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📝 Set up limits of integration (13 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 13 questions available

←Previous Topic:Type I and Type II regions double integrals📝 Set up limits of integrationReversing order of integration:Next Topic→

What is Set up limits of integration?

Definition:
To set up limits, sketch the region, identify boundary curves, and determine whether vertical strips (Type I) or horizontal strips (Type II) simplify the description. The outer limits are constants, while inner limits are functions.

Example:
For the circle $x^2+y^2 \le 1$ , using Type I: $x$ goes from -1 to 1, and $y$ goes from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$ .

Reason:
Correctly setting limits is crucial because incorrect bounds will integrate over the wrong area, leading to erroneous results in volume, mass, or probability calculations.

Easy

Medium

Hard

📝 All Set up limits of integration MCQs

Q1. For the region $R$ enclosed by $y=x^2$ , $y=2x$ , and the $x$ -axis, which iterated integral correctly represents $\iint_R f(x,y)\,dA$ using the order $dy\,dx$ ?

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

✅

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

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

\int_0^1\int_{x^2}^{2x} f(x,y)\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The curves $y=x^2$ and $y=2x$ intersect at $x=0$ and $x=2$ . For every $x$ between these values, the region lies between the parabola below and the line above. Therefore $y$ runs from $x^2$ to $2x$ , while $x$ runs from $0$ to $2$ .

Q2. A region is described by $0\le y\le 4$ and $y/2\le x\le \sqrt y$ . A student claims that the limits can be written as $0\le x\le2$ and $x^2\le y\le2x$ . Which statement best evaluates the student's setup?

A.It is correct because both descriptions use the same boundary curves ✅

B.It is incorrect because

x

must begin at

y/2

and end at

\sqrt y

C.It is correct only for

0\le x\le1

D.It is incorrect because the correct

y

-limits are

2x\le y\le x^2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student's description is equivalent to the original region after solving the inequalities for $y$ . From $x\ge y/2$ , we obtain $y\le2x$ , and from $x\le\sqrt y$ , we obtain $y\ge x^2$ . The curves intersect at $x=0$ and $x=2$ , so the proposed setup is valid.

Q3. A designer models a plate by the region between $y=x^2$ and $y=4$ . If vertical slices are used, which limits correctly describe the region without splitting it?

0\le x\le4,\ x^2\le y\le4

-2\le x\le2,\ x^2\le y\le4

✅

-4\le x\le4,\ 4\le y\le x^2

-2\le x\le2,\ 4\le y\le x^2

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The parabola $y=x^2$ meets $y=4$ when $x=\pm2$ . For every $x$ in that interval, the parabola is the lower boundary and the horizontal line is the upper boundary. Thus the region is completely captured by $-2\le x\le2$ and $x^2\le y\le4$ .

Q4. Consider the triangular region bounded by $x=0$ , $y=x+1$ , and $y=5-x$ . Which setup using horizontal slices represents the region with a single pair of iterated limits?

\int_1^5\int_0^{y-1}f(x,y)\,dx\,dy

\int_1^5\int_{y-1}^{5-y}f(x,y)\,dx\,dy

\int_1^3\int_0^{y-1}f(x,y)\,dx\,dy+\int_3^5\int_0^{5-y}f(x,y)\,dx\,dy

✅

\int_1^5\int_0^{5-y}f(x,y)\,dx\,dy

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The two slanted boundaries meet at $x=2,\ y=3$ , while the vertical boundary is $x=0$ . For $1\le y\le3$ , the right boundary is $x=y-1$ . For $3\le y\le5$ , it is $x=5-y$ . Because the right boundary changes formula, the horizontal description requires two integrals.

Q5. A farmer's irrigation plot is bounded by $y=x$ , $y=4-x$ , and $y=0$ . To calculate a quantity depending on both coordinates, the farmer wants vertical slices. What is the correct setup?

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

\int_0^2\int_0^x f(x,y)\,dy\,dx+\int_2^4\int_0^{4-x}f(x,y)\,dy\,dx

✅

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

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

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The two lines intersect at $x=2,\ y=2$ . From $x=0$ to $x=2$ , the upper boundary is $y=x$ . From $x=2$ to $x=4$ , the upper boundary becomes $y=4-x$ . Since the lower boundary remains $y=0$ , the vertical description must be split at $x=2$ .

Q6. A region is bounded by $y=x^2$ and $y=2-x$ . A student writes $\int_{-2}^{1}\int_{2-x}^{x^2}f(x,y)\,dy\,dx$ . What is the most important error?

A.The

x

-interval is incorrect

B.The lower and upper

y

-boundaries are reversed ✅

C.The region cannot be described using vertical slices

D.The integrand must be multiplied by

x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The curves intersect where $x^2=2-x$ , giving $x=-2$ and $x=1$ . Thus the student's $x$ -interval is correct. Between those intersections, the line $y=2-x$ lies above the parabola $y=x^2$ , so the lower limit must be $x^2$ and the upper limit must be $2-x$ .

