Multivariable Calculus Quiz: Common Pitfalls

17 questions · exam conditions

0:00

Common PitfallsQuestion 1 of 17

Let $S$ be the closed surface bounding the solid region $E$ defined by $z \ge 0$ and $z \le 4 - x^2 - y^2$ . $S$ is oriented with the outward-pointing normal. A student calculates the flux of the vector field $\vec{F} = \langle 0, 0, z \rangle$ across $S$ and gets an answer of $16\pi$ . What is the most likely error the student made?

The student made a sign error by using an inward normal vector instead of an outward normal vector for the surface.

The student calculated the volume of a bounding cylinder instead of the volume of the solid region

E

The student forgot to include the Jacobian term

r

when setting up the integral in cylindrical coordinates.

The student incorrectly applied the Divergence Theorem, as the surface

S

is not closed.

← Back to quizzes

Multivariable Calculus Quiz

Multivariable Calculus Quiz: Common Pitfalls

Practice Common Pitfalls in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Common Pitfalls, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The student made a sign error by using an inward normal vector instead of an outward normal vector for the surface.
The student calculated the volume of a bounding cylinder instead of the volume of the solid region $E$ . (correct answer)
The student forgot to include the Jacobian term $r$ when setting up the integral in cylindrical coordinates.
The student incorrectly applied the Divergence Theorem, as the surface $S$ is not closed.

Explanation: The flux of

\vec{F}

across the closed surface

S

can be found using the Divergence Theorem:

\iint_S \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} dV

. Here,

\nabla \cdot \vec{F} = \frac{\partial}{\partial z}(z) = 1

. So the flux is equal to the volume of the solid region

E

. The region

E

is a paraboloid. Its volume is given by

\int_0^{2\pi} \int_0^2 (4-r^2) r \, dr \, d\theta = 2\pi [2r^2 - \frac{r^4}{4}]_0^2 = 2\pi(8-4) = 8\pi

. The student's answer was

16\pi

. This value corresponds to the volume of the cylinder with radius 2 and height 4 that encloses the paraboloid:

V_{cyl} = \pi r^2 h = \pi (2^2)(4) = 16\pi

. This indicates a common bounds error where the student integrated over a simpler bounding box or cylinder rather than the correct region.

Question 2

A region $R$ in the $xy$ -plane is in the first quadrant, bounded by the circle $x^2 + y^2 = 4$ and the line $x=1$ . Which of the following integrals correctly represents the area of $R$ when converted to polar coordinates?

$\int_0^{\pi/3} \int_{1}^{2} r \, dr \, d\theta$
$\int_0^{\pi/2} \int_{\sec(\theta)}^{2} r \, dr \, d\theta$
$\int_0^{\pi/3} \int_{\sec(\theta)}^{2} r \, dr \, d\theta$ (correct answer)
$\int_0^{\pi/3} \int_{0}^{2} r \, dr \, d\theta$

Explanation: The region is bounded by a circle and a vertical line. In polar coordinates, the circle

x^2 + y^2 = 4

r=2

. The line

x=1

r \cos(\theta) = 1

, or

r = \sec(\theta)

. The region is in the first quadrant, so

\theta

starts at 0. To find the upper limit for

\theta

, we find the intersection of

x=1

and

x^2+y^2=4

. This occurs at

(1, \sqrt{3})

. In polar coordinates, this point has

\tan(\theta) = \sqrt{3}/1

, so

\theta = \pi/3

. For a given angle

\theta

between 0 and

\pi/3

, the radius

r

extends from the line

r=\sec(\theta)

to the circle

r=2

. The area element in polar coordinates is

r \, dr \, d\theta

. Therefore, the correct integral is

\int_0^{\pi/3} \int_{\sec(\theta)}^{2} r \, dr \, d\theta

. Distractor A incorrectly treats the bound

x=1

r=1

. Distractor B uses an incorrect upper limit for

\theta

. Distractor D ignores the boundary

x=1

completely.

Question 3

Let $C$ be the triangle with vertices $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ , oriented counter-clockwise when viewed from above. Calculate the line integral $\oint_C \vec{F} \cdot d\vec{r}$ for the vector field $\vec{F} = \langle z, x, y \rangle$ .

