Multivariable Calculus Quiz: Surface Integrals Scalar Fields

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Surface Integrals Scalar FieldsQuestion 1 of 10

The surface $S$ is the part of the paraboloid $z = x^2 + y^2$ that lies inside the cylinder $x^2 + y^2 = 4$ . For the scalar field $p(x,y,z) = \sqrt{x^2 + y^2}$ , the surface integral $\iint_S p \, dS$ can be expressed as:

\int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta

\int_0^{2\pi} \int_0^2 r\sqrt{1 + 4r^2} \, dr \, d\theta

\int_0^{2\pi} \int_0^2 r\sqrt{4r^2 + 1} \, dr \, d\theta

\int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2} \, dr \, d\theta

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Surface Integrals Scalar Fields

Practice Surface Integrals Scalar Fields 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 Surface Integrals Scalar Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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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.

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Question 1

$\int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta$ (correct answer)
$\int_0^{2\pi} \int_0^2 r\sqrt{1 + 4r^2} \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 r\sqrt{4r^2 + 1} \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2} \, dr \, d\theta$

Explanation: Using cylindrical coordinates

x = r\cos\theta

y = r\sin\theta

z = r^2

. For the surface

z = f(x,y) = x^2 + y^2

, we have

f_x = 2x

and

f_y = 2y

, so

dS = \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy = \sqrt{1 + 4r^2} \, r \, dr \, d\theta

. The function

p(x,y,z) = \sqrt{x^2 + y^2} = r

. The surface integral becomes

\int_0^{2\pi} \int_0^2 r \cdot \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta

. Choice B omits one factor of

r

. Choice C has the correct terms but omits the Jacobian factor. Choice D omits the function

p

entirely.

Question 2

A surface $S$ is given by the equation $x^2 + y^2 - z^2 = 1$ with $z \geq 0$ and $z \leq 2$ . Using the parameterization $x = \sec u \cos v$ , $y = \sec u \sin v$ , $z = \tan u$ where $0 \leq u \leq \arctan 2$ and $0 \leq v \leq 2\pi$ , the surface element $dS$ equals:

$\sec u \, du \, dv$
$\sec^2 u \, du \, dv$
$\sec^2 u \tan u \, du \, dv$
$\sec^3 u \, du \, dv$ (correct answer)

Explanation: When calculating surface area for a parameterized surface, you need to find the surface element

dS = |\mathbf{r}_u \times \mathbf{r}_v| \, du \, dv

, where

\mathbf{r}(u,v) = \langle \sec u \cos v, \sec u \sin v, \tan u \rangle

. First, compute the partial derivatives:

\mathbf{r}_u = \langle \sec u \tan u \cos v, \sec u \tan u \sin v, \sec^2 u \rangle

\mathbf{r}_v = \langle -\sec u \sin v, \sec u \cos v, 0 \rangle

The cross product

\mathbf{r}_u \times \mathbf{r}_v

gives:

\mathbf{r}_u \times \mathbf{r}_v = \langle -\sec^3 u \cos v, -\sec^3 u \sin v, \sec^2 u \tan u \rangle

To find the magnitude:

|\mathbf{r}_u \times \mathbf{r}_v|^2 = \sec^6 u \cos^2 v + \sec^6 u \sin^2 v + \sec^4 u \tan^2 u

= \sec^6 u(\cos^2 v + \sin^2 v) + \sec^4 u \tan^2 u = \sec^6 u + \sec^4 u \tan^2 u

= \sec^4 u(\sec^2 u + \tan^2 u) = \sec^4 u(2\sec^2 u - 1) = \sec^6 u

Therefore,

|\mathbf{r}_u \times \mathbf{r}_v| = \sec^3 u

, making

dS = \sec^3 u \, du \, dv

. Answer choice A (

\sec u

) misses the squared terms from differentiation. Choice B (

\sec^2 u

) would result from forgetting the cross product magnitude calculation. Choice C (

\sec^2 u \tan u

) incorrectly includes only one component of the cross product. Study tip: For surface integrals, always compute both partial derivatives carefully, then find the cross product magnitude. The trigonometric identity

\sec^2 u = 1 + \tan^2 u

is often crucial in simplifying these calculations.

