dt$.
5. Integrate: $[t + \frac{2}{3}t^3]_0^1 = 1 + \frac{2}{3} = \frac{5}{3}$.

Answer: \frac{5}{3}

1. Parameterize the path: $\mathbf{r}(t) = 2t\mathbf{i} + t\mathbf{j}$ for $t \in [0, 1]$.
2. Find the velocity vector: $\mathbf{r}'(t) = 2\mathbf{i} + 1\mathbf{j}$.
3. Substitute the path into the field: $\mathbf{F}(\mathbf{r}(t)) = 2(2t)\mathbf{i} + 3(t)\mathbf{j} = 4t\mathbf{i} + 3t\mathbf{j}$.
4. Compute the dot product: $\mathbf{F} \cdot \mathbf{r}' = (4t)(2) + (3t)(1) = 8t + 3t = 11t$.
5. Integrate from $0$ to $1$: $\int_0^1 11t \, dt = [5.5t^2]_0^1 = 5.5$.

Answer: 5.5

1. Parameterize the path: $\mathbf{r}(t) = 2\cos(t)\mathbf{i} + 2\sin(t)\mathbf{j}$ for $t \in [0, \pi]$.
2. Find the velocity vector: $\mathbf{r}'(t) = -2\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j}$.
3. Substitute into the field: $\mathbf{F}(\mathbf{r}(t)) = -2\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j}$.
4. Compute the dot product: $\mathbf{F} \cdot \mathbf{r}' = (-2\sin t)(-2\sin t) + (2\cos t)(2\cos t) = 4\sin^2 t + 4\cos^2 t = 4$.
5. Integrate from $0$ to $\pi$: $\int_0^\pi 4 \, dt = [4t]_0^\pi = 4\pi$.

Answer: 4\pi

1. Find the velocity vector: $\mathbf{r}'(t) = 2t\mathbf{i} + 3t^2\mathbf{j}$.
2. Substitute the path into the field: $\mathbf{F}(\mathbf{r}(t)) = t^3\mathbf{i} - t^2\mathbf{j}$.
3. Compute the dot product: $\mathbf{F} \cdot \mathbf{r}' = (t^3)(2t) + (-t^2)(3t^2) = 2t^4 - 3t^4 = -t^4$.
4. Integrate from $0$ to $1$: $\int_0^1 -t^4 \, dt = [-\frac{1}{5}t^5]_0^1 = -0.2$.

Answer: -0.2

1. Check conservativeness: $P_y = \frac{\partial}{\partial y}(2xy + 1) = 2x$ and $Q_x = \frac{\partial}{\partial x}(x^2 + 3y^2) = 2x$. Since $P_y = Q_x$, it is conservative.
2. Find potential $f$: $\int (2xy + 1) dx = x^2y + x + g(y)$.
3. Differentiate with respect to $y$: $x^2 + g'(y) = x^2 + 3y^2$, so $g'(y) = 3y^2$, which means $g(y) = y^3$.
4. Potential function is $f(x, y) = x^2y + x + y^3$.
5. Compute $f(1, 2) - f(0, 0) = (1^2(2) + 1 + 2^3) - 0 = 2 + 1 + 8 = 11$.

Answer: 11

1. Since the potential $f$ is given, we use the Fundamental Theorem directly.
2. Evaluate $f$ at the end point $(1, 0, 2)$: $f(1, 0, 2) = e^{1+0} + 2^2 = e + 4$.
3. Evaluate $f$ at the start point $(0, 0, 0)$: $f(0, 0, 0) = e^{0+0} + 0^2 = 1$.
4. Work $W = f(1, 0, 2) - f(0, 0, 0) = e + 4 - 1 = e + 3$.

Answer: e + 3

1. Check if $\mathbf{F}$ is conservative: $P = y^2$, so $P_y = 2y$. $Q = 2xy$, so $Q_x = 2y$.
2. Since $P_y = Q_x$, the field is conservative.
3. The unit circle is a closed loop, meaning the start and end points are identical.
4. By the Fundamental Theorem for Line Integrals, the integral of a conservative field over any closed loop is zero.

Answer: 0

1. Integrate $P = 2x + y$ with respect to $x$: $\int (2x + y) \, dx = x^2 + xy + g(y)$.
2. Differentiate with respect to $y$: $\frac{\partial}{\partial y}(x^2 + xy + g(y)) = x + g'(y)$.
3. Set equal to $Q = x + 2$: $x + g'(y) = x + 2$, so $g'(y) = 2$.
4. Integrate $g'(y)$: $g(y) = 2y + C$.

Answer: $f(x, y) = x^2 + xy + 2y + C$

1. Integrate $P = 2xy + 1$ with respect to $x$: $\int (2xy + 1) \, dx = x^2 y + x + g(y)$.
2. Differentiate with respect to $y$: $\frac{\partial}{\partial y}(x^2 y + x + g(y)) = x^2 + g'(y)$.
3. Set equal to $Q = x^2$: $x^2 + g'(y) = x^2$, so $g'(y) = 0$.
4. Integrate $g'(y)$: $g(y) = C$.

Answer: $f(x, y) = x^2 y + x + C$

1. Integrate $P = 2xz$ with respect to $x$: $f = x^2 z + g(y, z)$.
2. Differentiate with respect to $y$: $\frac{\partial f}{\partial y} = 0 + \frac{\partial g}{\partial y}$. Set equal to $Q = 2yz$: $\frac{\partial g}{\partial y} = 2yz$.
3. Integrate $\frac{\partial g}{\partial y}$ with respect to $y$: $g(y, z) = y^2 z + h(z)$.
4. Now $f = x^2 z + y^2 z + h(z)$. Differentiate with respect to $z$: $\frac{\partial f}{\partial z} = x^2 + y^2 + h'(z)$.
5. Set equal to $R = x^2 + y^2$: $x^2 + y^2 + h'(z) = x^2 + y^2$, so $h'(z) = 0$.