$3/2$ (correct answer)
$-3/2$
$3$
$0$

Explanation: By Stokes' Theorem,

\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}

. First, we compute the curl of

\vec{F}

\nabla \times \vec{F} = \langle \frac{\partial}{\partial y}(y) - \frac{\partial}{\partial z}(x), \frac{\partial}{\partial z}(z) - \frac{\partial}{\partial x}(y), \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(z) \rangle = \langle 1, 1, 1 \rangle

. The surface

S

is the part of the plane

x+y+z=1

bounded by

C

. The counter-clockwise orientation of

C

(viewed from above) induces an upward-pointing normal vector on

S

. The surface can be parameterized by

\vec{r}(x,y) = \langle x, y, 1-x-y \rangle

, and the normal vector is

d\vec{S} = \langle 1, 1, 1 \rangle dA

. The integral becomes

\iint_D \langle 1, 1, 1 \rangle \cdot \langle 1, 1, 1 \rangle dA = \iint_D 3 dA

, where

D

is the projection of

S

onto the

xy

-plane, a triangle with vertices

(0,0), (1,0), (0,1)

. The area of

D

1/2

. The value of the integral is

3 \times (\text{Area of } D) = 3(1/2) = 3/2

. An answer of

-3/2

results from an orientation error (using a downward normal). An answer of

3

results from an error in calculating the area of the projected region.

Question 4

Let $R$ be the parallelogram with vertices $(0,0)$ , $(3,1)$ , $(1,2)$ , and $(4,3)$ . A student uses the transformation $x=3u+v$ , $y=u+2v$ to evaluate an integral over $R$ . What is the absolute value of the Jacobian determinant, $|\frac{\partial(x,y)}{\partial(u,v)}|$ , for this transformation?

$5$ (correct answer)
$1/5$
$7$
$1/7$

Explanation: The Jacobian determinant for the transformation from

(u,v)

(x,y)

is given by the determinant of the matrix of partial derivatives:

J = \frac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}

. From the given transformation, we have:

\frac{\partial x}{\partial u} = 3

\frac{\partial x}{\partial v} = 1

\frac{\partial y}{\partial u} = 1

\frac{\partial y}{\partial v} = 2

. Plugging these into the determinant formula:

J = \det \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5

. The absolute value is

|5|=5

. A common mistake is to calculate the inverse Jacobian,

\frac{\partial(u,v)}{\partial(x,y)}

, which would yield

1/5

. Another error is to miscalculate the determinant, for example as

3+2=5

3*1+1*2=5

which happen to work, but the process is wrong. For instance

3*2+1*1=7

could be a distractor.

Question 5

Let $\vec{F} = \langle y, -x, z^2 \rangle$ and let $S$ be the part of the cone $z=\sqrt{x^2+y^2}$ below the plane $z=1$ , oriented with an upward-pointing normal vector. A student uses Stokes' Theorem to evaluate $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$ . Which line integral must the student calculate?

The integral of $\vec{F}$ over the circle $x^2+y^2=1$ in the $xy$ -plane, oriented counter-clockwise.
The integral of $\vec{F}$ over the circle $x^2+y^2=1$ in the plane $z=1$ , oriented clockwise when viewed from above.
The integral of $\nabla \times \vec{F}$ over the circle $x^2+y^2=1$ in the plane $z=1$ , oriented counter-clockwise when viewed from above.
The integral of $\vec{F}$ over the circle $x^2+y^2=1$ in the plane $z=1$ , oriented counter-clockwise when viewed from above. (correct answer)

Explanation: When you encounter a surface integral of a curl and need to apply Stokes' Theorem, remember that this theorem converts surface integrals into line integrals around the boundary curve. Stokes' Theorem states:

\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{\partial S} \vec{F} \cdot d\vec{r}