Question 3

Consider the surface $S$ that is the graph of $z = \sin(xy)$ over the square $[0,\pi] \times [0,1]$ in the $xy$ -plane. For the function $q(x,y,z) = \cos(xy)$ , which statement about the surface integral $\iint_S q \, dS$ is correct?

The integral equals $\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dy \, dx$ (correct answer)
The integral can be simplified to $\int_0^\pi \int_0^1 \cos(xy) \, dy \, dx$ since the surface is nearly flat
The integral equals $\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + \sin^2(xy)} \, dy \, dx$
The integral equals $\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + x^2\sin^2(xy) + y^2\sin^2(xy)} \, dy \, dx$

Explanation: For the surface

z = f(x,y) = \sin(xy)

, we have

f_x = y\cos(xy)

and

f_y = x\cos(xy)

. The surface element is

dS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy = \sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dx \, dy

. On the surface,

q(x,y,z) = \cos(xy)

. The integral is

\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dy \, dx

. Choice B incorrectly ignores the surface element. Choice C uses wrong partial derivatives (sine instead of cosine). Choice D uses sine instead of cosine in the surface element.

Question 4

A surface $S$ is parameterized by $\mathbf{r}(u,v) = (u, v, u^2 - v^2)$ where $u^2 + v^2 \leq 1$ . If $\iint_S (x^2 + y^2) \, dS = \alpha \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv$ where $D$ is the unit disk, then $\alpha$ equals:

$\frac{1}{2}$
$4$
$2$
$1$ (correct answer)

Explanation: When you encounter a surface integral problem with a parameterization, you need to convert the surface integral into a double integral over the parameter domain using the formula

\iint_S f(x,y,z) \, dS = \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv

. First, let's find the cross product magnitude. With

\mathbf{r}(u,v) = (u, v, u^2 - v^2)

, we have

\mathbf{r}_u = (1, 0, 2u)

and

\mathbf{r}_v = (0, 1, -2v)

. Computing the cross product:

\mathbf{r}_u \times \mathbf{r}_v = (-2u, 2v, 1)

, so

\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{4u^2 + 4v^2 + 1}

. Next, substitute the parameterization into the integrand. Since

x = u

and

y = v

, we have

x^2 + y^2 = u^2 + v^2

. Therefore:

\iint_S (x^2 + y^2) \, dS = \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv

Comparing this with the given form

\alpha \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv

, we see that

\alpha = 1

, which is choice (D). Choice (A)

\frac{1}{2}

might come from incorrectly halving something in the cross product calculation. Choice (B)

4

could result from mistakenly using

4u^2 + 4v^2

instead of the correct magnitude. Choice (C)

2

might arise from errors in computing the cross product components. Strategy tip: Always work systematically through parameterized surface integrals: compute partial derivatives, find the cross product magnitude, then substitute the parameterization into your integrand. Double-check that your final integral matches the given form exactly.

Question 5

Consider the surface $S$ that is the portion of the cylinder $x^2 + z^2 = 4$ with $0 \leq y \leq 3$ and $z \geq 0$ . For the scalar field $g(x,y,z) = yz$ , which expression correctly represents the surface integral $\iint_S g \, dS$ ?

$\int_0^3 \int_{-2}^2 yz \sqrt{1 + \left(\frac{-x}{\sqrt{4-x^2}}\right)^2} \, dx \, dy$
$\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy$ (correct answer)
$\int_0^3 \int_{-2}^2 yz \, dx \, dy$ where $z = \sqrt{4-x^2}$
$\int_0^\pi \int_0^3 2y\sin\theta \, dy \, d\theta$

Explanation: Using cylindrical coordinates with

x = 2\cos\theta

z = 2\sin\theta

for

0 \leq \theta \leq \pi

and

0 \leq y \leq 3

. The parameterization is

\mathbf{r}(\theta,y) = (2\cos\theta, y, 2\sin\theta)

. We get

\mathbf{r}_\theta = (-2\sin\theta, 0, 2\cos\theta)

and

\mathbf{r}_y = (0, 1, 0)

. The magnitude

|\mathbf{r}_\theta \times \mathbf{r}_y| = 2

. On the surface,

g = yz = y \cdot 2\sin\theta

. The integral becomes

\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy

. Choice A uses the wrong surface element calculation. Choice C omits the surface element entirely. Choice D has the correct integrand but wrong bounds and missing factor of 2.

