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📝 Double Integrals over Nonrectangular Regions (15 MCQs)

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

←Previous Topic:Properties of Double Integrals📝 Double Integrals over Nonrectangular RegionsIterated integrals with variable limits:Next Topic→

What is Double Integrals over Nonrectangular Regions?

Definition:
For nonrectangular regions, the limits of integration become functions of the other variable. Type I regions are bounded by $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$ ; Type II regions are bounded by $c \le y \le d$ and $h_1(y) \le x \le h_2(y)$ .

Example:
To integrate over the triangle bounded by $y=0, x=1, y=x$ , we use Type I: $\int_0^1 \int_0^x f(x,y) \, dy \, dx$ .

Reason:
Real-world regions are rarely perfect rectangles; adapting limits to the boundary curves allows us to apply double integrals to arbitrary shapes found in physics and engineering.

Easy

Medium

Hard

📝 All Double Integrals over Nonrectangular Regions MCQs

Q1. Let $R=\{(x,y):0\le x\le1,\ x^2\le y\le x\}$ . Which iterated integral correctly represents $\iint_R (x+y)\,dA$ when integrating with respect to $y$ first?

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

\int_0^1\int_y^{\sqrt y}(x+y)\,dx\,dy

✅

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

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

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The region lies between $y=x^2$ and $y=x$ , with $0\le x\le1$ . Reversing the order requires solving these boundaries for $x$ : $x=y$ and $x=\sqrt y$ . For $0\le y\le1$ , $y\le\sqrt y$ , so the correct bounds are $0\le y\le1$ and $y\le x\le\sqrt y$ .

Q2. A student says that for $R=\{(x,y):0\le x\le2,\ 0\le y\le x\}$ , the integral $\iint_R f(x,y)\,dA$ can be written as $\int_0^2\int_0^2 f(x,y)\,dy\,dx$ because both variables range from $0$ to $2$ . What is the fundamental flaw?

A.The integrand must be continuous

B.The region is triangular rather than rectangular, so

y

cannot independently range to

2

for every

x

✅

C.The

x

-interval must begin at

1

D.Double integrals cannot be evaluated over triangular regions

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The variables are not independent throughout the region. For each fixed $x$ , the allowable values satisfy $0\le y\le x$ , not $0\le y\le2$ . Using $0\le y\le2$ includes points outside the triangular region, specifically points above $y=x$ , and therefore changes the integral being evaluated.

Q3. A region is enclosed by $y=x$ , $y=4-x$ , and the $x$ -axis. Which description gives a correct decomposition for integrating vertically?

0\le x\le4,\ 0\le y\le\min(x,4-x)

0\le x\le2,\ 0\le y\le x

, together with

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

✅

0\le x\le2,\ 0\le y\le4-x

only

0\le x\le4,\ 0\le y\le x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The two sloping lines meet at $x=2$ , $y=2$ . To the left of $x=2$ , the upper boundary is $y=x$ ; to the right, it is $y=4-x$ . Therefore the region must be split into two vertical pieces. This illustrates why nonrectangular regions sometimes require piecewise bounds.

Q4. For $R=\{(x,y):0\le y\le1,\ y\le x\le1\}$ , a student reverses the order and writes $\int_0^1\int_0^y f(x,y)\,dx\,dy$ . Which correction is appropriate?

A.Replace the bounds by

0\le x\le1,\ 0\le y\le x

✅

B.Replace the bounds by

0\le x\le1,\ x\le y\le1

C.Replace the bounds by

0\le x\le1,\ 0\le y\le1-x

D.The student's bounds are already correct

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original region consists of points satisfying $0\le y\le1$ and $y\le x\le1$ . Describing it horizontally means $x$ ranges from $0$ to $1$ , while for each $x$ , $y$ ranges from $0$ to $x$ . Thus the reversed-order representation is $\int_0^1\int_0^x f(x,y)\,dy\,dx$ .

Q5. A rectangular plot is modified so that only the portion below the curve $y=\sqrt{x}$ , above the $x$ -axis, and between $x=0$ and $x=4$ remains. Which setup correctly models the area of the remaining plot?

