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📝 Double Integrals in Calculus (12 MCQs)

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

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What is Double Integrals in Calculus?

Definition:
A double integral extends the concept of a single integral to functions of two variables, representing the accumulation of quantities over a two-dimensional region.

Example:
The expression $\iint_R f(x,y) \, dA$ represents the integration of function $f$ over region $R$ .

Reason:
This allows us to calculate quantities like volume, mass, and average value for objects defined in two dimensions.

Easy

Medium

Hard

📝 All Double Integrals in Calculus MCQs

Q1. A region $R$ has area $6$ , and a continuous function satisfies $2\leq f(x,y)\leq5$ throughout $R$ . Which conclusion about $\iint_R f(x,y)\,dA$ must be true?

A.It must equal

21

B.It lies between

12

and

30

✅

C.It lies between

6

and

15

D.It must be greater than

30

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since $f(x,y)$ is at least $2$ and at most $5$ over a region of area $6$ , its double integral must lie between $2(6)=12$ and $5(6)=30$ . Equality occurs only when the function is constant at the corresponding bound throughout the region.

Q2. A student evaluates $\iint_R (x+y)\,dA$ over a region symmetric about the origin and concludes that the integral is zero because both $x$ and $y$ change sign. What is the best assessment?

A.The conclusion is always correct because the region is bounded

B.The conclusion is incorrect because

x+y

is always positive

C.The conclusion is correct only if the region is symmetric about both coordinate axes

D.The conclusion is correct whenever the region is symmetric under

(x,y)\mapsto(-x,-y)

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Under the transformation $(x,y)\mapsto(-x,-y)$ , the function $x+y$ changes to $-(x+y)$ , while the region and area element remain unchanged. Therefore, central symmetry about the origin is sufficient for cancellation, even if the region is not separately symmetric about either coordinate axis.

Q3. A rectangular metal plate occupies $0\leq x\leq2$ and $0\leq y\leq3$ . Its surface density is $\rho(x,y)=x+2y$ . Which expression correctly models the total mass?

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

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

C.Both A and B represent the same mass ✅

6\int_0^2\int_0^3(x+2y)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Both iterated integrals describe integration of the same density over the same rectangular region. Changing the order of integration does not change the value because the density is continuous on the rectangle. The extra factor $6$ in option D would incorrectly count the area once again.

Q4. The graph of a nonnegative function $f(x,y)$ lies above a rectangular region $R$ . Another function $g(x,y)$ is everywhere at least as large as $f(x,y)$ on $R$ . Which statement is necessarily true?

\iint_R g\,dA<\iint_R f\,dA

\iint_R g\,dA=\iint_R f\,dA

only when

g=f

at every point

\iint_R g\,dA\geq\iint_R f\,dA

✅

D.The relationship cannot be determined from the graphs

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Because $g(x,y)\geq f(x,y)$ at every point of the same region, the volume represented by $g$ above the region cannot be smaller than the volume represented by $f$ . Therefore, the double integral of $g$ must be at least as large as that of $f$ .

Q5. A region is bounded by $y=x^2$ , $y=4$ , and the vertical axis. A student writes $\int_0^4\int_{x^2}^4 f(x,y)\,dy\,dx$ . What is the error?

A.The outer limits must be

0

2

, not

0

4

✅

B.The inner limits must be

0

x^2

C.The order of integration cannot be reversed

D.The region cannot be represented using an iterated integral

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The curves $y=x^2$ and $y=4$ intersect when $x^2=4$ , giving $x=2$ in the first-quadrant region. Thus $x$ ranges from $0$ to $2$ . Using $0$ to $4$ extends the integration beyond the actual region and includes points that do not belong to it.

Q6. A region $R$ is split into two non-overlapping subregions $R_1$ and $R_2$ , except possibly along their common boundary. Which modelling principle allows $\iint_R f\,dA$ to be evaluated by adding the two separate integrals?

