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📝 Double integrals in Simple Polar Regions (13 MCQs)

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

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What is Double integrals in Simple Polar Regions?

Definition:
A simple polar region is defined by $\alpha \le \theta \le \beta$ and $g_1(\theta) \le r \le g_2(\theta)$ . These regions are bounded by rays from the origin and curves defined in polar form.

Example:
The region inside the cardioid $r = 1+\cos\theta$ is described by $0 \le \theta \le 2\pi$ and $0 \le r \le 1+\cos\theta$ .

Reason:
Many natural shapes (petals, spirals, circles off-center) are naturally described in polar form, making this coordinate system essential for accurate modeling and integration.

Easy

Medium

Hard

📝 All Double integrals in Simple Polar Regions MCQs

Q1. A region in the plane is described by $0\le r\le 3$ and $0\le\theta\le\pi/2$ . Which geometric description best matches this region?

A.A semicircle of radius 3

B.A quarter-disk of radius 3 in the first quadrant ✅

C.A circular ring between radii 0 and 3

D.A triangle bounded by two radial lines and a curved side

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The condition $0\le r\le3$ includes every point from the origin outward to the circle of radius 3, while $0\le\theta\le\pi/2$ restricts the points to the first quadrant. Their intersection is therefore a quarter-disk, not merely its boundary or a triangular region.

Q2. For a region described by $1\le r\le4$ and $\pi/6\le\theta\le\pi/3$ , what feature distinguishes it from a sector containing the origin?

A.The angular interval is too small

B.The radial lower bound excludes the origin ✅

C.The radial upper bound is greater than 1

D.The angles are both positive

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A sector extending to the origin must contain $r=0$ . Here $r$ begins at 1, so all points with $0\le r<1$ are excluded. The region is therefore a sector-shaped annular region bounded by two circles and two radial rays, rather than a sector reaching the origin.

Q3. Suppose a planar region is bounded by $r=2$ , $r=5$ , $\theta=0$ , and $\theta=\pi/4$ . Which description correctly identifies its geometry?

A.A quarter-disk centered at the origin

B.A circular sector extending from the origin

C.An annular sector between two concentric circles ✅

D.A rectangular region in Cartesian coordinates

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The two equations $r=2$ and $r=5$ represent concentric circles, while the two constant-angle equations represent radial boundaries. Because the region lies between two different positive radii, it does not include the origin. Thus it is an annular sector, sometimes viewed as a sector with its central disk removed.

Q4. A designer models a flower bed using $0\le r\le2+2\cos\theta$ for $-\pi/2\le\theta\le\pi/2$ . What should be checked before using this as a simple polar description of the entire bed?

A.Whether the radial expression remains nonnegative over the stated angular interval ✅

B.Whether

r

is replaced by

x

everywhere

C.Whether

\theta

must always be measured in degrees

D.Whether the region must be rectangular

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A simple polar description with $0\le r\le f(\theta)$ requires careful attention to the sign of $f(\theta)$ . On the stated interval, $2+2\cos\theta$ is nonnegative, reaching zero at the endpoints. This makes the radial interval geometrically meaningful throughout the specified angular range.

Q5. A student claims that $0\le r\le4$ and $3\pi/4\le\theta\le5\pi/4$ describes the entire disk $x^2+y^2\le16$ . What is the student's main error?

A.The radial bound should be

r\ge4

B.The angular restriction covers only the left half of the disk ✅

C.The radius 4 corresponds to a square

D.The angular values must be positive

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The condition $0\le r\le4$ correctly describes all radii inside the circle of radius 4. However, the angular interval from $3\pi/4$ to $5\pi/4$ covers only directions centered around the negative $x$ -axis, so only the left half of the disk is included. The angle restriction is therefore the error.

Q6. A region consists of all points inside $r=6$ and between the rays $\theta=\pi/4$ and $\theta=3\pi/4$ . If its area is to be computed directly in polar coordinates, which setup is appropriate?