\int_0^4\int_0^{\sqrt{x}}1\,dy\,dx

✅

\int_0^4\int_{\sqrt{x}}^4 1\,dy\,dx

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

\int_0^4\int_x^4 1\,dy\,dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Area can be represented by integrating the constant function $1$ over the specified region. Since the plot is below $y=\sqrt{x}$ , above $y=0$ , and $0\le x\le4$ , each vertical slice has height $\sqrt{x}$ . Therefore the correct model is $\int_0^4\int_0^{\sqrt{x}}1\,dy\,dx$ .

Q6. A sensor covers the region bounded by $y=x^2$ , $y=2x$ , and $x=0$ . The sensor's intensity is $I(x,y)=x+y$ . Which setup correctly computes the total intensity?

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

✅

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

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

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

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The curves intersect where $x^2=2x$ , giving $x=0$ and $x=2$ . Between these values, $2x$ lies above $x^2$ , so vertical slices satisfy $x^2\le y\le2x$ . Multiplying the intensity by the differential area and integrating over these bounds gives the total sensor intensity.

Q7. Consider the region $R$ enclosed by $y=x^2$ and $y=2-x$ . Which strategy is generally more efficient for integrating a function that depends mainly on $y$ , such as $f(x,y)=y^3$ ?

A.Always integrate with respect to

x

first

B.Choose the order that gives the simpler boundary description, potentially reversing the order after solving the curves for

x

✅

C.Convert every boundary into polar coordinates

D.Use rectangular bounds enclosing the entire region

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For nonrectangular regions, the most efficient order depends on how easily the boundaries can be described. Since $y^3$ is simple with respect to $x$ , integrating horizontally may be advantageous if the curves can be solved cleanly for $x$ . Choosing bounds that minimize splitting and algebraic complexity is an important modeling decision.

Q8. A region is described by $0\le x\le3$ and $x\le y\le3$ . A student claims that reversing the order gives $0\le y\le3$ and $0\le x\le3-y$ . Evaluate the student's reasoning.

A.Correct; the region is unchanged

B.Incorrect; the correct horizontal bounds are

0\le y\le3

and

0\le x\le y

✅

C.Incorrect; the correct bounds are

0\le y\le3

and

3-y\le x\le3

D.Correct only when the integrand is symmetric

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The original inequalities imply $x\le y$ , so for a fixed $y$ , $x$ begins at $0$ and ends at $y$ . The student's expression $x\le3-y$ describes a different triangular region. A reliable reversal method is to identify the actual boundary inequality rather than merely replacing variables in the original limits.

Q9. A designer models a triangular region with vertices $(0,0)$ , $(3,0)$ , and $(3,2)$ . Which vertical-slice description correctly represents the region?

0\le x\le3,\ 0\le y\le\frac{2}{3}x

✅

0\le x\le3,\ 0\le y\le2-\frac{2}{3}x

0\le x\le2,\ 0\le y\le\frac{3}{2}x

0\le x\le3,\ \frac{2}{3}x\le y\le2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The slanted edge connects $(0,0)$ to $(3,2)$ , so its equation is $y=\frac{2}{3}x$ . The region lies below this line and above the $x$ -axis, with $x$ ranging from $0$ to $3$ . Thus each vertical slice satisfies $0\le y\le\frac{2}{3}x$ .

Q10. A graph shows a region bounded above by $y=4$ , below by $y=x^2$ , and horizontally between the two intersection points. Which statement correctly identifies the horizontal limits after reversing the order?

0\le y\le4,\ -\sqrt y\le x\le\sqrt y

✅

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

0\le y\le2,\ -\sqrt y\le x\le\sqrt y

0\le y\le4,\ -4+y\le x\le4-y

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The parabola $y=x^2$ can be solved as $x=\pm\sqrt y$ . Since the region extends vertically from the parabola up to $y=4$ , every horizontal slice has $0\le y\le4$ , with $x$ ranging from the left branch $-\sqrt y$ to the right branch $\sqrt y$ .

