← Back to 15. Multiple Integrals Calculus

📝 Volume under surface double integral (12 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 12 questions available

←Previous Topic:Double Integrals in Calculus📝 Volume under surface double integralDouble integral definition Riemann sums:Next Topic→

What is Volume under surface double integral?

Definition:
The volume of a solid region bounded above by a continuous surface $z = f(x, y)$ and below by a region $R$ in the xy-plane is given by the double integral $V = \iint_R f(x, y) \, dA$ .

Example:
To find the volume under the plane $z = 4 - x - y$ over the rectangle $R = [0, 1] \times [0, 2]$ , we evaluate $\int_0^2 \int_0^1 (4 - x - y) \, dx \, dy$ .

Reason:
This geometric interpretation connects the abstract concept of integration to physical space, allowing us to calculate quantities like mass or charge when density varies.

Easy

Medium

Hard

📝 All Volume under surface double integral MCQs

Q1. A solid occupies the region $0\le x\le 2$ , $0\le y\le 3$ , and $0\le z\le 4-x-y$ . Which double integral correctly represents its volume after integrating with respect to $z$ first?

\int_0^2\int_0^3(4-x-y)\,dy\,dx

\int_0^2\int_0^{4-x}(4-x-y)\,dy\,dx

✅

\int_0^4\int_0^{3-x}(4-x-y)\,dy\,dx

\int_0^2\int_0^{4-x}1\,dy\,dx

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For the solid to exist, the upper surface $z=4-x-y$ must be nonnegative, so $x+y\le4$ . With $0\le x\le2$ , this gives $0\le y\le4-x$ . Integrating the vertical height $4-x-y$ over this triangular projection produces the correct volume representation.

Q2. A solid lies above the $xy$ -plane and below $z=9-x^2-y^2$ . Which description best explains why polar coordinates are more natural for computing its volume?

A.The boundary in the

xy

-plane is a circle, so the projection is described simply by

0\le r\le3

. ✅

B.The surface contains

z

, so polar coordinates eliminate

z

automatically.

C.Polar coordinates always make every triple integral easier.

D.The

xy

-projection is rectangular, so polar coordinates remove the need for limits.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The projection onto the $xy$ -plane is determined by $9-x^2-y^2\ge0$ , giving $x^2+y^2\le9$ , a disk. In polar coordinates this becomes $0\le r\le3$ , while the height becomes $9-r^2$ . Thus the geometry directly favors polar coordinates.

Q3. A storage tank has vertical height $z=12-x^2-y^2$ above the $xy$ -plane. An engineer proposes using a rectangular projection $0\le x\le3$ , $0\le y\le3$ . What is the main flaw in this model?

A.The height should be

12+x^2+y^2

B.The projection includes points where

x^2+y^2>12

, making the proposed upper height negative. ✅

C.The tank must have a cylindrical rather than rectangular projection.

D.The

z

-coordinate cannot be integrated first.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The solid exists only where $12-x^2-y^2\ge0$ , so its projection satisfies $x^2+y^2\le12$ . The proposed square includes points such as $(3,3)$ , where the height is negative. A volume cannot be represented by integrating negative vertical heights over those points.

Q4. A solid is bounded below by $z=0$ and above by $z=6-x-y$ , with $x\ge0$ and $y\ge0$ . A student says the volume can be found by integrating over the entire first-quadrant $xy$ -plane because the formula gives the height everywhere. How should this reasoning be evaluated?

A.It is correct because the height remains positive everywhere.

B.It is correct only if

x

and

y

are integers.

C.It is incorrect because the projection must be restricted to

x+y\le6

. ✅

D.It is incorrect because

z=0

cannot be used as a lower boundary.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The upper surface meets the $xy$ -plane when $6-x-y=0$ , so only points satisfying $x+y\le6$ belong to the solid. Outside that triangular region the proposed upper height becomes negative, which cannot represent a physical volume. Therefore the projection must be restricted.

Q5. Consider the solid below $z=4-x^2-y^2$ and above $z=0$ . A student computes $\int_0^{2\pi}\int_0^2(4-r^2)\,dr\,d\theta$ and obtains a value. What important factor is missing?

A.A factor of

r

from the area element is missing. ✅

B.A factor of

r^2

must be added.

C.The upper limit for

r

should be

4

D.The angular interval should be

0\le\theta\le\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When Cartesian coordinates are transformed to polar coordinates, the area element becomes $dA=r\,dr\,d\theta$ . The radial factor accounts for the changing area of circular strips. Therefore the correct volume integral must contain $r(4-r^2)$ , not merely $4-r^2$ .

Q6. A solid is bounded by $z=2x+3y$ , $z=0$ , $x=0$ , $y=0$ , and $x+y=2$ . Which strategy is most efficient for finding its volume?

A.Integrate over a triangular

xy

-projection using the height

2x+3y

. ✅

B.Integrate over the entire square

0\le x,y\le2

C.Use only a one-dimensional integral because the solid has a planar top.

