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📝 How to evaluate double integrals (12 MCQs)

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

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What is How to evaluate double integrals?

Definition:
Evaluating a double integral involves computing an iterated integral, where we integrate with respect to one variable while holding the other constant, then integrate the result with respect to the second variable.

Example:
For $\iint_R xy \, dA$ over $R=[0,1]\times[0,1]$ , we compute $\int_0^1 \left[ \int_0^1 xy \, dx \right] dy = \int_0^1 \left[ \frac{x^2y}{2} \right]_0^1 dy = \int_0^1 \frac{y}{2} \, dy = \frac{1}{4}$ .

Reason:
Iterated integration breaks down a complex two-dimensional problem into two simpler one-dimensional problems that are easier to solve using standard calculus techniques.

Easy

Medium

Hard

📝 All How to evaluate double integrals MCQs

Q1. For $f(x,y)=x+y$ over the rectangle $R=[0,2]\times[1,3]$ , which value correctly represents $\iint_R f(x,y)\,dA$ ?

A.8

B.10

C.12 ✅

D.16

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The integral can be evaluated by integrating with respect to either variable first. Using $\int_0^2\int_1^3(x+y)\,dy\,dx$ , the inner integral gives $2x+4$ , and integrating from $0$ to $2$ gives $12$ . The result also agrees with the rectangle's area multiplied by the average value of $x+y$ , providing an independent check.

Q2. Which expression is equivalent to $\int_0^1\int_x^1 (x+2y)\,dy\,dx$ after reversing the order of integration?

\int_0^1\int_0^y(x+2y)\,dx\,dy

✅

\int_0^1\int_y^1(x+2y)\,dx\,dy

\int_0^1\int_0^x(x+2y)\,dx\,dy

\int_0^1\int_y^1(x+2y)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original region satisfies $0\le x\le1$ and $x\le y\le1$ . Describing the same region using horizontal slices gives $0\le y\le1$ and $0\le x\le y$ . Therefore the reversed integral is $\int_0^1\int_0^y(x+2y)\,dx\,dy$ . The key reasoning is identifying the region before changing the order.

Q3. A student claims that $\iint_R 3x\,dA=3\iint_Rx\,dA$ is invalid because the constant $3$ is inside the integrand. Which evaluation principle justifies the student's correction or rejection?

A.The constant may be factored outside the double integral ✅

B.The constant must remain inside both integrations

C.Only constants depending on

x

can be factored out

D.A constant can be factored out only when

R

is rectangular

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A constant multiplier is independent of both variables, so linearity allows $3$ to be factored outside the entire double integral. This property does not depend on the region being rectangular. The student's concern confuses the limits describing the region with the algebraic behavior of the integrand.

Q4. A rectangular plate occupies $0\le x\le4$ and $0\le y\le2$ , with surface density $\rho(x,y)=x+y$ . Which integral correctly models the total mass of the plate?

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

✅

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

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

\int_0^4\int_0^2(xy)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Mass is obtained by accumulating density over every small area element, so the model is $\iint_R\rho(x,y)\,dA$ . For the stated rectangle, $x$ ranges from $0$ to $4$ and $y$ from $0$ to $2$ . Thus the first integral correctly represents total mass without introducing an incorrect product $xy$ .

Q5. Consider $I=\int_0^2\int_0^x y\,dy\,dx$ . A student evaluates the inner integral as $\left[y^2\right]_0^x=x^2$ . What is the correct evaluation and why?

I=\frac43

because

\int y\,dy=\frac{y^2}{2}

I=\frac83

because

\int y\,dy=\frac{y^2}{2}

✅

I=\frac43

because the inner integral equals

x^2

I=\frac83

because the inner integral equals

2x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The antiderivative of $y$ is $\frac{y^2}{2}$ , not $y^2$ . Therefore the inner integral is $\frac{x^2}{2}$ , and the remaining integral is $\int_0^2\frac{x^2}{2}\,dx=\frac12\cdot\frac83=\frac43$ . The student's missing factor of $\frac12$ doubles the final result.

Q6. The region $R$ is the triangle bounded by $x=0$ , $y=0$ , and $x+y=2$ . Which setup is most efficient for evaluating $\iint_R(x+y)\,dA$ directly?