The surface

S

is the cone

z = \sqrt{x^2 + y^2}

below

z = 1

, which means it's bounded by the circle where the cone intersects the plane

z = 1

. Since

z = \sqrt{x^2 + y^2} = 1

, this boundary is the circle

x^2 + y^2 = 1

in the plane

z = 1

. For orientation, since

S

has an upward-pointing normal, when you curl the fingers of your right hand in the direction of the boundary curve, your thumb should point upward. This means the boundary must be oriented counter-clockwise when viewed from above. Therefore, you need the line integral of

\vec{F}

over the circle

x^2 + y^2 = 1

in the plane

z = 1

, oriented counter-clockwise from above. This is answer D. A is wrong because the boundary isn't in the

xy

-plane—it's where the cone meets

z = 1

. B has the correct curve but wrong orientation (clockwise instead of counter-clockwise). C incorrectly suggests integrating

\nabla \times \vec{F}

instead of

\vec{F}

itself. Study tip: Always identify the boundary curve first, then use the right-hand rule to match the surface orientation with the correct boundary orientation.

Question 6

Evaluate the integral $\int_0^4 \int_{x/2}^2 e^{y^2} \, dy \, dx$ . A student who tries to evaluate this integral directly will fail because $e^{y^2}$ has no elementary antiderivative. What is the correct value of the integral after reversing the order of integration?

$e^4$
$2(e^4 - 1)$
$e^2 - 1$
$e^4 - 1$ (correct answer)

Explanation: When you encounter a double integral where the integrand has no elementary antiderivative, reversing the order of integration often transforms an impossible problem into a manageable one. This is a key strategy in multivariable calculus. To reverse the integration order, you must first understand the region of integration. The original bounds describe the region where

0 \leq x \leq 4

and

\frac{x}{2} \leq y \leq 2

. Sketching this region, you'll see it's bounded by

x = 0

x = 4

y = 2

, and

y = \frac{x}{2}

(or equivalently,

x = 2y

). For the reversed order, you integrate with respect to

x

first, then

y

. The

y

-values range from

0

2

, and for each fixed

y

x

ranges from

0

2y

. This gives us:

\int_0^2 \int_0^{2y} e^{y^2} \, dx \, dy

Now you can evaluate this step by step. The inner integral is simply:

\int_0^{2y} e^{y^2} \, dx = e^{y^2} \cdot 2y = 2ye^{y^2}

The outer integral becomes:

\int_0^2 2ye^{y^2} \, dy

Using substitution with

u = y^2

du = 2y \, dy

\int_0^4 e^u \, du = e^4 - e^0 = e^4 - 1

This confirms answer D is correct. Answer A (

e^4

) forgets to subtract 1. Answer B (

2(e^4-1)

) incorrectly doubles the result. Answer C (

e^2-1

) uses wrong integration bounds. Always sketch the region when reversing integration order—visualization prevents bound errors and ensures you set up the integral correctly.

Question 7

To evaluate $\iint_R y^2 dA$ where $R$ is the region in the first quadrant bounded by $xy=1$ , $xy=4$ , $y=x$ , and $y=9x$ , a student uses the transformation $u=xy$ and $v=y/x$ . Which of the following represents the correctly transformed integral?

$\int_1^9 \int_1^4 uv (2v) \, du \, dv$
$\int_1^9 \int_1^4 \frac{u}{2v} \, du \, dv$
$\int_1^9 \int_1^4 \frac{u}{2} \, du \, dv$ (correct answer)
$\int_1^3 \int_1^2 y^2 \, du \, dv$

Explanation: First, transform the bounds:

xy=1 \implies u=1

xy=4 \implies u=4

y=x \implies y/x=1 \implies v=1

y=9x \implies y/x=9 \implies v=9

. The region in the

uv

-plane is a rectangle

[1,4] \times [1,9]

. Next, transform the integrand

y^2

. From

u=xy

and

v=y/x

, we multiply them:

uv = (xy)(y/x) = y^2

. Finally, calculate the Jacobian of the transformation from

(u,v)

(x,y)

. It is easier to first find the Jacobian of

(u,v)

with respect to

(x,y)

J' = \frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} y & x \\ -y/x^2 & 1/x \end{pmatrix} = y(1/x) - x(-y/x^2) = y/x + y/x = 2y/x = 2v