Question 6

A surface $S$ is parameterized by $\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2)$ for $0 \leq u \leq 2$ and $0 \leq v \leq \pi$ . If $f(x,y,z) = x^2 + y^2$ , what is the value of the surface integral $\iint_S f \, dS$ ?

$\frac{2\pi}{3}(17\sqrt{17} - 1)$
$\frac{\pi}{6}(17\sqrt{17} - 1)$ (correct answer)
$\frac{2\pi}{3}(9\sqrt{17} - 1)$
$\frac{\pi}{3}(17\sqrt{17} - 1)$

Explanation: We have

\mathbf{r}_u = (\cos v, \sin v, 2u)

and

\mathbf{r}_v = (-u\sin v, u\cos v, 0)

. Computing the cross product:

\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u)

. Thus

|\mathbf{r}_u \times \mathbf{r}_v| = u\sqrt{4u^2 + 1}

. On the surface,

f(x,y,z) = u^2

. The integral becomes

\int_0^\pi \int_0^2 u^2 \cdot u\sqrt{4u^2 + 1} \, du \, dv = \pi \int_0^2 u^3\sqrt{4u^2 + 1} \, du

. Using substitution

w = 4u^2 + 1

, this evaluates to

\frac{\pi}{6}(17\sqrt{17} - 1)

. Choice A has the wrong coefficient (too large by factor of 4). Choice C uses the wrong upper limit in evaluation. Choice D has the wrong coefficient (too large by factor of 2).

Question 7

Consider the surface $S$ parameterized by $\mathbf{r}(s,t) = (s + t, s - t, s^2 + t^2)$ where $0 \leq s \leq 1$ and $0 \leq t \leq 1$ . If $m(x,y,z) = x + y + z$ , which of the following is closest to the numerical value of $\iint_S m \, dS$ ?

$2.1$
$2.8$ (correct answer)
$3.4$
$4.2$

Explanation: We compute

\mathbf{r}_s = (1, 1, 2s)

and

\mathbf{r}_t = (1, -1, 2t)

. The cross product is

\mathbf{r}_s \times \mathbf{r}_t = (2t + 2s, -2t + 2s, -2)

, so

|\mathbf{r}_s \times \mathbf{r}_t| = 2\sqrt{(t+s)^2 + (s-t)^2 + 1} = 2\sqrt{2s^2 + 2t^2 + 1}

. On the surface,

m = (s+t) + (s-t) + (s^2+t^2) = 2s + s^2 + t^2

. The integral is

\int_0^1 \int_0^1 (2s + s^2 + t^2) \cdot 2\sqrt{2s^2 + 2t^2 + 1} \, ds \, dt

. This evaluates numerically to approximately

2.83

. Choice A underestimates by neglecting the square root factor properly. Choice C overestimates by incorrect calculation of the magnitude. Choice D significantly overestimates the integral value.

Question 8

The surface $S$ is defined by $z = xy$ over the triangular region $R$ in the $xy$ -plane with vertices at $(0,0)$ , $(2,0)$ , and $(0,3)$ . If $h(x,y,z) = z^2 + 1$ , what is the value of $\iint_S h \, dS$ ?

$\int_0^2 \int_0^{3-\frac{3x}{2}} (x^2y^2 + 1)\sqrt{x^2 + y^2 + 1} \, dy \, dx$
$\int_0^3 \int_0^{2-\frac{2y}{3}} (x^2y^2 + 1)\sqrt{x^2 + y^2 + 1} \, dx \, dy$
$\int_0^2 \int_0^{3(1-\frac{x}{2})} (x^2y^2 + 1)\sqrt{x^2 + y^2 + 1} \, dy \, dx$ (correct answer)
$\int_0^3 \int_0^{\frac{2(3-y)}{3}} (x^2y^2 + 1)\sqrt{x^2 + y^2 + 1} \, dx \, dy$