Q11. Suppose a region is bounded by $y=x$ , $y=3x$ , and $x=2$ , with $x\ge0$ . Which description is most useful if the integrand is $f(x,y)=e^x$ ?

A.Integrate horizontally because the integrand contains

x

B.Integrate vertically because

x

is directly bounded and the

y

-bounds are simple ✅

C.Convert to polar coordinates because the boundaries are lines

D.Split the region into four rectangles

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a vertical slice, $x$ runs directly from $0$ to $2$ , while $y$ lies between $x$ and $3x$ . Since $e^x$ has an especially simple antiderivative with respect to $y$ , integrating vertically first also reduces the inner integration to multiplication by the slice height.

Q12. Let $R$ be the region between $y=x$ and $y=x^2$ for $0\le x\le1$ . Which expression gives the same integral after reversing the order, and why is the reversal valid?

\int_0^1\int_y^{\sqrt y}f(x,y)\,dx\,dy

, because the same points are described using horizontal slices ✅

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

, because the upper curve becomes the lower curve

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

, because

x^2=y

\int_0^1\int_0^y f(x,y)\,dx\,dy

, because

x=y

is one boundary

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The original region has $x^2\le y\le x$ . Solving these equations for $x$ gives $x=\sqrt y$ and $x=y$ . For $0\le y\le1$ , $y\le\sqrt y$ , so horizontal slices run from $x=y$ to $x=\sqrt y$ . Reversing the order preserves the same geometric set.

Q13. A quantity is modeled by $\iint_R (x^2+y^2)\,dA$ , where $R$ is the triangular region $0\le y\le x\le1$ . A second analyst instead integrates over $0\le x\le1,\ 0\le y\le1$ . Without performing either integral, what can be concluded?

A.Both values are equal because the integrand is symmetric

B.The second value is larger because it includes points outside

R

, where the integrand is nonnegative ✅

C.The second value is smaller because

R

has greater density

D.No comparison is possible because the regions overlap

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The square $0\le x\le1,\ 0\le y\le1$ contains the triangular region $y\le x$ as only part of it. The integrand $x^2+y^2$ is nonnegative everywhere, so adding the extra portion cannot decrease the integral. In fact, the second integral is strictly larger because the omitted triangular portion has positive area and positive integrand except at the origin.

Q14. Let $R=\{(x,y):x\ge0,\ y\ge0,\ x+y\le1\}$ . Which substitution-style reasoning correctly describes the geometry before evaluating an integral over $R$ ?

A.The region is a square because both variables are at most

1

B.The region is a right triangle, and for each

x\in[0,1]

y

ranges from

0

1-x

✅

C.The region is a semicircle of radius

1

D.The region is a rectangle with

0\le x\le1

and

0\le y\le1+x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The inequality $x+y\le1$ , together with $x,y\ge0$ , places the points below the line $x+y=1$ in the first quadrant. Its intercepts are $(1,0)$ and $(0,1)$ , producing a right triangle. A vertical slice therefore has $0\le x\le1$ and $0\le y\le1-x$ .

Q15. For $R=\{(x,y):0\le x\le1,\ x^3\le y\le x\}$ , a researcher wants to reverse the order without splitting the region. Which setup is correct?

\int_0^1\int_y^{\sqrt[3]{y}}f(x,y)\,dx\,dy

✅

\int_0^1\int_{\sqrt[3]{y}}^y f(x,y)\,dx\,dy

\int_0^1\int_{y^3}^{y}f(x,y)\,dx\,dy

\int_0^1\int_0^{\sqrt[3]{y}}f(x,y)\,dx\,dy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The boundaries $y=x^3$ and $y=x$ become $x=\sqrt[3]{y}$ and $x=y$ . For $0\le y\le1$ , $y\le\sqrt[3]{y}$ , so the horizontal slice runs from $x=y$ to $x=\sqrt[3]{y}$ . Because the two curves intersect only at $0$ and $1$ , no splitting is necessary.

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