. The required Jacobian is

J = \frac{\partial(x,y)}{\partial(u,v)} = (J')^{-1} = \frac{1}{2v}

. The integral is

\iint_R y^2 dA = \int_1^9 \int_1^4 (uv) |\frac{1}{2v}| \, du \, dv = \int_1^9 \int_1^4 \frac{u}{2} \, du \, dv

Question 8

Consider the integral $I = \int_0^1 \int_{\arcsin(y)}^{\pi/2} f(x,y) \, dx \, dy$ . After reversing the order of integration, the integral becomes $\int_a^b \int_{g(x)}^{h(x)} f(x,y) \, dy \, dx$ . What is the function $h(x)$ ?

$\sin(x)$ (correct answer)
$\arcsin(x)$
$1$
$\cos(x)$

Explanation: The original bounds are

0 \le y \le 1

and

\arcsin(y) \le x \le \pi/2

. The equation

x = \arcsin(y)

is equivalent to

y = \sin(x)

for

x

[-\pi/2, \pi/2]

. The region of integration is bounded by the curve

y=\sin(x)

, the vertical line

x=\pi/2

, and the horizontal line

y=0

(since

x=\arcsin(0)=0

). To reverse the order of integration, we fix

x

and let

y

vary. The

x

values range from

0

\pi/2

. For a fixed

x

in this interval,

y

goes from the lower boundary

y=0

to the upper boundary

y=\sin(x)

. Therefore, the new bounds are

0 \le x \le \pi/2

and

0 \le y \le \sin(x)

. The function

h(x)

is the upper bound for

y

, which is

\sin(x)

Question 9

For the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ where $\vec{F} = \langle x, y, z \rangle$ and $S$ is the hemisphere $x^2 + y^2 + z^2 = 1$ with $z \geq 0$ , a student parameterizes using $\vec{r}(\phi, \theta) = \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$ with $0 \leq \phi \leq \pi/2$ and $0 \leq \theta \leq 2\pi$ . They compute $\vec{r}_\phi \times \vec{r}_\theta = \sin\phi \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$ . What is problematic about this approach?

The parameterization bounds are wrong; $\phi$ should range from $0$ to $\pi$ for a hemisphere
The cross product has the wrong orientation; it points inward instead of outward
The cross product calculation is missing the factor $\sin^2\phi$ in the first two components (correct answer)
The parameterization uses spherical coordinates incorrectly; $z$ should be $\sin\phi$

Explanation: Computing

\vec{r}_\phi = \langle \cos\phi\cos\theta, \cos\phi\sin\theta, -\sin\phi \rangle

and

\vec{r}_\theta = \langle -\sin\phi\sin\theta, \sin\phi\cos\theta, 0 \rangle

, the cross product

\vec{r}_\phi \times \vec{r}_\theta

should be

\langle \sin^2\phi\cos\theta, \sin^2\phi\sin\theta, \sin\phi\cos\phi \rangle

. The student wrote

\sin\phi

instead of

\sin^2\phi

in the first two components. Choice A is wrong;

\phi \in [0, \pi/2]

is correct for the upper hemisphere. Choice B is incorrect about orientation. Choice D is wrong about the spherical coordinate conversion.

Question 10

A student attempts to use the divergence theorem to evaluate $\iiint_E \nabla \cdot \vec{F} \, dV$ where $\vec{F} = \langle x^2, y^2, z^2 \rangle$ and $E$ is the region between two concentric spheres: $1 \leq x^2 + y^2 + z^2 \leq 4$ . They write this as $\iint_{S_2} \vec{F} \cdot d\vec{S} - \iint_{S_1} \vec{F} \cdot d\vec{S}$ where $S_2$ has radius 2 and $S_1$ has radius 1, both with outward normal. What is the error in this setup?

