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📝 Triple Integral Evaluation over General Regions (16 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 16 questions available

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What is Triple Integral Evaluation over General Regions?

Definition:
For general regions, limits become functions. Type 1: $z$ between surfaces, $(x,y)$ in projection. Type 2: $y$ between surfaces, $(x,z)$ in projection. Type 3: $x$ between surfaces, $(y,z)$ in projection.

Example:
For a tetrahedron bounded by planes, we might use $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f \, dz \, dy \, dx$ .

Reason:
Real-world solids have irregular shapes; adapting limits to the bounding surfaces allows accurate integration over any 3D region.

Easy

Medium

Hard

📝 All Triple Integral Evaluation over General Regions MCQs

Q1. A solid $E$ is bounded below by $z=0$ , above by $z=4-x^2-y^2$ , and projects onto the disk $x^2+y^2\le4$ . Which setup correctly represents $\iiint_E z\,dV$ using cylindrical coordinates?

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

✅

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

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

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The projection is a disk of radius $2$ , so $0\le r\le2$ and $0\le\theta\le2\pi$ . The upper surface becomes $z=4-r^2$ , while the lower surface is $z=0$ . Because cylindrical coordinates have volume element $dV=r\,dz\,dr\,d\theta$ , the factor $r$ must be included.

Q2. Why is changing from Cartesian to cylindrical coordinates especially advantageous when a solid and its projection are described using expressions involving $x^2+y^2$ ?

A.It always eliminates every variable from the integrand.

B.It converts radial symmetry into simpler bounds and often simplifies the integrand. ✅

C.It guarantees that the resulting integral has only one variable.

D.It changes every curved boundary into a planar boundary.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Cylindrical coordinates are designed around the quantity $x^2+y^2=r^2$ . Thus circular or rotationally symmetric projections can often be described with constant angular bounds and simple radial limits. The transformation does not necessarily reduce the number of variables, but it can make both the region and integrand substantially easier to represent.

Q3. A region is described by $0\le x\le1$ , $0\le y\le1-x$ , and $0\le z\le x+y$ . A student claims that reversing the order to $dz,dx,dy$ gives $0\le y\le1$ , $0\le x\le1-y$ , and $0\le z\le x+y$ . Which evaluation best describes the student's reasoning?

A.The reasoning is correct because reversing two variables never changes the region.

B.The reasoning is incorrect because the

z

-bound must become

0\le z\le y

C.The reasoning is correct because the triangular projection is symmetric in

x

and

y

. ✅

D.The reasoning is incorrect because

x

and

y

cannot be reversed in a triple integral.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The student's reasoning is correct. The original $xy$ -projection is the triangle $x\ge0$ , $y\ge0$ , and $x+y\le1$ . Rewriting it as $0\le y\le1$ and $0\le x\le1-y$ describes exactly the same triangle. The vertical condition $0\le z\le x+y$ remains unchanged.

Q4. A solid has $z$ between $x^2+y^2$ and $6$ , with $x^2+y^2\le6$ . Which strategy is most efficient for evaluating $\iiint_E 1\,dV$ , and why?

A.Use spherical coordinates because every paraboloid becomes a sphere.

B.Use cylindrical coordinates because both the surfaces and projection depend naturally on

r^2=x^2+y^2

. ✅

C.Use Cartesian coordinates because the circular projection is rectangular.

D.Use spherical coordinates because the upper boundary is a constant.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Cylindrical coordinates match the geometry directly: the lower surface becomes $z=r^2$ , the upper surface is $z=6$ , and the projection is $0\le r\le\sqrt6$ . This produces simple nested bounds. Spherical coordinates are not naturally aligned with the paraboloid, while Cartesian coordinates would require less convenient circular bounds.

Q5. A solid lies inside the cylinder $x^2+y^2\le9$ , above $z=0$ , and below the plane $z=x+y+5$ . Before integrating, a student notices that the plane becomes negative for some points of the disk. What must be done to obtain a valid description of the solid?

A.Ignore the negative portion because volume cannot be negative.

B.Replace the lower surface by

z=x+y+5

everywhere.

C.Restrict the projection to points where

x+y+5\ge0

, because the solid requires

0\le z\le x+y+5

. ✅

D.Use

z=|x+y+5|

as the upper surface.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The condition $0\le z\le x+y+5$ is possible only where $x+y+5\ge0$ . Therefore, the projection cannot automatically remain the entire disk. The correct region must be split or restricted according to where the plane lies above $z=0$ . Simply integrating a negative height would not represent the intended solid.