Answer: $f(x, y, z) = (x^2 + y^2)z + C$

1. Identify $P = 3x^2 + y$ and $Q = 2xy + e^y$.
2. Compute $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3x^2 + y) = 1$.
3. Compute $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy + e^y) = 2y$.
4. Compare the results: $1 \neq 2y$.

Answer: Not conservative

1. Identify $P = 2x + e^y$ and $Q = x e^y + 3y^2$.
2. Compute $\frac{\partial P}{\partial y} = e^y$.
3. Compute $\frac{\partial Q}{\partial x} = e^y$.
4. Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ and the domain is simply connected, the field is conservative.

Answer: Conservative

1. Identify $P = \frac{-y}{x^2+y^2}$ and $Q = \frac{x}{x^2+y^2}$.
2. Compute $\frac{\partial P}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2}$ using the quotient rule.
3. Compute $\frac{\partial Q}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2}$ using the quotient rule.
4. The partials are equal, but the domain is not simply connected. A line integral around the unit circle yields $2\pi$, not $0$, so the field is not conservative.

Answer: Not conservative

1. Identify $P(x, y) = x^3 + y^2$ and $Q(x, y) = \cos(xy)$.
2. Compute $\frac{\partial P}{\partial x} = 3x^2$.
3. Compute $\frac{\partial Q}{\partial y} = -x\sin(xy)$ using the chain rule.
4. Sum the partials: $\text{div } \mathbf{F} = 3x^2 - x\sin(xy)$.

Answer: 3x^2 - x\sin(xy)

1. Identify components: $P = yz$, $Q = xz$, $R = xy$.
2. Compute $\frac{\partial P}{\partial x} = 0$.
3. Compute $\frac{\partial Q}{\partial y} = 0$.
4. Compute $\frac{\partial R}{\partial z} = 0$.
5. Sum the partials: $0 + 0 + 0 = 0$.

Answer: Yes, it is solenoidal.

1. Compute $\frac{\partial P}{\partial x} = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$.
2. Compute $\frac{\partial Q}{\partial y} = 2e^{2y}$.
3. The general divergence is $\text{div } \mathbf{F} = \frac{2x}{x^2+1} + 2e^{2y}$.
4. Evaluate at $(1, 0)$: $\frac{2(1)}{1^2+1} + 2e^{2(0)} = \frac{2}{2} + 2(1) = 1 + 2 = 3$.

Answer: 3

1. Identify $P = 3xy$ and $Q = x^2$.
2. Compute $\frac{\partial Q}{\partial x} = 2x$.
3. Compute $\frac{\partial P}{\partial y} = 3x$.
4. Subtract them: $\text{curl } \mathbf{F} = 2x - 3x = -x$.

Answer: \text{curl } \mathbf{F} = -x; \text{ not irrotational}

1. Identify $P = e^x$ and $Q = \sin(y)$.
2. Compute $\frac{\partial Q}{\partial x} = 0$ since $\sin(y)$ does not depend on $x$.
3. Compute $\frac{\partial P}{\partial y} = 0$ since $e^x$ does not depend on $y$.
4. Subtract them: $0 - 0 = 0$.

Answer: \text{curl } \mathbf{F} = 0

1. Identify $P = y^3$ and $Q = 3xy^2$.
2. Compute $\frac{\partial Q}{\partial x} = 3y^2$.
3. Compute $\frac{\partial P}{\partial y} = 3y^2$.
4. Subtract them: $3y^2 - 3y^2 = 0$.

Answer: \text{curl } \mathbf{F} = 0; \text{ no rotation at } (1, 2)

1. Compute the curl of $\mathbf{F}$.
2. The x-component of the curl is $\frac{\partial}{\partial y}(x) - \frac{\partial}{\partial z}(x^2) = 0 - 0 = 0$.
3. The y-component of the curl is $\frac{\partial}{\partial z}(2xy + z) - \frac{\partial}{\partial x}(x) = 1 - 1 = 0$.
4. The z-component of the curl is $\frac{\partial}{\partial x}(x^2) - \frac{\partial}{\partial y}(2xy + z) = 2x - 2x = 0$.
5. Since the curl is the zero vector, the field is irrotational and can be a gradient.

Answer: Yes, it can be a gradient because $\text{curl } \mathbf{F} = \mathbf{0}$.

1. Identify that $\mathbf{G}$ is defined as the curl of another vector field $\mathbf{A}$.
2. Apply the identity $\text{div}(\text{curl } \mathbf{F}) = 0$.
3. Since $\mathbf{G} = \text{curl } \mathbf{A}$, then $\text{div } \mathbf{G} = \text{div}(\text{curl } \mathbf{A}) = 0$.

Answer: 0

1. Compute the divergence of $\mathbf{V}$.
2. $\text{div } \mathbf{V} = \frac{\partial}{\partial x}(x^2 + y^2) + \frac{\partial}{\partial y}(2xy) + \frac{\partial}{\partial z}(z^2)$.
3. $\text{div } \mathbf{V} = 2x + 2x + 2z = 4x + 2z$.
4. Since the divergence is not zero everywhere, the field is not solenoidal.
5. By the identity $\text{div}(\text{curl } \mathbf{G}) = 0$, a field must have zero divergence to be a curl.

Answer: No, it cannot be a curl because $\text{div } \mathbf{V} \neq 0$.