The normal for $S_1$ should point inward (toward the origin) relative to region $E$ (correct answer)
Both surface integrals should use inward normals for consistency
The divergence theorem doesn't apply since $\vec{F}$ is not differentiable everywhere
The order should be reversed: $\iint_{S_1} \vec{F} \cdot d\vec{S} - \iint_{S_2} \vec{F} \cdot d\vec{S}$

Explanation: When applying the divergence theorem to a region between two surfaces, you must carefully consider which direction is "outward" from your region of integration. The divergence theorem states that

\iiint_E \nabla \cdot \vec{F} \, dV = \iint_{\partial E} \vec{F} \cdot d\vec{S}

, where the surface integral uses the outward normal relative to region

E

. For the region

E

between concentric spheres, the boundary consists of both

S_1

and

S_2

. The outward normal from

E

points away from the region itself. On the outer sphere

S_2

, "outward from

E

" means radially outward from the origin. However, on the inner sphere

S_1

, "outward from

E

" means radially inward toward the origin, since you're exiting the region

E

by moving toward the center. The student used outward normals for both spheres relative to their individual surfaces, not relative to the region

E

. This is why (A) is correct — the normal for

S_1

should point inward toward the origin to be outward relative to region

E

. (B) is wrong because consistency isn't the issue; both normals should be outward from

E

, which means different directions for each sphere. (C) is wrong because

\vec{F} = \langle x^2, y^2, z^2 \rangle

is differentiable everywhere. (D) is wrong because changing the order doesn't fix the fundamental normal direction error. Study tip: Always identify what region you're integrating over first, then determine which direction is truly "outward" from that specific region, not from individual surfaces.

Question 11

When evaluating the triple integral $\iiint_E f(x,y,z) \, dV$ where $E$ is the region bounded by $x^2 + y^2 = 4$ , $z = 0$ , and $z = x + y + 2$ , a student sets up the integral in cylindrical coordinates as $\int_0^{2\pi} \int_0^2 \int_0^{r\cos\theta + r\sin\theta + 2} f(r\cos\theta, r\sin\theta, z) \, dz \, dr \, d\theta$ . What is the primary error in this setup?

The bounds for $z$ should be $0$ to $r\cos\theta + r\sin\theta - 2$ instead
The Jacobian $r$ is missing from the integrand (correct answer)
The bounds for $\theta$ should be $0$ to $\pi$ since $x + y \geq 0$ is required
The bounds for $r$ should be $0$ to $\sqrt{4 - z^2}$ instead of constant

Explanation: The correct integral should include the Jacobian determinant

r

when converting to cylindrical coordinates. The integrand should be

f(r\cos\theta, r\sin\theta, z) \cdot r

. All the bounds are correctly set up:

\theta

goes from

0

2\pi

r

from

0

2

, and

z

from

0

r\cos\theta + r\sin\theta + 2

. Choice A incorrectly changes the upper bound for

z

. Choice C incorrectly restricts

\theta

when no such restriction is needed. Choice D confuses the order of integration.

Question 12

A student is computing $\oint_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = \langle y^2, -x^2 \rangle$ and $C$ is the boundary of the square with vertices at $(0,0)$ , $(1,0)$ , $(1,1)$ , and $(0,1)$ , traversed counterclockwise. Using Green's theorem, they calculate $\iint_D \left(\frac{\partial(-x^2)}{\partial x} - \frac{\partial(y^2)}{\partial y}\right) dA = \iint_D (-2x - 2y) \, dA$ . What error did they make?

They used the wrong orientation; the integral should be $\iint_D (2x + 2y) \, dA$
They incorrectly computed $\frac{\partial(-x^2)}{\partial x} = -2x$ instead of $-x$
They applied Green's theorem incorrectly; it should be $\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}$
They mixed up the components; $P = -x^2$ and $Q = y^2$ for this vector field (correct answer)

Explanation: For

\vec{F} = \langle y^2, -x^2 \rangle

, we have

P = y^2

and

Q = -x^2

. Green's theorem states

\oint_C \vec{F} \cdot d\vec{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA

. So we need

\frac{\partial(-x^2)}{\partial x} - \frac{\partial(y^2)}{\partial y} = -2x - 2y

. The student incorrectly identified

P

and

Q

. Choice A is wrong about orientation. Choice B incorrectly states the derivative. Choice C gives the wrong form of Green's theorem.