A.The function must be linear

B.The two subregions must have equal area

C.The integral over the whole region equals the sum of the integrals over the subregions ✅

D.The function must be positive everywhere

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Double integration is additive over regions that partition the domain, provided the pieces do not overlap except along boundaries of zero area. Therefore, the integral over $R$ equals the sum of the integrals over $R_1$ and $R_2$ , regardless of whether their areas are equal.

Q7. A contour-style graph shows a function whose values increase steadily from the lower-left corner of a square region toward the upper-right corner. Without performing calculations, which conclusion is most reasonable about the average value of the function?

A.It must equal the minimum value

B.It must equal the maximum value

C.It must lie between the minimum and maximum values ✅

D.It must always be zero

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For a continuous function on a bounded region, the average value is constrained by the smallest and largest function values. A graph showing increasing values does not by itself determine the exact average, but it guarantees that the average lies between the minimum and maximum values.

Q8. Suppose $R$ is a disk centered at the origin and $f(x,y)=x^2+y^2$ . A student argues that symmetry makes $\iint_R f\,dA=0$ . Which diagnosis is correct?

A.The student is correct because both variables change sign

B.The student is wrong because

f

is nonnegative and generally positive ✅

C.The student is correct because the disk has equal area in all quadrants

D.The student is wrong only if the radius is greater than

1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Although $x$ and $y$ individually change sign under reflections, the function $x^2+y^2$ does not. It remains nonnegative and is positive everywhere except at the origin. Therefore, its integral over a disk of positive area must be positive rather than zero.

Q9. A rectangular region has dimensions $4$ by $5$ , and the graph of $f(x,y)$ over the region appears nearly flat at height $3$ , with small variations above and below $3$ . Which estimate is most reasonable for $\iint_R f\,dA$ ?

A.Approximately

3

B.Approximately

12

C.Approximately

20

D.Approximately

60

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The double integral represents accumulated height over area. The rectangle has area $4(5)=20$ , and the function is approximately $3$ across the region. Hence the integral should be close to $3(20)=60$ . Small variations around height $3$ may change the exact value but not this basic estimate.

Q10. A region is described by $0\leq y\leq2$ and $y\leq x\leq4$ . Which integral represents the integral of $f(x,y)$ over this region when integrating with respect to $y$ first?

\int_0^4\int_0^x f(x,y)\,dy\,dx

\int_0^2\int_y^4 f(x,y)\,dx\,dy

✅

\int_0^2\int_0^y f(x,y)\,dy\,dx

\int_0^4\int_x^2 f(x,y)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The description directly gives the horizontal bounds: $y$ ranges from $0$ to $2$ , while for each fixed $y$ , $x$ ranges from $y$ to $4$ . Therefore, the natural iterated integral is obtained by integrating with respect to $x$ first, despite the wording about the region; option B correctly represents the region.

Q11. For a region $R$ , a continuous function satisfies $f(x,y)=4-x^2-y^2$ . The region lies entirely inside the circle $x^2+y^2\leq4$ . Which reasoning gives the strongest conclusion about the integral?

A.The integral must be negative because of the squared terms

B.The integral is nonnegative because

f\geq0

throughout

R

✅

C.The integral is zero because the region is circular

D.The integral cannot be compared with the area of

R

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Inside the circle $x^2+y^2\leq4$ , we have $4-x^2-y^2\geq0$ . Thus the function contributes no negative values anywhere in the region, so its double integral must be nonnegative. If the region has positive area inside the circle, the integral is actually positive.

Q12. A square region $R=[-1,1]\times[-1,1]$ is used to compare two methods for evaluating $\iint_R (x^2+y^2+xy)\,dA$ . One student integrates directly; another removes the $xy$ term using symmetry and then integrates. Which assessment is correct?

A.Only direct integration is valid

B.Only symmetry is valid

C.Both methods are valid, and symmetry simplifies the

xy

term ✅

D.Neither method works because the integrand is not separable

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The square is symmetric under $x\mapsto-x$ and $y\mapsto-y$ . The term $xy$ changes sign under either reflection, so its integral over the square is zero. The remaining $x^2+y^2$ terms can then be integrated directly, making the symmetry-based method efficient while preserving correctness.

🔗 Related Topics (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 📝 Properties of Double Integrals MCQs

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