1. Check for conservativity: $\frac{\partial P}{\partial y} = 2x$ and $\frac{\partial Q}{\partial x} = 2x$. The field is conservative.
2. Find the potential function $f(x,y)$: $\int 2xy dx = x^2 y + g(y)$. Differentiating with respect to $y$ gives $x^2 + g'(y) = x^2 + 1$, so $g'(y) = 1$ and $g(y) = y$.
3. The potential function is $f(x,y) = x^2 y + y$.
4. Evaluate at endpoints: $f(1,1) = 1^2(1) + 1 = 2$ and $f(0,0) = 0$.
5. Work is $2 - 0 = 2$.

Answer: 2

1. Identify $P = y^2$ and $Q = 3xy$.
2. Compute the curl: $\frac{\partial Q}{\partial x} = 3y$ and $\frac{\partial P}{\partial y} = 2y$.
3. The integrand is $3y - 2y = y$.
4. Set up the double integral: $\int_0^1 \int_0^2 y dy dx$.
5. Inner integral: $[\frac{1}{2}y^2]_0^2 = 2$.
6. Outer integral: $\int_0^1 2 dx = 2$.

Answer: 2

1. Compute divergence: $\frac{\partial}{\partial x}(e^x \sin y) + \frac{\partial}{\partial y}(-e^x \cos y) = e^x \sin y + e^x \sin y = 2e^x \sin y$. Not zero, so not solenoidal.
2. Compute curl: $\frac{\partial}{\partial x}(-e^x \cos y) - \frac{\partial}{\partial y}(e^x \sin y) = -e^x \cos y - e^x \cos y = -2e^x \cos y$. Not zero, so not irrotational.
3. The field is neither.

Answer: Neither solenoidal nor irrotational

1. Identify $P = x+y$ and $Q = x^2$.
2. Compute partial derivatives: $\frac{\partial Q}{\partial x} = 2x$ and $\frac{\partial P}{\partial y} = 1$.
3. The integrand is $2x - 1$.
4. Set up the double integral over the square: $\int_0^1 \int_0^1 (2x - 1) dy dx$.
5. Integrate with respect to $y$: $\int_0^1 (2x - 1) dx$.
6. Integrate with respect to $x$: $[x^2 - x]_0^1 = (1-1) - (0-0) = 0$.

Answer: 0

1. Identify $P = y^2$ and $Q = 3xy$.
2. Compute partial derivatives: $\frac{\partial Q}{\partial x} = 3y$ and $\frac{\partial P}{\partial y} = 2y$.
3. The integrand is $3y - 2y = y$.
4. Set up the double integral over the disk $R$ of radius 2: $\iint_R y dA$.
5. By symmetry, the integral of $y$ over a disk centered at the origin is 0.

Answer: 0

1. Identify $P = e^x - y$ and $Q = e^y + x$.
2. Compute partial derivatives: $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = -1$.
3. The integrand is $1 - (-1) = 2$.
4. The region $R$ is a rectangle with area $2 \times 1 = 2$.
5. The integral is $\iint_R 2 dA = 2 \times \text{Area}(R) = 2 \times 2 = 4$.

Answer: 4

1. Identify $P = y + x^2$ and $Q = x + y^2$.
2. Compute the 2D curl: $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = 1$.
3. The integrand is $1 - 1 = 0$.
4. The double integral of $0$ over the square is $0$.

Answer: 0

1. Identify $P = y^2$ and $Q = 3xy$.
2. Compute the 2D curl: $\frac{\partial Q}{\partial x} = 3y$ and $\frac{\partial P}{\partial y} = 2y$.
3. The integrand is $3y - 2y = y$.
4. Set up the double integral in polar coordinates: $\int_0^\pi \int_0^1 (r \sin \theta) r dr d\theta$.
5. Integrate $r^2$ to get $1/3$ and $\sin \theta$ from $0$ to $\pi$ to get $2$.
6. The result is $2/3$.

Answer: 2/3

1. Identify $P = e^x - y$ and $Q = e^y + x$.
2. Compute the 2D curl: $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = -1$.
3. The integrand is $1 - (-1) = 2$.
4. The double integral over the disk of radius $2$ is $2 \times \text{Area} = 2 \times (\pi \cdot 2^2) = 8\pi$.
5. Since the orientation is clockwise, multiply by $-1$.

Answer: -8\pi

1. Compute differentials: $dx = -2 \sin t dt$ and $dy = \cos t dt$.
2. Substitute into the integrand: $x dy - y dx = (2 \cos t)(\cos t dt) - (\sin t)(-2 \sin t dt)$.
3. Simplify the expression: $2 \cos^2 t dt + 2 \sin^2 t dt = 2(\cos^2 t + \sin^2 t) dt = 2 dt$.
4. Integrate from $0$ to $2\pi$: $\int_0^{2\pi} 2 dt = 4\pi$.
5. Multiply by the factor of $1/2$: $\text{Area} = \frac{1}{2}(4\pi) = 2\pi$.

Answer: 2\pi

1. Compute differentials: $dx = 2t dt$ and $dy = (3t^2 - 3) dt$.
2. Substitute into $x dy - y dx$: $(t^2)(3t^2 - 3) dt - (t^3 - 3t)(2t dt)$.
3. Expand and simplify: $(3t^4 - 3t^2) dt - (2t^4 - 6t^2) dt = (t^4 + 3t^2) dt$.
4. Integrate from $-\sqrt{3}$ to $\sqrt{3}$: $\int_{-\sqrt{3}}^{\sqrt{3}} (t^4 + 3t^2) dt = [\frac{1}{5}t^5 + t^3]_{-\sqrt{3}}^{\sqrt{3}}$.
5. Evaluate: $(\frac{9\sqrt{3}}{5} + 3\sqrt{3}) - (-\frac{9\sqrt{3}}{5} - 3\sqrt{3}) = \frac{18\sqrt{3}}{5} + 6\sqrt{3} = \frac{48\sqrt{3}}{5}$.
6. Multiply by $1/2$: $\text{Area} = \frac{24\sqrt{3}}{5}$.