Question 13

For the line integral $\int_C (x^2 + y^2) \, ds$ where $C$ is the curve parameterized by $\vec{r}(t) = \langle t^2, t^3 \rangle$ for $-1 \leq t \leq 1$ , a student computes $\int_{-1}^1 (t^4 + t^6) \sqrt{(2t)^2 + (3t^2)^2} \, dt = \int_{-1}^1 (t^4 + t^6) \sqrt{4t^2 + 9t^4} \, dt$ . They then factor to get $\int_{-1}^1 (t^4 + t^6) |t| \sqrt{4 + 9t^2} \, dt$ and remove the absolute value to get $\int_{-1}^1 (t^4 + t^6) t \sqrt{4 + 9t^2} \, dt$ . What is the issue with this approach?

Removing the absolute value changes the sign of the integrand for $t < 0$ , which affects the result (correct answer)
The factoring is incorrect; $\sqrt{4t^2 + 9t^4} = |t|\sqrt{4 + 9t^2}$ only for $t \neq 0$
The parameterization derivatives are wrong; $\frac{dx}{dt} = 2t^2$ and $\frac{dy}{dt} = 3t^3$
The integral bounds should be $0$ to $1$ since negative $t$ values are not physical

Explanation: When evaluating line integrals with respect to arc length, you must be careful with absolute values that arise from the differential element

ds = \|\vec{r}'(t)\| \, dt

. The student correctly computed

\|\vec{r}'(t)\| = \sqrt{4t^2 + 9t^4}

and factored this as

|t|\sqrt{4 + 9t^2}

. However, when they removed the absolute value to get just

t

, they made a critical error. For

t < 0

, we have

|t| = -t

, not

t

. This means that for negative values of

t

, removing the absolute value changes the sign of the integrand, which will affect the final numerical result of the integral. Let's examine why the other options are incorrect. Option B is wrong because the factoring

\sqrt{4t^2 + 9t^4} = |t|\sqrt{4 + 9t^2}

is mathematically correct for all

t

in the domain, including

t = 0

(where both sides equal 0). Option C is incorrect since the derivatives are computed correctly:

\frac{dx}{dt} = 2t

and

\frac{dy}{dt} = 3t^2

. Option D is wrong because the parameterization domain

[-1,1]

is given in the problem, and negative parameter values are perfectly valid in mathematics. The correct answer is A because removing the absolute value fundamentally changes the integrand's sign for

t < 0

. Study tip: Always preserve absolute values in arc length calculations until after integration, or split the integral at points where the expression inside the absolute value changes sign.

Question 14

When parameterizing the surface $z = x^2 + y^2$ over the region $x^2 + y^2 \leq 1$ for a surface integral, a student uses $\vec{r}(u,v) = \langle u\cos v, u\sin v, u^2 \rangle$ with $0 \leq u \leq 1$ and $0 \leq v \leq 2\pi$ . They compute $\vec{r}_u \times \vec{r}_v = \langle -2u^2\cos v, -2u^2\sin v, u \rangle$ . If the surface integral evaluates to a negative value, what is the most likely cause?

The parameterization bounds are incorrect; $u$ should range from $0$ to $2\pi$
The cross product calculation is wrong; the $z$ -component should be $-u$
The orientation is upward instead of downward due to the parameterization choice (correct answer)
The parameterization should use $z = u^2\cos^2 v + u^2\sin^2 v$ instead

Explanation: The cross product

\vec{r}_u \times \vec{r}_v = \langle -2u^2\cos v, -2u^2\sin v, u \rangle

has a positive

z

-component (

u > 0

), indicating upward orientation. If this gives a negative surface integral, the issue is likely that the problem expects downward orientation, but this parameterization gives upward orientation. Choice A confuses the bounds. Choice B is incorrect about the cross product. Choice D unnecessarily complicates the parameterization which is already correct.

Question 15

When evaluating $\int_0^1 \int_0^x \int_0^{x+y} f(x,y,z) \, dz \, dy \, dx$ by changing the order of integration to $dz \, dx \, dy$ , a student determines the region as $\{(x,y,z) : 0 \leq x \leq 1, 0 \leq y \leq x, 0 \leq z \leq x+y\}$ and rewrites it as $\int_0^1 \int_y^1 \int_0^{x+y} f(x,y,z) \, dz \, dx \, dy$ . What error did they make?