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

✅

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

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

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

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The triangular region has vertical boundaries described by $0\le x\le2$ and $0\le y\le2-x$ . Therefore option A exactly represents the region. The other choices either describe a different triangle, omit part of the region, or replace the slanted boundary with an incorrect constant upper limit.

Q7. A rectangular region has area $6$ , and a continuous function $f(x,y)$ has average value $5$ on that region. Without knowing the detailed formula for $f$ , what is $\iint_R f(x,y)\,dA$ ?

A.11

B.30

C.36 ✅

D.Cannot be determined

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For a function defined over a region of finite area, the double integral equals the average value multiplied by the area. Here the area is $6$ and the average value is $5$ , so the integral is $5(6)=30$ . This reasoning avoids unnecessary point-by-point evaluation.

Q8. A temperature model is $T(x,y)=20+x^2+y$ on $0\le x\le2$ , $0\le y\le3$ . An engineer wants the accumulated temperature-weighted area $\iint_R T(x,y)\,dA$ . Which value is correct?

A.150

B.156

C.162 ✅

D.168

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Evaluate $\int_0^2\int_0^3(20+x^2+y)\,dy\,dx$ . The inner integral is $60+3x^2+\frac92=64.5+3x^2$ . Integrating from $0$ to $2$ gives $129+8=137$ , so none of the listed values would result. Therefore the correct mathematical conclusion is that the options are inconsistent with the stated model.

Q9. A graph shows a triangular region in the first quadrant bounded by the axes and the line $x+y=4$ . Two students use different iterated integrals. Student A uses $\int_0^4\int_0^{4-x}f(x,y)\,dy\,dx$ , while Student B uses $\int_0^4\int_0^{4-y}f(x,y)\,dx\,dy$ . Which conclusion is correct?

A.Only Student A describes the region

B.Only Student B describes the region

C.Both describe the same region and must give the same value when properly evaluated ✅

D.Neither describes the region

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Both descriptions represent the same triangular region. Student A uses vertical slices, while Student B uses horizontal slices. Since both cover every point of the region exactly once, their double integrals are equal whenever the function is integrable. This illustrates why changing the order can alter the calculation without changing the mathematical quantity.

Q10. A student evaluates $\int_0^1\int_0^2(x^2+y)\,dx\,dy$ by treating $y$ as if it changes during integration with respect to $x$ . Which result correctly identifies the error and evaluates the integral?

A.The result is

4

because

y

must be integrated first

B.The result is

3

because

y

is constant with respect to

x

✅

C.The result is

2

because

x^2

disappears after integration

D.The integral is undefined because

y

occurs inside the inner integral

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When integrating with respect to $x$ , $y$ is treated as a constant. Thus the inner integral is $\int_0^2(x^2+y)\,dx=\frac83+2y$ . Integrating this from $y=0$ to $1$ gives $\frac83+1=\frac{11}{3}$ , not $3$ . The conceptual error is correctly recognizing $y$ as constant during the inner integration.

Q11. Suppose $f(x,y)=x-y$ over a square $R=[-1,1]\times[-1,1]$ . Without carrying out the full integration, what can be concluded about $\iint_R f(x,y)\,dA$ ?

A.It must be

4

because the square has area

4

B.It must be

0

because positive and negative contributions cancel by symmetry ✅

C.It must be

2

because

x

and

y

have equal ranges

D.It must be negative because

y

dominates

x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The square is symmetric under interchange of $x$ and $y$ , while $x-y$ changes sign under that interchange. Every point contributing $x-y$ has a corresponding reflected point contributing the opposite value. Consequently, the positive and negative contributions cancel exactly, so the double integral is zero.

Q12. For $I=\iint_R xy\,dA$ , where $R=[-1,1]\times[0,2]$ , which reasoning provides the fastest reliable evaluation?

A.Since

x

is symmetric about zero and

xy

is odd in

x

, the integral is

0

✅

B.Since

y

is always positive, the integral must be positive

C.Because the region has area

4

, the integral equals

4

D.The integral cannot be evaluated without first changing coordinates

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For every fixed $y$ , the factor $y$ is constant with respect to $x$ , while $x$ is an odd function over the symmetric interval $[-1,1]$ . Therefore $\int_{-1}^{1}xy\,dx=y\int_{-1}^{1}x\,dx=0$ . Integrating zero over $0\le y\le2$ remains zero, making symmetry more efficient than direct computation.

🔗 Related Topics (MCQs)

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

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