Q6. A region in the $xy$ -plane is the first-quadrant portion between the circles $r=2$ and $r=5$ , and $z$ ranges from $r$ to $10-r$ . Which observation is essential before setting up the integral?

A.The radial bounds are

0\le r\le5

B.The angular bounds must be

0\le\theta\le2\pi

C.The vertical interval is nonempty only when

r\le5

, which is consistent with the annular projection. ✅

D.The upper surface must be

z=10

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The vertical interval is $r\le z\le10-r$ , so it exists only when $r\le10-r$ , giving $r\le5$ . This compatibility condition matters when interpreting a three-dimensional region. Since the projection already has $2\le r\le5$ , the bounds are consistent, but recognizing this prevents an invalid setup.

Q7. A solid is bounded by $z=0$ , $z=4-x^2-y^2$ , and the vertical cylinder $x^2+y^2=1$ . Which statement correctly distinguishes the two possible descriptions of the solid?

A.The cylinder determines the upper

z

-bound.

B.The cylinder restricts the radial projection to

r\le1

, while the paraboloid determines the vertical height. ✅

C.The paraboloid determines the angular range.

D.The cylinder and paraboloid must be converted into Cartesian coordinates before integration.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The vertical cylinder $x^2+y^2=1$ becomes $r=1$ and therefore controls the radial extent of the projection. The paraboloid $z=4-r^2$ supplies the upper vertical boundary, while $z=0$ supplies the lower one. Separating the roles of boundaries is a key step in constructing correct triple-integral limits.

Q8. Imagine a graph of a solid whose projection onto the $xy$ -plane is a semicircular region $x^2+y^2\le16$ with $y\ge0$ . The solid extends vertically from $z=1$ to $z=5-x^2-y^2/4$ . Which cylindrical-coordinate angular interval matches the graph?

0\le\theta\le\pi

✅

\pi\le\theta\le2\pi

0\le\theta\le2\pi

-\pi/2\le\theta\le\pi/2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The condition $y\ge0$ selects the upper half of the circular projection. In cylindrical coordinates, $y=r\sin\theta$ , so $y\ge0$ corresponds to $0\le\theta\le\pi$ . The full interval $0\le\theta\le2\pi$ would include the lower half and therefore describe a larger projection than the graph.

Q9. For a solid bounded by $z=x^2+y^2$ , $z=8$ , and $x^2+y^2=4$ , a student uses $0\le r\le\sqrt8$ . Another uses $0\le r\le2$ . Which conclusion is correct?

A.The first student is correct because

z=8

gives

r=\sqrt8

B.The second student is correct because the cylinder explicitly restricts the projection to

r\le2

. ✅

C.Both are correct because the cylinder affects only

z

D.Neither is correct because

r

must be negative on part of the region.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The cylinder $x^2+y^2=4$ is equivalent to $r=2$ , so it directly restricts the radial projection. Although the surfaces $z=r^2$ and $z=8$ meet at $r=\sqrt8$ , the additional cylindrical boundary removes the portion with $2<r\le\sqrt8$ . Thus the actual projection has $0\le r\le2$ .

Q10. A modeler wants the volume of the region inside $x^2+y^2+z^2\le25$ but above the cone $z=\sqrt{x^2+y^2}$ . Which coordinate system most directly captures both boundaries, and what is the key angular restriction?

A.Cylindrical coordinates;

0\le\theta\le\pi

B.Spherical coordinates;

0\le\phi\le\pi/4

✅

C.Cartesian coordinates;

0\le x\le5

D.Spherical coordinates;

0\le\phi\le\pi/2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The sphere is naturally represented by $\rho=5$ , while the cone $z=r$ corresponds to a constant polar angle in spherical coordinates. Since $z=\rho\cos\phi$ and $r=\rho\sin\phi$ , the cone gives $\tan\phi=1$ , so $\phi=\pi/4$ . Above the cone means $0\le\phi\le\pi/4$ .

Q11. Suppose a region is symmetric about the $z$ -axis and the integrand is $f(x,y,z)=x^2+y^2+z$ . A student says symmetry allows replacing the integral of $f$ by zero. What is the best assessment?

A.Correct, because every function integrated over a rotationally symmetric region is zero.

B.Incorrect, because

x^2+y^2

is rotationally symmetric and generally contributes positively, while only an appropriate odd part could cancel. ✅

C.Correct, because

z

is always an odd function.

D.Incorrect, because symmetry can never simplify a triple integral.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Symmetry must be applied to the integrand's individual components, not automatically to the entire expression. The terms $x^2+y^2$ are nonnegative and rotationally symmetric, so they do not cancel. If the region is symmetric with respect to the $xy$ -plane, the $z$ -term may cancel, but that still leaves the $x^2+y^2$ contribution.

