← Back to 15. Multiple Integrals Calculus

📝 Triple Integrals in Calculus (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

←Previous Topic:Surface area of parametric surfaces📝 Triple Integrals in CalculusTriple integral definition:Next Topic→

What is Triple Integrals in Calculus?

Definition:
A triple integral extends the concept of double integrals to three dimensions, integrating a function $f(x,y,z)$ over a 3D region $E$ : $\iiint_E f(x,y,z) \, dV$ .

Example:
The mass of a solid with density $\rho(x,y,z)$ is $M = \iiint_E \rho(x,y,z) \, dV$ .

Reason:
Triple integrals allow us to calculate physical quantities like mass, charge, and average value over 3D volumes, essential for physics and engineering applications.

Easy

Medium

Hard

📝 All Triple Integrals in Calculus MCQs

Q1. A cylindrical region has density $\rho(r,\theta,z)=z$ . If the total mass is computed as $M=\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{h} z \, r \, dz \, dr \, d\theta$ , a student claims mass is independent of $h$ because $z$ cancels. Which error analysis is correct?

A.The student integrated

z

incorrectly; the inner integral is

h^2/2

, so mass depends on

h^2

. ✅

B.The student forgot the Jacobian

r

; without it mass is independent of

h

C.The student used wrong limits;

z

should go from 0 to

2\pi

D.The student is correct; the

r \, dz

makes mass independent of

h

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The inner integral $\int_0^h z \, dz = h^2/2$ , so $M = \pi h^2$ . The misconception arises from treating $z$ as a constant. The Jacobian $r$ is correctly included; the error is purely in evaluating the $z$ -integral. Thus mass strongly depends on $h^2$ .

Q2. A solid occupies the region bounded by $z=4-x^2-y^2$ and $z=0$ . A student sets up $\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{4-x^2-y^2} dz \, dy \, dx$ . What is the most efficient method to verify the volume?

A.Convert to cylindrical:

\int_0^{2\pi}\int_0^2\int_0^{4-r^2} r \, dz \, dr \, d\theta

✅

B.Convert to spherical:

\int_0^{2\pi}\int_0^{\pi/2}\int_0^2 \rho^2\sin\phi \, d\rho \, d\phi \, d\theta

C.Keep Cartesian but swap

dz

and

dx

D.Use divergence theorem on the surface

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The region is a paraboloid of revolution. Cylindrical symmetry makes the $r$ -integral $\int_0^2 r(4-r^2)dr = 4$ , times $2\pi$ gives $8\pi$ . Spherical is messy because the surface is not a sphere. Cartesian works but is more tedious. The question tests recognition of coordinate system efficiency.

Q3. A tetrahedron has vertices at $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$ . The integral $I=\int_0^1\int_0^{1-x}\int_0^{1-x-y} f(x,y,z) \, dz \, dy \, dx$ is given. If the order is changed to $dz \, dx \, dy$ , what are the new limits for $x$ ?

x

from 0 to

1-y

✅

x

from 0 to

1-y-z

x

from 0 to

1-y

and

z

from 0 to

1-x-y

x

from 0 to

1-y

and

z

from 0 to

1-x-y

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The tetrahedron is defined by $x\ge0,y\ge0,z\ge0,x+y+z\le1$ . For order $dz \, dx \, dy$ , outer $y$ from 0 to 1, middle $x$ from 0 to $1-y$ , inner $z$ from 0 to $1-x-y$ . So $x$ limits are $0$ to $1-y$ . Option C is same as A but duplicates; A is correct. This requires reconstructing the region from limits.

Q4. Consider the integral $\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} dz \, dy \, dx$ . Which graph correctly describes the region?

A.A quarter of a sphere of radius 1 in the first octant ✅

B.A full sphere of radius 1

C.A cylinder of radius 1 and height 1

D.A cone with vertex at origin

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The upper limit on $z$ is $\sqrt{1-x^2-y^2}$ , and $x,y$ are in the first quadrant quarter circle. This is exactly the portion of the unit sphere where $x\ge0, y\ge0, z\ge0$ . The graph would show a spherical octant. This tests visualization of 3D regions from integral limits.

Q5. A student evaluates $\int_0^1 \int_0^x \int_0^{x+y} dz \, dy \, dx$ and gets $1/6$ . Another student changes order to $dz \, dx \, dy$ and gets $1/3$ . Which is correct and why?