Answer: \frac{24\sqrt{3}}{5}

1. The boundary consists of two parts: $C_1$ from $(0,0)$ to $(1,1)$ along $y=x^2$, and $C_2$ from $(1,1)$ back to $(0,0)$ along $y=x$.
2. For $C_1$: $x=t, y=t^2, dy=2t dt$ for $t \in [0,1]$. $\int_{C_1} x dy = \int_0^1 t(2t dt) = \int_0^1 2t^2 dt = [\frac{2}{3}t^3]_0^1 = \frac{2}{3}$.
3. For $C_2$: $x=t, y=t, dy=dt$ for $t$ from $1$ to $0$. $\int_{C_2} x dy = \int_1^0 t dt = [\frac{1}{2}t^2]_1^0 = 0 - \frac{1}{2} = -\frac{1}{2}$.
4. Sum the integrals: $\text{Area} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.

Answer: 1/6

1. Identify $P = 3x$ and $Q = 2y$.
2. Compute the divergence: $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = 3 + 2 = 5$.
3. Apply the flux form of Green's Theorem: $\text{Flux} = \iint_R 5 dA$.
4. The area of the disk with radius 3 is $\pi(3)^2 = 9\pi$.
5. The total flux is $5 \times 9\pi = 45\pi$.

Answer: 45\pi

1. Identify $P = x+y$ and $Q = x-y$.
2. Compute the divergence: $\frac{\partial P}{\partial x} = 1$ and $\frac{\partial Q}{\partial y} = -1$.
3. The divergence is $1 + (-1) = 0$.
4. The integral of zero over any region is zero.

Answer: 0

1. Identify $P = e^y$ and $Q = x^2$.
2. Compute the divergence: $\frac{\partial P}{\partial x} = 0$ and $\frac{\partial Q}{\partial y} = 0$.
3. The divergence is $0 + 0 = 0$.
4. The total flux is the double integral of zero, which is zero.

Answer: 0

1. Rewrite the surface as $z = 1 - x - y$.
2. Compute partial derivatives: $f_x = -1$ and $f_y = -1$.
3. Find the area element: $\sqrt{1 + (-1)^2 + (-1)^2} = \sqrt{3}$.
4. Identify the domain $D$ in the $xy$-plane: the triangle bounded by $x=0, y=0,$ and $x+y=1$.
5. Integrate: $\int_0^1 \int_0^{1-x} \sqrt{3} \, dy dx = \sqrt{3} \int_0^1 (1-x) dx = \sqrt{3} [x - x^2/2]_0^1 = \sqrt{3}/2$.

Answer: $\frac{\sqrt{3}}{2}$

1. Compute partial derivatives: $f_x = y$ and $f_y = x$.
2. The area element is $\sqrt{1 + y^2 + x^2}$.
3. Set up the integral: $\int_0^1 \int_0^1 \sqrt{1 + x^2 + y^2} \, dx dy$.

Answer: $\int_0^1 \int_0^1 \sqrt{1 + x^2 + y^2} \, dx dy$

1. Compute $\mathbf{r}_\theta = \langle -R\sin\theta, R\cos\theta, 0 \rangle$.
2. Compute $\mathbf{r}_z = \langle 0, 0, 1 \rangle$.
3. Compute the cross product: $\mathbf{r}_\theta \times \mathbf{r}_z = \langle R\cos\theta, R\sin\theta, 0 \rangle$.
4. Find the magnitude: $\sqrt{R^2\cos^2\theta + R^2\sin^2\theta + 0} = R$.
5. Integrate over $0 \le \theta \le 2\pi$ and $0 \le z \le H$: $\int_0^H \int_0^{2\pi} R \, d\theta dz = R(2\pi)(H) = 2\pi RH$.

Answer: $2\pi RH$

1. Parameterise the surface as $z = 1 - x - y$ over the domain $D: 0 \le x \le 1, 0 \le y \le 1 - x$.
2. Compute the surface area element: $z_x = -1, z_y = -1$, so $dS = \sqrt{1 + (-1)^2 + (-1)^2} dx dy = \sqrt{3} dx dy$.
3. Substitute $z$ into the function: $f = x + y + (1 - x - y) = 1$.
4. Set up the integral: $\int_0^1 \int_0^{1-x} 1 \cdot \sqrt{3} dy dx$.
5. Evaluate: $\sqrt{3} \int_0^1 (1 - x) dx = \sqrt{3} [x - x^2/2]_0^1 = \sqrt{3}/2$.

Answer: $\frac{\sqrt{3}}{2}$

1. Use spherical coordinates with $\rho = 2$. The area element is $dS = R^2 \sin \phi d\theta d\phi = 4 \sin \phi d\theta d\phi$.
2. The density is $\rho = z = 2 \cos \phi$.
3. Set up the integral: $\int_0^{2\pi} \int_0^{\pi/2} (2 \cos \phi)(4 \sin \phi) d\phi d\theta$.
4. Simplify the integrand: $8 \sin \phi \cos \phi = 4 \sin(2\phi)$.
5. Evaluate: $2\pi \int_0^{\pi/2} 4 \sin(2\phi) d\phi = 2\pi [-2 \cos(2\phi)]_0^{\pi/2} = 2\pi [2 - (-2)] = 8\pi$.