Explanation: For the surface

z = f(x,y) = xy

, we have

f_x = y

and

f_y = x

. The surface element is

dS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy = \sqrt{1 + y^2 + x^2} \, dx \, dy

. On the surface,

h = z^2 + 1 = x^2y^2 + 1

. The triangular region has the line from

(2,0)

(0,3)

given by

\frac{x}{2} + \frac{y}{3} = 1

, so

y = 3(1 - \frac{x}{2})

. Thus the correct bounds are

0 \leq x \leq 2

and

0 \leq y \leq 3(1-\frac{x}{2})

. Choice A has incorrect upper bound for

y

. Choice B uses wrong order of integration with incorrect bounds. Choice D has wrong bounds for

x

Question 9

A surface $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 9$ that lies above the plane $z = \frac{3}{2}$ . For the function $k(x,y,z) = z$ , the surface integral $\iint_S k \, dS$ equals:

$\frac{27\pi}{4}$ (correct answer)
$\frac{81\pi}{8}$
$\frac{27\pi}{2}$
$\frac{9\pi}{2}$

Explanation: Using spherical coordinates:

x = 3\sin\phi\cos\theta

y = 3\sin\phi\sin\theta

z = 3\cos\phi

. The condition

z \geq \frac{3}{2}

means

3\cos\phi \geq \frac{3}{2}

, so

\cos\phi \geq \frac{1}{2}

, giving

0 \leq \phi \leq \frac{\pi}{3}

. The surface element is

dS = 9\sin\phi \, d\phi \, d\theta

. The integral becomes

\int_0^{2\pi} \int_0^{\pi/3} 3\cos\phi \cdot 9\sin\phi \, d\phi \, d\theta = 27 \cdot 2\pi \int_0^{\pi/3} \cos\phi\sin\phi \, d\phi = 27\pi \int_0^{\pi/3} \frac{1}{2}\sin(2\phi) \, d\phi = \frac{27\pi}{4}[-\cos(2\phi)]_0^{\pi/3} = \frac{27\pi}{4}

. Choice B doubles the result incorrectly. Choice C omits the factor of

\frac{1}{2}

. Choice D uses wrong radius.

Question 10

The surface $S$ is the portion of the cone $z = \sqrt{x^2 + y^2}$ between the planes $z = 1$ and $z = 3$ . For the scalar field $w(x,y,z) = \frac{1}{z}$ , the value of $\iint_S w \, dS$ is:

$4\pi$
$2\pi\sqrt{2}$
$4\pi\sqrt{2}$ (correct answer)
$8\pi$

Explanation: When you encounter a surface integral over a portion of a cone, you're dealing with parametric surfaces and need to carefully set up both the parameterization and the surface area element. The cone

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

between

z = 1

and

z = 3

is best parameterized using cylindrical coordinates. Since

z = r

on this cone, you can write:

\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r)

where

1 \leq r \leq 3

and

0 \leq \theta \leq 2\pi

. To find

dS

, compute the cross product of partial derivatives:

\mathbf{r}_r = (\cos\theta, \sin\theta, 1)

and

\mathbf{r}_\theta = (-r\sin\theta, r\cos\theta, 0)

. The magnitude

|\mathbf{r}_r \times \mathbf{r}_\theta| = r\sqrt{2}

, so

dS = r\sqrt{2} \, dr \, d\theta

. The integral becomes:

\iint_S \frac{1}{z} \, dS = \int_0^{2\pi} \int_1^3 \frac{1}{r} \cdot r\sqrt{2} \, dr \, d\theta = \int_0^{2\pi} \int_1^3 \sqrt{2} \, dr \, d\theta = \sqrt{2} \cdot 2\pi \cdot (3-1) = 4\pi\sqrt{2}

Answer A (

4\pi

) results from forgetting the

\sqrt{2}

factor in the surface area element. Answer B (

2\pi\sqrt{2}

) comes from integrating over only half the cone (

0 \leq \theta \leq \pi

). Answer D (

8\pi

) represents both missing the

\sqrt{2}

factor and doubling the height incorrectly. Study tip: For cone surface integrals, always remember that the surface area element includes a

\sqrt{2}

factor due to the cone's slope, and double-check your parameter bounds.