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