Answer: $8\pi$

1. Parameterise the cylinder: $x = \cos \theta, y = \sin \theta, z = z$.
2. Compute $dS$: $\mathbf{r}_\theta = (-\sin \theta, \cos \theta, 0)$ and $\mathbf{r}_z = (0, 0, 1)$. The cross product is $(\cos \theta, \sin \theta, 0)$, and its magnitude is $1$. So $dS = d\theta dz$.
3. Substitute into $f$: $f = \cos^2 \theta + \sin^2 \theta = 1$.
4. Compute the total integral: $\int_0^2 \int_0^{2\pi} 1 \cdot 1 d\theta dz = 4\pi$.
5. Compute the surface area: $\text{Area}(S) = \int_0^2 \int_0^{2\pi} 1 d\theta dz = 4\pi$.
6. Average value: $f_{\text{avg}} = 4\pi / 4\pi = 1$.

Answer: $1$

1. The surface is $z=0$, so the parameterisation is $\mathbf{r}(u, v) = \langle u, v, 0 \rangle$ with $u^2 + v^2 \le 1$.
2. The normal vector is $\mathbf{r}_u \times \mathbf{r}_v = \langle 0, 0, 1 \rangle$.
3. Substitute the surface into the field: $\mathbf{F}(\mathbf{r}(u, v)) = \langle 0, u, v \rangle$.
4. Compute the dot product: $\langle 0, u, v \rangle \cdot \langle 0, 0, 1 \rangle = v$.
5. Integrate $v$ over the unit disk: $\iint_D v \, du dv$.
6. By symmetry, the integral of $v$ over a disk centered at the origin is zero.

Answer: 0

1. Parameterise the cylinder: $\mathbf{r}(\theta, z) = \langle \cos\theta, \sin\theta, z \rangle$ for $0 \le \theta \le 2\pi, 0 \le z \le 2$.
2. Compute tangents: $\mathbf{r}_\theta = \langle -\sin\theta, \cos\theta, 0 \rangle$ and $\mathbf{r}_z = \langle 0, 0, 1 \rangle$.
3. Cross product: $\mathbf{r}_\theta \times \mathbf{r}_z = \langle \cos\theta, \sin\theta, 0 \rangle$. This points outwards.
4. Substitute into field: $\mathbf{F} = \langle 0, 0, \cos^2\theta + \sin^2\theta \rangle = \langle 0, 0, 1 \rangle$.
5. Dot product: $\langle 0, 0, 1 \rangle \cdot \langle \cos\theta, \sin\theta, 0 \rangle = 0$.
6. The integral of zero is zero.

Answer: 0

1. Use the graph form $z = \sqrt{1 - x^2 - y^2}$.
2. Compute partials: $z_x = -x/\sqrt{1-x^2-y^2}$ and $z_y = -y/\sqrt{1-x^2-y^2}$.
3. The normal element is $\mathbf{n} dS = \langle -z_x, -z_y, 1 \rangle dx dy = \langle x/z, y/z, 1 \rangle dx dy$.
4. Dot product: $\mathbf{F} \cdot \mathbf{n} dS = \langle x, y, 0 \rangle \cdot \langle x/z, y/z, 1 \rangle dx dy = (x^2/z + y^2/z) dx dy = \frac{x^2+y^2}{z} dx dy$.
5. Convert to polar: $\iint_D \frac{r^2}{\sqrt{1-r^2}} r dr d\theta$.
6. Integrate $r^3/\sqrt{1-r^2}$ from $0$ to $1$ using $u = 1-r^2$. The result is $2/3$.
7. Multiply by $2\pi$ from the $\theta$ integral to get $4\pi/3$.

Answer: 4\pi/3

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 1 + 2 + 3 = 6$.
2. Apply the Divergence Theorem: $\Phi = \iiint_V 6 dV$.
3. The volume of the sphere is $\frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi$.
4. The flux is $6 \cdot \frac{4}{3}\pi = 8\pi$.

Answer: 8\pi

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2$.
2. Set up the triple integral: $\int_0^1 \int_0^1 \int_0^1 (3x^2 + 3y^2 + 3z^2) dx dy dz$.
3. Integrate $3x^2$ over the cube: $\int_0^1 3x^2 dx = [x^3]_0^1 = 1$.
4. By symmetry, the integrals of $3y^2$ and $3z^2$ are also 1.
5. Total flux is $1 + 1 + 1 = 3$.

Answer: 3

1. Use the Divergence Theorem: $\Phi = \iiint_V (\nabla \cdot \mathbf{F}) dV$.
2. Substitute the known values: $-100\pi = \iiint_V -5 dV$.
3. Pull the constant out: $-100\pi = -5 \cdot \text{Volume}(V)$.
4. Solve for volume: $\text{Volume}(V) = \frac{-100\pi}{-5} = 20\pi$.

Answer: 20\pi

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 1 + 2 + 3 = 6$.
2. Apply the Divergence Theorem: $\Phi = \iiint_V 6 dV$.
3. The volume of a sphere is $\frac{4}{3}\pi r^3$. For $r=2$, $V = \frac{4}{3}\pi(8) = \frac{32\pi}{3}$.
4. The flux is $6 \cdot \frac{32\pi}{3} = 64\pi$.

Answer: 64\pi

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2$.
2. Set up the triple integral: $\int_0^1 \int_0^1 \int_0^1 (3x^2 + 3y^2 + 3z^2) dx dy dz$.
3. Integrate each term: $\int_0^1 3x^2 dx = 1$. By symmetry, the other two terms are also $1$.
4. The total flux is $1 + 1 + 1 = 3$.