D.Use cylindrical coordinates because the boundary contains

x+y=2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The vertical height is $2x+3y$ , while the projection onto the $xy$ -plane is the triangle bounded by $x=0$ , $y=0$ , and $x+y=2$ . A double integral of this height over that triangular region directly models the volume without introducing unnecessary coordinate transformations.

Q7. A manufacturing process creates a solid whose vertical height over a point $(x,y)$ is $h(x,y)=10-2x-y$ . The base is the triangular region $x\ge0$ , $y\ge0$ , $2x+y\le10$ . Which modeling choice correctly represents the volume?

A.Integrate

10-2x-y

over the stated triangular base. ✅

B.Integrate

2x+y-10

over the stated triangular base.

C.Integrate

10-2x-y

over a square of side

10

D.Integrate

10-2x-y

over the entire first quadrant.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The volume is obtained by accumulating vertical height over the base. Since the base is explicitly restricted by $2x+y\le10$ , the height $10-2x-y$ is nonnegative throughout that triangular region. Extending the domain beyond the triangle would introduce negative heights and incorrectly model the solid.

Q8. A graph of a solid shows a circular base centered at the origin with radius $2$ , while the upper surface rises according to $z=5-x^2-y^2$ . Which integral most faithfully translates the graph into a volume calculation?

\int_0^{2\pi}\int_0^2\int_0^{5-r^2}r\,dz\,dr\,d\theta

✅

\int_0^{2\pi}\int_0^5\int_0^2r\,dz\,dr\,d\theta

\int_0^\pi\int_0^2\int_0^{5-r^2}r\,dz\,dr\,d\theta

\int_0^{2\pi}\int_0^2\int_0^5(5-r^2)\,dz\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The graph indicates a circular projection with $0\le r\le2$ and a complete revolution, so $0\le\theta\le2\pi$ . The vertical coordinate ranges from $0$ to $5-r^2$ . Because cylindrical coordinates are used, the volume element is $r\,dz\,dr\,d\theta$ .

Q9. Two students evaluate the volume under $z=16-x^2-y^2$ and above $z=0$ . Student A uses Cartesian coordinates over a disk and Student B uses polar coordinates. Both correctly describe the same region. Which conclusion is most justified?

A.Student A must obtain a larger volume because Cartesian coordinates use rectangular elements.

B.Student B must obtain a larger volume because polar coordinates are exact.

C.Both should obtain the same volume if their limits and area elements are correct. ✅

D.Only Student A can obtain the volume because the surface is written in Cartesian form.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A coordinate transformation does not change the geometric volume. Cartesian and polar descriptions partition the same solid differently, but correctly accounting for their respective area elements must produce identical results. Any disagreement indicates an error in the limits, integrand, Jacobian factor, or evaluation rather than a difference in the actual volume.

Q10. A student wants to find the volume of the region under $z=xy$ and above the rectangle $0\le x\le2$ , $1\le y\le3$ . The student writes $\int_0^2\int_1^3xy\,dx\,dy$ . What is the most important issue with this setup?

A.The limits are reversed in both variables.

B.The order of integration is impossible because

x

appears in the integrand.

C.The expression correctly represents the volume; only the order of differentials should be reversed. ✅

D.The lower surface

z=0

must be included explicitly as a subtraction.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The integral $\int_0^2\int_1^3xy\,dx\,dy$ correctly accumulates the vertical height $xy$ over the given rectangular base. The lower surface is $z=0$ , so no separate subtraction is necessary. Although the notation places $dx$ inside while its associated bounds are $0$ to $2$ , the setup is mathematically consistent.

Q11. A designer models a solid above the $xy$ -plane and below $z=12-r^2$ , but only where the upper surface is nonnegative. The designer then compares two approaches: Method I integrates the height over the circular base, while Method II integrates directly in cylindrical coordinates. Why should both methods agree?

A.Both methods use the same vertical height and cover exactly the same three-dimensional region. ✅

B.Method II counts each point multiple times, but symmetry cancels the error.

C.Method I ignores the curved boundary, but the error disappears after integration.

D.They agree only because the maximum height happens to be

12

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The condition $12-r^2\ge0$ determines the same circular projection used by the height-based method. Method I integrates the vertical height over that base, while Method II includes the radial and angular dimensions directly. Since both describe every point of the same solid exactly once, their volumes must agree.

Q12. For $a>0$ , a solid is bounded above by $z=a^2-x^2-y^2$ and below by $z=0$ . Without evaluating a full integral, which expression best describes how the volume changes when $a$ is replaced by $2a$ ?

A.The volume becomes

2

times as large.

B.The volume becomes

4

times as large.

C.The volume becomes

8

times as large.

D.The volume becomes

16

times as large. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The projection satisfies $x^2+y^2\le a^2$ , giving radius $a$ , so its area scales as $a^2$ . The vertical height $a^2-r^2$ also scales as $a^2$ . Therefore the total volume scales as $a^4$ . When $a$ is replaced by $2a$ , the volume is multiplied by $16$ .

🔗 Related Topics (MCQs)

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

←Previous Topic:Double Integrals in Calculus📝 Volume under surface double integralDouble integral definition Riemann sums:Next Topic→