The bounds for $y$ should be $0$ to $\min(1,z)$ when integrating with respect to $y$ last
The bounds for $x$ should be $\max(y,z-y)$ to $1$ to account for the constraint $z \leq x+y$ (correct answer)
They correctly identified the region but used the wrong integration order $dx \, dy$ instead of $dy \, dx$
The upper bound for $z$ should be $1+y$ instead of $x+y$ in the new integral

Explanation: To change the order to

dz \, dx \, dy

, we need to express the bounds in terms of the constraints. The region is

0 \leq y \leq x \leq 1

and

0 \leq z \leq x+y

. For fixed

y

and

z

, we need

x \geq y

and

x \geq z-y

(from

z \leq x+y

), so

x \geq \max(y, z-y)

. Also

x \leq 1

. The student forgot the constraint from

z \leq x+y

. Choice A confuses the variable relationships. Choice C incorrectly states they got the region right. Choice D incorrectly changes the

z

bound.

Question 16

The integral $\int_{-2}^{2} \int_{0}^{\sqrt{4-x^2}} (x^2+y^2) \, dy \, dx$ is converted to polar coordinates. Which of the following represents the correct setup?

$\int_{0}^{2\pi} \int_{0}^{2} r^3 \, dr \, d\theta$
$\int_{0}^{\pi} \int_{0}^{2} r^3 \, dr \, d\theta$ (correct answer)
$\int_{0}^{\pi} \int_{0}^{2} r^2 \, dr \, d\theta$
$\int_{-2}^{2} \int_{0}^{2} r^3 \, dr \, d\theta$

Explanation: The bounds of integration in Cartesian coordinates are

-2 \le x \le 2

and

0 \le y \le \sqrt{4-x^2}

. The equation

y=\sqrt{4-x^2}

represents the top half of a circle centered at the origin with radius 2. Since

y \ge 0

, the region is the upper semi-disk of radius 2. In polar coordinates, this corresponds to

0 \le r \le 2

and

0 \le \theta \le \pi

. The integrand

x^2+y^2

becomes

r^2

. The area element

dy \, dx

becomes

r \, dr \, d\theta

. Therefore, the integrand in polar coordinates is

(r^2)(r) = r^3

. Combining these gives the integral

\int_{0}^{\pi} \int_{0}^{2} r^3 \, dr \, d\theta

. Distractor A uses an incorrect range for

\theta

(a full disk). Distractor C misses a factor of

r

from the Jacobian. Distractor D incorrectly translates the

x

bounds directly to

\theta

bounds.

Question 17

Let $C$ be the boundary of the rectangle with vertices $(0,0), (2,0), (2,1), (0,1)$ , traversed clockwise. Compute the line integral $\oint_C (y^2 + e^x) \, dx + (3xy + \cos(y)) \, dy$ .

$-1$ (correct answer)
$1$
$0$
$-2$

Explanation: We use Green's Theorem. Let

P(x,y) = y^2 + e^x

and

Q(x,y) = 3xy + \cos(y)

. Then

\frac{\partial Q}{\partial x} = 3y

and

\frac{\partial P}{\partial y} = 2y

. Green's Theorem states

\oint_C P \, dx + Q \, dy = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA

for a positively (counter-clockwise) oriented curve

C

. Here, the curve is traversed clockwise, which is a negative orientation. Thus, we must negate the result of the double integral.

\oint_C (y^2 + e^x) \, dx + (3xy + \cos(y)) \, dy = -\iint_R (3y - 2y) \, dA = -\iint_R y \, dA

. The region

R

is the rectangle

[0,2] \times [0,1]

. The integral is

- \int_0^2 \int_0^1 y \, dy \, dx = - \int_0^2 [\frac{y^2}{2}]_0^1 \, dx = - \int_0^2 \frac{1}{2} \, dx = - (\frac{1}{2})(2) = -1

. Choosing

1

is a classic orientation error.