A.First is correct;

1/6

is correct because region is

0\le y\le x\le1, 0\le z\le x+y

B.Second is correct;

1/3

is correct because limits for

x

should be

y

to 1

C.Both are incorrect; correct value is

1/4

✅

D.Both are correct because Fubini's theorem allows any order

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The direct computation gives $\int_0^1 \int_0^x (x+y) dy dx = \int_0^1 (3x^2/2) dx = 1/2$ . The second student's order also gives 1/2 if limits are correctly transformed. The first student's 1/6 and second's 1/3 are arithmetic errors. This question tests the ability to detect and correct errors in order switching.

Q6. The volume of a solid is given by $V=\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$ . What is the shape of the solid?

A.A spherical cone with radius 2 and angle

\pi/4

✅

B.A hemisphere of radius 2

C.A cylinder of radius 2 and height 2

D.A cone of height 2 and base radius 2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Spherical coordinates: $\rho$ from 0 to 2, $\phi$ from 0 to $\pi/4$ , $\theta$ full circle. This is a spherical sector (a cone-shaped region) with apex at origin, radius 2, and half-angle $\pi/4$ . Not a full hemisphere because $\phi$ stops at $\pi/4$ . This tests recognition of spherical coordinate limits.

Q7. A solid has density $\delta(x,y,z)=x^2+y^2$ . The moment of inertia about z-axis is $I_z=\int\int\int (x^2+y^2)\delta \, dV$ . Which integral in cylindrical coordinates is correct?

\int_0^{2\pi}\int_0^R\int_0^H r^2 \cdot r \cdot r \, dz \, dr \, d\theta

\int_0^{2\pi}\int_0^R\int_0^H r^2 \cdot r \, dz \, dr \, d\theta

\int_0^{2\pi}\int_0^R\int_0^H r^2 \cdot r^2 \, dz \, dr \, d\theta

\int_0^{2\pi}\int_0^R\int_0^H r^2 \cdot r^3 \, dz \, dr \, d\theta

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Moment of inertia about z-axis requires $(x^2+y^2)$ times density $\delta$ . In cylindrical, $x^2+y^2=r^2$ , $\delta=r^2$ , and $dV=r\,dz\,dr\,d\theta$ . So total integrand is $r^2 \cdot r^2 \cdot r = r^5$ . Option D correctly represents this as $r^2 \cdot r^3$ .

Q8. The integral $\int_0^1 \int_0^{1-z} \int_0^{1-y-z} dx \, dy \, dz$ represents the volume of a tetrahedron. What is the volume?

1/6

✅

1/3

1/2

1

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The region is $x\ge0,y\ge0,z\ge0, x+y+z\le1$ . The volume of this standard tetrahedron is $1/6$ . This is a classic result. The integral evaluates to $\int_0^1 \int_0^{1-z} (1-y-z) dy dz = \int_0^1 (1-z)^2/2 dz = 1/6$ .

Q9. A student sets up a triple integral for the volume of a sphere of radius R using spherical coordinates as $\int_0^{2\pi}\int_0^{\pi}\int_0^R \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ . They then claim the volume is $\frac{4}{3}\pi R^3$ . Another student uses $\int_0^{2\pi}\int_0^{\pi}\int_0^R \rho^2 \, d\rho \, d\phi \, d\theta$ and gets $2\pi R^3/3$ . Which is correct and what is the error?

A.First is correct; second forgot

\sin\phi

Jacobian ✅

B.Second is correct; first has extra

\sin\phi

C.Both are correct because they use different coordinate systems

D.Neither; volume of sphere is

\pi R^3

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Spherical volume element is $\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ . Omitting $\sin\phi$ is a common mistake. The first integral correctly includes it and yields $4/3\pi R^3$ . The second yields $2\pi R^3/3$ , which is the volume of a cylinder in spherical coordinates without the angular correction. This tests the crucial Jacobian factor.

Q10. A solid is bounded by $z=0$ , $z=4-y^2$ , and $x^2+y^2=4$ . Which of the following correctly sets up the volume integral in cylindrical coordinates?

\int_0^{2\pi}\int_0^2\int_0^{4-r^2\sin^2\theta} r \, dz \, dr \, d\theta

✅

\int_0^{2\pi}\int_0^2\int_0^{4-r^2} r \, dz \, dr \, d\theta

\int_0^{2\pi}\int_0^4\int_0^{4-r^2} r \, dz \, dr \, d\theta

\int_0^{2\pi}\int_0^2\int_0^{4-y^2} dz \, dy \, dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The cylinder $x^2+y^2=4$ becomes $r=2$ . The paraboloid $z=4-y^2$ becomes $z=4-r^2\sin^2\theta$ . So the integrand is $r$ , limits: $r:0\to2$ , $\theta:0\to2\pi$ , $z:0\to4-r^2\sin^2\theta$ . Option A is correct. Option B incorrectly uses $4-r^2$ which would be a paraboloid of revolution, but here it's parabolic in y only.