Answer: 3

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 1 + 1 + e^z = 2 + e^z$.
2. Use cylindrical coordinates: $\int_0^{2\pi} \int_0^1 \int_0^1 (2 + e^z) r dz dr d\theta$.
3. Integrate with respect to $z$: $[2z + e^z]_0^1 = 2 + e - 1 = 1 + e$.
4. Integrate with respect to $r$ and $\theta$: $(1 + e) \cdot \pi(1)^2 = \pi(1 + e)$.

Answer: \pi(1 + e)

1. Step 1: Compute the curl of $\mathbf{F}$. $\nabla \times \mathbf{F} = \langle z-x, x-y, y-z \rangle$.
2. Step 2: Identify the surface $S$. The simplest surface is the flat triangle in the $xy$-plane where $z=0$.
3. Step 3: Determine the normal vector. For the $xy$-plane, $\mathbf{n} = \langle 0, 0, 1 \rangle$.
4. Step 4: Compute the dot product $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = y-z$. Since $z=0$ on the surface, this is just $y$.
5. Step 5: Integrate $y$ over the triangle region $D$ bounded by $x=0, y=0, x+y=2$. $\int_0^2 \int_0^{2-x} y dy dx = \int_0^2 \frac{1}{2}(2-x)^2 dx = [-\frac{1}{6}(2-x)^3]_0^2 = 0 - (-\frac{8}{6}) = \frac{4}{3}$.

Answer: 4/3

1. Step 1: A closed sphere has no boundary curve $C$.
2. Step 2: According to Stokes' Theorem, the surface integral of the curl equals the line integral around the boundary.
3. Step 3: Since the boundary is empty (or a single point), the line integral is zero.
4. Step 4: Therefore, the surface integral of the curl over any closed surface must be zero.

Answer: The boundary curve is empty, so the line integral is zero.

1. Step 1: Compute the curl. $\nabla \times \mathbf{F} = \langle 0, 0, 1 - (-1) \rangle = \langle 0, 0, 2 \rangle$.
2. Step 2: Choose the surface $S$ as the flat disk $x^2 + y^2 \le 9$ at $z=2$.
3. Step 3: The normal vector is $\mathbf{n} = \langle 0, 0, 1 \rangle$.
4. Step 4: The dot product is $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = 2$.
5. Step 5: Integrate 2 over the area of the disk with radius 3. $\iint_S 2 dS = 2 \times (\pi \cdot 3^2) = 18\pi$.

Answer: 18\pi

1. Compute the curl: $\nabla \times \mathbf{F} = [z-x, 0-z, z-y] = [z-x, -z, z-y]$.
2. The surface is the triangle in the $xy$-plane, so $z=0$ and $\mathbf{n} = [0,0,1]$.
3. The dot product is $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = z-y$. Since $z=0$ on the surface, the integrand is $-y$.
4. Integrate $-y$ over the triangle: $\int_0^1 \int_0^{1-x} -y dy dx = \int_0^1 [-\frac{1}{2}y^2]_0^{1-x} dx = \int_0^1 -\frac{1}{2}(1-x)^2 dx$.
5. Evaluating the integral: $[\frac{1}{6}(1-x)^3]_0^1 = 0 - \frac{1}{6} = -\frac{1}{6}$.

Answer: -\frac{1}{6}

1. Compute the curl: $\nabla \times \mathbf{F} = [2y-0, 2z-0, 2x-0] = [2y, 2z, 2x]$.
2. The surface is a flat disk at $z=3$ with $\mathbf{n} = [0,0,1]$.
3. The dot product is $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = 2x$.
4. Integrate $2x$ over the disk $x^2 + y^2 \le 4$.
5. By symmetry, the integral of $x$ over a disk centered at the origin is zero.

Answer: 0

1. Line integral: $C$ is the unit circle in $xy$-plane. $\mathbf{r}(t) = [\cos t, \sin t, 0]$, $\mathbf{r}'(t) = [-\sin t, \cos t, 0]$. $\mathbf{F} \cdot \mathbf{r}' = [-\sin t, \cos t, 0] \cdot [-\sin t, \cos t, 0] = 1$. $\int_0^{2\pi} 1 dt = 2\pi$.
2. Surface integral: $\nabla \times \mathbf{F} = [0, 0, 2]$.
3. Using the flat disk $S$ as the surface: $\mathbf{n} = [0,0,1]$. $\iint_S [0,0,2] \cdot [0,0,1] dS = 2 \times \text{Area}(\text{disk}) = 2\pi$.
4. Both sides equal $2\pi$.

Answer: 2\pi

1. Compute the divergence: $\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$.
2. Identify the volume of the cube: $V = 2 \times 2 \times 2 = 8$.
3. Apply the Divergence Theorem: $\iint_S \mathbf{F} \cdot \mathbf{n} dS = \iiint_V 3 dV = 3 \times 8 = 24$.

Answer: 24

1. By the unified view (Stokes' Theorem), the line integral around $C$ equals the surface integral of the curl of $\mathbf{F}$ over a surface $S$ bounded by $C$.
2. The curl of a gradient field is always zero: $\nabla \times (\nabla f) = 0$.
3. Integrating zero over the surface $S$ results in zero.

Answer: 0

1. The integral is $3 \times \text{Volume of sphere} = 3 \times (\frac{4}{3}\pi 2^3) = 3 \times \frac{32\pi}{3} = 32\pi$.
2. We need $\nabla \cdot \mathbf{F} = 3$.
3. A simple field satisfying this is $\mathbf{F} = \langle x, y, z \rangle$.