Q7. Suppose $R$ is enclosed by $x=y^2$ and $x=2y$ . Which description is most efficient if the inner integral is with respect to $x$ ?

0\le y\le2,\ y^2\le x\le2y

✅

-2\le y\le2,\ 2y\le x\le y^2

0\le x\le4,\ x/2\le y\le\sqrt x

-2\le y\le2,\ y^2\le x\le2y

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because both boundaries are already expressed as $x$ in terms of $y$ , horizontal slices are natural. The curves meet when $y^2=2y$ , giving $y=0$ and $y=2$ . For this interval, $y^2$ is the left boundary and $2y$ is the right boundary.

Q8. A region is bounded by $y=x$ , $y=2x$ , and $x=3$ . A student argues that the limits $\int_0^6\int_{y/2}^{y}f(x,y)\,dx\,dy$ describe the entire region. Which conclusion is correct?

A.The setup is correct

B.The

y

-range should be

0\le y\le3

C.The setup covers the region only for

0\le y\le3

, so it must be split or corrected ✅

D.The inner limits should be

y\le x\le2y

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For horizontal slices, the inequalities $y=x$ and $y=2x$ give $x=y$ and $x=y/2$ . However, the vertical boundary $x=3$ changes which slanted line determines the right endpoint. For $0\le y\le3$ , the right endpoint is $x=y$ ; for $3\le y\le6$ , it is $x=3$ .

Q9. A graph shows a region lying above $y=x^2$ , below $y=2$ , and between $x=-1$ and $x=1$ . Which expression correctly represents the region using vertical slices?

\int_{-1}^{1}\int_2^{x^2}f(x,y)\,dy\,dx

\int_{-1}^{1}\int_{x^2}^{2}f(x,y)\,dy\,dx

✅

\int_0^1\int_{x^2}^{2}f(x,y)\,dy\,dx

\int_{-1}^{1}\int_{-2}^{x^2}f(x,y)\,dy\,dx

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The graph indicates that the parabola $y=x^2$ forms the lower boundary while the horizontal line $y=2$ forms the upper boundary. The stated vertical extent is from $x=-1$ to $x=1$ . Therefore the inner limits are $x^2$ to $2$ , producing the second expression.

Q10. A region is bounded by $y=x^2$ , $y=4$ , and $x=0$ . A student chooses $0\le y\le4$ and $0\le x\le\sqrt y$ . Another student chooses $0\le x\le2$ and $x^2\le y\le4$ . What is the best interpretation?

A.Only the first description is valid

B.Only the second description is valid

C.Both descriptions represent the same region, using different slice directions ✅

D.They represent different regions because one uses a square root

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The first description uses horizontal slices: solving $y=x^2$ for the right boundary gives $x=\sqrt y$ , with $0\le y\le4$ . The second uses vertical slices directly. Both trace exactly the same region, demonstrating that a region may have multiple valid iterated-integral descriptions.

Q11. An engineer needs to integrate over the region bounded by $y=x^2$ , $y=4-x^2$ . Which approach is preferable if the goal is to avoid splitting the integral?

A.Use vertical slices because the curves are both written as functions of

x

B.Use horizontal slices because solving for

x

produces symmetric left and right boundaries ✅

C.Either direction necessarily requires two integrals

D.Use only the positive-

x

half because the region is symmetric

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The curves intersect when $x^2=4-x^2$ , giving $x=\pm\sqrt2$ . Vertical slices would require determining which curve is above, but they can still work. Horizontal slices are especially efficient because $x=\pm\sqrt{(y/2)}$ , giving symmetric left and right boundaries across one continuous $y$ -range.

Q12. A region is defined by $0\le x\le1$ , $x\le y\le2-x$ . Which setup correctly computes $\iint_R (x+y)\,dA$ without changing the region?

\int_0^1\int_x^{2-x}(x+y)\,dy\,dx

✅

\int_0^1\int_{2-x}^{x}(x+y)\,dy\,dx

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

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

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The region is explicitly given in vertical-slice form: $x$ ranges from $0$ to $1$ , while $y$ starts at the lower line $y=x$ and ends at the upper line $y=2-x$ . Therefore the first integral reproduces the region exactly and preserves the given integrand.

Q13. Consider the region bounded by $y=x^2$ and $y=2|x|$ . An analyst wants a single integral with horizontal slices. Which setup is correct?

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

✅

\int_0^4\int_{-\sqrt y}^{\sqrt y}f(x,y)\,dx\,dy

\int_0^4\int_{-y/2}^{\sqrt y}f(x,y)\,dx\,dy

\int_{-2}^{2}\int_{x^2}^{2|x|}f(x,y)\,dy\,dx

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The boundaries intersect when $x^2=2|x|$ , giving $|x|=2$ and $y=4$ , while the lowest point is at $y=0$ . For a fixed $y$ , the inequality $y\le2|x|$ gives $|x|\ge y/2$ , while being above $x^2$ gives $|x|\le\sqrt y$ . This actually produces two separated horizontal intervals, so option A is not valid. The correct conclusion is that no listed single-integral setup represents the entire region.

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