Q11. The average value of $f(x,y,z)=x$ over the unit sphere is 0. A student argues that because the sphere is symmetric, the average of $x^2$ is also 0. Which is correct?

A.The student is wrong;

x^2

is always positive so average is positive ✅

B.The student is correct by symmetry

C.The average of

x^2

is 0 only if the sphere is centered at origin

D.The average of

x^2

is 0 because

x^2

is odd

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The average of $x$ is 0 because $x$ is an odd function over a symmetric domain. However $x^2$ is even and non-negative, so its average is $1/3$ (by symmetry, $\bar{x^2}=\bar{y^2}=\bar{z^2}$ and sum = 1). The student confuses oddness with evenness. This tests symmetry arguments and expectation values.

Q12. A solid is formed by intersecting the cylinder $x^2+y^2=1$ and the planes $z=0$ and $z=x+2$ . A student computes volume as $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{x+2} dz \, dy \, dx$ . Is this correct? If not, what is the error?

A.Correct ✅

B.Incorrect:

z

lower limit should be

x+2

and upper 0

C.Incorrect: the cylinder radius is 2, not 1

D.Incorrect: the order of integration should be

dz \, dx \, dy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The cylinder is $x^2+y^2=1$ , so $x$ from -1 to 1, $y$ from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$ , and $z$ from 0 to $x+2$ . Since $x+2 \ge 1$ for all $x\in[-1,1]$ , the lower limit 0 is always below the upper, so the setup is correct. The integral gives volume $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x+2) dy dx = \int_{-1}^{1} 2\sqrt{1-x^2}(x+2) dx = 2\int_{-1}^{1} x\sqrt{1-x^2} dx + 4\int_{-1}^{1}\sqrt{1-x^2} dx = 0 + 4\cdot(\pi/2) = 2\pi$ . So correct. This is a multi-step reasoning problem combining geometry and integration.

Q13. The integral $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f(x,y,z) \, dz \, dy \, dx$ is given. If $f(x,y,z)=1$ , the value is 1/6. If $f(x,y,z)=x$ , what is the value?

1/24

✅

1/12

1/6

1/8

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The tetrahedron region is $x+y+z\le1$ . We need $\int\int\int x \, dV$ . By symmetry, $\int x = \int y = \int z$ , and sum $\int (x+y+z) = \int 1 = 1/6$ . Since each is equal, each is $(1/6)/3 = 1/18$ . Wait, that gives 1/18, not in options. Let's compute directly: $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} x \, dz \, dy \, dx = \int_0^1 \int_0^{1-x} x(1-x-y) dy dx = \int_0^1 x[(1-x)^2/2] dx = \frac{1}{2}\int_0^1 x(1-2x+x^2) dx = \frac{1}{2}[\frac{1}{2} - \frac{2}{3} + \frac{1}{4}] = \frac{1}{2}[\frac{6-8+3}{12}] = \frac{1}{2}\cdot\frac{1}{12} = \frac{1}{24}$ . So correct is 1/24. This tests expectation values and symmetry.

Q14. A student attempts to find the volume of the region inside the sphere $x^2+y^2+z^2=4$ and above the cone $z=\sqrt{x^2+y^2}$ using spherical coordinates. They set $\phi$ from 0 to $\pi/4$ and $\rho$ from 0 to 2. What is the error?

A.They should set

\phi

from 0 to

\pi/4

and

\rho

from 0 to

2/\cos\phi

B.They should set

\phi

from

\pi/4

\pi/2

C.They should set

\phi

from 0 to

\pi/4

and

\rho

from 0 to 2 is correct ✅

D.They should set

\phi

from 0 to

\pi/2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The cone $z=\sqrt{x^2+y^2}$ has spherical equation $\phi=\pi/4$ . The region inside the sphere and above the cone corresponds to $0\le\phi\le\pi/4$ and $0\le\rho\le2$ . So the student's limits are correct. No error. This tests the correct interpretation of 'above' in spherical coordinates.

🔗 Related Topics (MCQs)

📝 Double Integrals in Calculus MCQs 📝 Volume under surface double integral MCQs 📝 Double integral definition Riemann sums MCQs 📝 How to evaluate double integrals MCQs 📝 Fubini's theorem for double integrals MCQs

←Previous Topic:Surface area of parametric surfaces📝 Triple Integrals in CalculusTriple integral definition:Next Topic→