Answer: 32\pi, \mathbf{F} = \langle x, y, z \rangle

1. By the Divergence Theorem, the flux is $\iiint_V (\nabla \cdot \mathbf{E}) dV$.
2. Substitute Maxwell's first equation: $\iiint_V \frac{\rho}{\epsilon_0} dV$.
3. Compute the integral: $\frac{1}{\epsilon_0} \int_0^1 \int_0^1 \int_0^1 (3x^2 + 2y^2 + z^2) dx dy dz$.
4. The integral is $[x^3]_0^1 + [\frac{2}{3}y^3]_0^1 + [\frac{1}{3}z^3]_0^1 = 1 + \frac{2}{3} + \frac{1}{3} = 2$.
5. The total flux is $2 / \epsilon_0$.

Answer: $2 / \epsilon_0$

1. Compute the divergence: $\nabla \cdot \mathbf{B} = \frac{\partial}{\partial x}(y^2) + \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial z}(0)$.
2. The result is $0 + 0 + 0 = 0$.
3. Since the divergence is zero, it satisfies the equation $\nabla \cdot \mathbf{B} = 0$.
4. Physically, this means there are no magnetic monopoles in this field; every field line that enters a region must leave it.

Answer: The divergence is $0$, so it satisfies the equation; no monopoles are present.

1. By Stokes' Theorem, $\oint_C \mathbf{E} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{E}) \cdot \mathbf{n} dS$.
2. The normal vector to the $xy$-plane is $\mathbf{n} = (0, 0, 1)$.
3. The dot product is $(0, 0, -2t) \cdot (0, 0, 1) = -2t$.
4. The integral is $\iint_S -2t dS = -2t \times \text{Area of circle}$.
5. Area is $\pi R^2 = \pi (2^2) = 4\pi$.
6. The voltage is $-2t(4\pi) = -8\pi t$.

Answer: $-8\pi t$

1. Compute the divergence: $\nabla \cdot \mathbf{F} = 2x + 2y + 2z$.
2. Set up the triple integral over the cube: $\int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) dx dy dz$.
3. Integrate $2x$ over the cube: $\int_0^1 2x dx = 1$, and the other dimensions contribute $1 \times 1$. Total for $2x$ is $1$.
4. By symmetry, the integrals of $2y$ and $2z$ are also $1$.
5. Sum the results: $1 + 1 + 1 = 3$.

Answer: 3

1. Identify $P = y^2$ and $Q = 3xy$.
2. Compute the curl: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3y - 2y = y$.
3. Set up the double integral over the semi-disk: $\iint_D y dA$.
4. Use polar coordinates: $\int_0^\pi \int_0^1 (r \sin\theta) r dr d\theta$.
5. Integrate $r^2$ from $0$ to $1$ to get $1/3$.
6. Integrate $\sin\theta$ from $0$ to $\pi$ to get $2$.
7. Multiply: $1/3 \times 2 = 2/3$.

Answer: 2/3

1. Compute the curl: $\nabla \times \mathbf{F} = \langle 1-0, 1-0, 1-0 \rangle = \langle 1, 1, 1 \rangle$.
2. The surface $S$ is the plane $x + y + z = 1$. The unit normal vector is $\mathbf{n} = \frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle$.
3. Compute the dot product: $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = \langle 1, 1, 1 \rangle \cdot \frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle = \frac{3}{\sqrt{3}} = \sqrt{3}$.
4. The area of the triangle is $\frac{1}{2} |\mathbf{AB} \times \mathbf{AC}|$. $\mathbf{AB} = \langle -1, 1, 0 \rangle, \mathbf{AC} = \langle -1, 0, 1 \rangle$.
5. Cross product is $\langle 1, 1, 1 \rangle$, magnitude is $\sqrt{3}$. Area is $\sqrt{3}/2$.
6. Integral is $\sqrt{3} \times \frac{\sqrt{3}}{2} = 3/2$.

Answer: 3/2

1. Compute the time derivative of density: $\frac{\partial \rho}{\partial t} = -3 e^{-3t}(x^2 + y^2 + z^2)$.
2. Compute the divergence of $\mathbf{J}$: $\nabla \cdot \mathbf{J} = e^{-3t}(2 + 2 + 2) = 6 e^{-3t}$.
3. Sum them: $-3 e^{-3t}(x^2 + y^2 + z^2) + 6 e^{-3t}$.
4. Observe that this sum is not zero for all $x, y, z$.

Answer: Not satisfied.

1. Substitute $\rho = 1$ into the continuity equation: $\frac{\partial (1)}{\partial t} + \nabla \cdot (1 \mathbf{v}) = 0$.
2. The time derivative of a constant is $0$.
3. The equation simplifies to $\nabla \cdot \mathbf{v} = 0$.

Answer: The velocity field must be solenoidal (divergence-free), $\nabla \cdot \mathbf{v} = 0$.

1. Compute the divergence of $\mathbf{J}$: $\nabla \cdot \mathbf{J} = \frac{\partial}{\partial x}(t) + 0 + 0 = 0$.
2. The continuity equation becomes $\frac{\partial \rho}{\partial t} + 0 = 0$.
3. This means $\rho$ does not depend on $t$.
4. Therefore, $\rho$ can be any function of space only: $\rho = f(x, y, z)$.

Answer: \rho = f(x, y, z)

1. Compute the gradient: $\nabla u = [2x, -2y, 2z]$.
2. Compute the divergence of the gradient: $\nabla \cdot [2x, -2y, 2z] = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(-2y) + \frac{\partial}{\partial z}(2z)$.
3. Simplify the derivatives: $2 - 2 + 2 = 2$.
4. Since $2 \neq 0$, the function does not satisfy Laplace's equation.