Q12. A solid is described by $0\le z\le4-x^2-y^2$ and $x^2+y^2\le4$ . An analyst evaluates its volume in cylindrical coordinates and obtains $16\pi$ . Which diagnostic is most useful for detecting a likely mistake?

A.The volume should be negative because

z

decreases with

r

B.The maximum height is

4

and the base has area

4\pi

, so

16\pi

is merely an upper bound; the actual volume must be smaller because the height decreases away from the center. ✅

C.The volume must equal the base area because the solid is rotationally symmetric.

D.Rotational symmetry proves the volume is

8\pi

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The base is a disk of area $4\pi$ , and the height varies from $4$ at the center to $0$ at the edge. Therefore $16\pi$ , obtained by multiplying maximum height by base area, is an upper bound rather than the exact volume. This comparison is a useful reasonableness check before trusting a computation.

Q13. A solid has projection $D=\{(x,y):0\le x\le2,\ x^2\le y\le2x\}$ , with $0\le z\le y$ . Which approach best simplifies the volume calculation if the goal is to avoid unnecessarily complicated triple-integral bounds?

A.Integrate with respect to

z

first, reducing the problem to a double integral over

D

. ✅

B.Convert immediately to spherical coordinates.

C.Integrate

x

first while treating

y

as unrestricted.

D.Use cylindrical coordinates because the projection is circular.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Because $z$ runs from $0$ to $y$ , integrating with respect to $z$ first immediately contributes a factor of $y$ . The remaining problem is a double integral over the explicitly described planar region $D$ . This avoids introducing an unnecessary coordinate transformation when the existing bounds are already manageable.

Q14. Consider a region inside the sphere $x^2+y^2+z^2\le9$ and above the plane $z=1$ . Which setup best reflects the geometry in spherical coordinates?

0\le\rho\le3,\ 0\le\phi\le\pi/2,\ 0\le\theta\le2\pi

0\le\rho\le3,\ 0\le\phi\le\pi/3,\ 0\le\theta\le2\pi

✅

0\le\rho\le3,\ 0\le\phi\le\pi,\ 0\le\theta\le\pi

1\le\rho\le3,\ 0\le\phi\le\pi/2,\ 0\le\theta\le2\pi

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The sphere gives $0\le\rho\le3$ . The plane $z=1$ becomes $\rho\cos\phi=1$ , so for a given direction the radial variable begins at $\rho=\sec\phi$ . An equivalent description can use angular restrictions and radial bounds, but among the listed choices, $0\le\phi\le\pi/3$ captures the directions that can intersect the plane within the sphere. The listed option is the closest geometric characterization, though a complete integral would still require the appropriate $\rho$ -lower bound.

Q15. A positive function $f$ is integrated over a solid $E$ . Two students use different coordinate systems. Student A obtains simple bounds but a complicated transformed integrand; Student B obtains complicated bounds but a simple transformed integrand. Which principle gives the better basis for choosing between them?

A.Always choose the coordinate system with the simpler integrand.

B.Always choose the coordinate system with the simpler bounds.

C.Compare the overall complexity of both the transformed integrand and the region, including the Jacobian, and choose the setup that minimizes total work. ✅

D.Both methods must produce different answers because coordinate systems change volume.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A coordinate transformation should be judged globally rather than by a single feature. A simple integrand may be outweighed by extremely complicated bounds, while simple bounds may not compensate for an unwieldy integrand. The Jacobian must also be included. The best setup is the one that makes the complete integral easiest to describe and evaluate accurately.

Q16. Let $E$ be the portion of the sphere $\rho\le3$ lying inside the cone $\phi\le\alpha$ , where $0<\alpha<\pi/2$ . For the integral of $z^2$ over $E$ , which structural observation leads to the most efficient setup?

A.Since

z^2=\rho^2\cos^2\phi

, the integrand and spherical Jacobian separate into a radial factor and an angular factor. ✅

B.The angular variable can be removed because

z^2

contains no

\theta

C.The radial bound must depend on

\theta

because the cone is curved.

D.The sphere cannot be represented using spherical coordinates when a cone is present.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: In spherical coordinates, $z=\rho\cos\phi$ , so $z^2=\rho^2\cos^2\phi$ . Multiplying by the Jacobian $\rho^2\sin\phi$ gives $\rho^4\cos^2\phi\sin\phi$ , while the bounds separate as $0\le\rho\le3$ , $0\le\phi\le\alpha$ , and $0\le\theta\le2\pi$ . This separation turns a three-variable problem into products of simpler one-variable integrals.

🔗 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

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