Answer: No, $\nabla^2 u = 2$

1. Find the first partials: $u_x = -2x$ and $u_y = -4y$.
2. Find the second partials: $u_{xx} = -2$ and $u_{yy} = -4$.
3. Compute the Laplacian: $\nabla^2 u = -2 - 4 = -6$.
4. Apply the heat equation: $u_t = 0.5 \cdot (-6) = -3$.

Answer: -3

1. Compute the time derivative: $u_t = -e^{-t} \sin(x)$.
2. Compute the spatial second derivative: $u_x = e^{-t} \cos(x)$, so $u_{xx} = -e^{-t} \sin(x)$.
3. Set up the equation: $-e^{-t} \sin(x) = k (-e^{-t} \sin(x))$.
4. Divide both sides by $-e^{-t} \sin(x)$ to find $k = 1$.

Answer: Yes, $k = 1$

1. The 1-form is constructed by replacing the components of the vector field with coefficients of $dx$ and $dy$: $\omega = 3x^2 dx + 2y dy$.
2. The exterior derivative is $d\omega = (\frac{\partial(2y)}{\partial x} - \frac{\partial(3x^2)}{\partial y}) dx \wedge dy$.
3. Computing the partials: $\frac{\partial(2y)}{\partial x} = 0$ and $\frac{\partial(3x^2)}{\partial y} = 0$.
4. Thus, $d\omega = (0 - 0) dx \wedge dy = 0$.

Answer: d\omega = 0

1. The exterior derivative is $d\omega = (\frac{\partial(x^2)}{\partial x} - \frac{\partial(y^2)}{\partial y}) dx \wedge dy$.
2. Computing the partials: $\frac{\partial(x^2)}{\partial x} = 2x$ and $\frac{\partial(y^2)}{\partial y} = 2y$.
3. The result is $d\omega = (2x - 2y) dx \wedge dy$.
4. In vector calculus, this represents the 2D curl (or the k-component of the curl) of the vector field $\mathbf{F} = \langle y^2, x^2 \rangle$.

Answer: d\omega = (2x - 2y) dx \wedge dy

1. The exterior derivative of a 2-form is $d\omega = (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) dx \wedge dy \wedge dz$, where $P, Q, R$ are the coefficients of $dy \wedge dz, dz \wedge dx, dx \wedge dy$ respectively.
2. Here $P = xz$, $Q = yz$, and $R = xy$.
3. Compute partials: $\frac{\partial(xz)}{\partial x} = z$, $\frac{\partial(yz)}{\partial y} = z$, and $\frac{\partial(xy)}{\partial z} = 0$.
4. Summing them: $z + z + 0 = 2z$.
5. The result is $d\omega = 2z dx \wedge dy \wedge dz$.

Answer: d\omega = 2z dx \wedge dy \wedge dz

1. Step 1: Compute the divergence of $\mathbf{F}$. $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3$.
2. Step 2: Use the Divergence Theorem. The flux is the triple integral of the divergence over the volume of the cube.
3. Step 3: The volume of a cube with side length $2$ is $2^3 = 8$.
4. Step 4: The flux is $\iiint_V 3 dV = 3 \times 8 = 24$.

Answer: 24

1. Step 1: Use the area formula $\text{Area} = \oint_C \frac{1}{2}(x dy - y dx)$.
2. Step 2: Parameterize the ellipse: $x = a \cos t$, $y = b \sin t$ for $t \in [0, 2\pi]$.
3. Step 3: Compute differentials: $dx = -a \sin t dt$, $dy = b \cos t dt$.
4. Step 4: Substitute into the integral: $\int_0^{2\pi} \frac{1}{2}( (a \cos t)(b \cos t) - (b \sin t)(-a \sin t) ) dt$.
5. Step 5: Simplify: $\int_0^{2\pi} \frac{1}{2} ab (\cos^2 t + \sin^2 t) dt = \int_0^{2\pi} \frac{1}{2} ab dt$.
6. Step 6: Evaluate: $\frac{1}{2} ab (2\pi) = \pi ab$.

Answer: \pi ab

1. Step 1: Check the curl. $\nabla \times \mathbf{F} = \langle \frac{\partial}{\partial y}(2yz) - \frac{\partial}{\partial z}(x^2 + z^2), \frac{\partial}{\partial z}(2xy) - \frac{\partial}{\partial x}(2yz), \frac{\partial}{\partial x}(x^2 + z^2) - \frac{\partial}{\partial y}(2xy) \rangle$.
2. Step 2: Compute components: $\langle 2z - 2z, 0 - 0, 2x - 2x \rangle = \langle 0, 0, 0 \rangle$. The field is conservative.
3. Step 3: Integrate $F_x$ with respect to $x$: $\phi = \int 2xy dx = x^2 y + g(y, z)$.
4. Step 4: Differentiate $\phi$ with respect to $y$: $\frac{\partial \phi}{\partial y} = x^2 + \frac{\partial g}{\partial y}$. Match with $F_y = x^2 + z^2$, so $\frac{\partial g}{\partial y} = z^2$.
5. Step 5: Integrate $\frac{\partial g}{\partial y}$ with respect to $y$: $g(y, z) = y z^2 + h(z)$. Thus $\phi = x^2 y + y z^2 + h(z)$.
6. Step 6: Differentiate $\phi$ with respect to $z$: $\frac{\partial \phi}{\partial z} = 2yz + h'(z)$. Match with $F_z = 2yz$, so $h'(z) = 0$.
7. Step 7: Final potential: $\phi(x, y, z) = x^2 y + y z^2 + C$.

Answer: \phi(x, y, z) = x^2 y + y z^2 + C