\int_{\pi/4}^{3\pi/4}\int_0^6 r\,dr\,d\theta

✅

\int_0^6\int_{\pi/4}^{3\pi/4} r\,d\theta\,dr

\int_{\pi/4}^{3\pi/4}\int_0^6 1\,dr\,d\theta

\int_0^{\pi}\int_0^6 r^2\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The region is a sector with radius 6 and angular bounds $\pi/4$ and $3\pi/4$ . In polar coordinates, the area element is $r\,dr\,d\theta$ . Therefore the radius should be integrated from 0 to 6, and the angle from $\pi/4$ to $3\pi/4$ , giving the first setup.

Q7. A circular irrigation zone occupies the part of the disk $r\le5$ lying between $\theta=-\pi/6$ and $\theta=\pi/3$ . Which expression gives its area?

\int_{-\pi/6}^{\pi/3}\int_0^5 r\,dr\,d\theta

✅

\int_0^5\int_{-\pi/6}^{\pi/3} r^2\,d\theta\,dr

\int_{-\pi/6}^{\pi/3}\int_0^5 dr\,d\theta

\int_0^{5}\int_{-\pi/6}^{\pi/3} r\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The irrigation zone is a sector of a circle centered at the origin. Its radial coordinate ranges from 0 to 5, while its angular coordinate ranges from $-\pi/6$ to $\pi/3$ . Since the polar area element contains the factor $r$ , the first double integral correctly models the area.

Q8. A student converts the region $2\le r\le5,\ 0\le\theta\le\pi/3$ into Cartesian coordinates and says it is bounded only by $x^2+y^2=4$ and $x^2+y^2=25$ . Why is this description incomplete?

A.The circles are not centered at the origin

B.The angular restrictions create two additional radial boundaries ✅

C.The region must also contain

r=0

D.The equations for the circles are incorrect

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The radial equations become the circles $x^2+y^2=4$ and $x^2+y^2=25$ , but the angular restrictions also matter. The conditions $\theta=0$ and $\theta=\pi/3$ correspond to two rays from the origin. Omitting those rays describes the entire annulus rather than only the specified annular sector.

Q9. A graph shows a shaded region inside the circle $r=4$ , entirely above the $x$ -axis, with its left and right boundaries lying on the rays $\theta=\pi/6$ and $\theta=5\pi/6$ . Which polar description matches the shaded region?

0\le r\le4,\ 0\le\theta\le\pi

0\le r\le4,\ \pi/6\le\theta\le5\pi/6

✅

0\le r\le4,\ -\pi/6\le\theta\le5\pi/6

0\le r\le4,\ \pi/6\le\theta\le7\pi/6

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The circle condition $r\le4$ supplies the radial boundary. The two visible straight boundaries are rays at $\pi/6$ and $5\pi/6$ , so those must be the angular limits. Because the shaded region lies between those rays and includes the origin, $r$ starts at zero.

Q10. A student evaluates $\iint_R (x^2+y^2)\,dA$ over $R=\{(r,\theta):0\le r\le3,\ 0\le\theta\le\pi/2\}$ and writes $\int_0^{\pi/2}\int_0^3 r^2\,dr\,d\theta$ . What factor is missing?

A.A factor of

r

from the polar area element ✅

B.A factor of

2

from symmetry

C.A factor of

\sin\theta

D.No factor is missing

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Since $x^2+y^2=r^2$ , the integrand becomes $r^2$ . However, changing to polar coordinates also changes the area element to $dA=r\,dr\,d\theta$ . Therefore the complete integrand is $r^3$ , and the missing factor is $r$ . Forgetting this Jacobian factor is a common modeling error.

Q11. Two methods are proposed for finding the area of the region $0\le r\le3,\ -\pi/4\le\theta\le\pi/4$ . Method A uses polar coordinates directly. Method B converts every boundary into Cartesian equations before integrating. Which conclusion is most reasonable?

A.Method B is always superior because Cartesian coordinates are more general

B.Method A is more natural because the boundaries are already constant-radius and constant-angle curves ✅

C.Both methods are invalid because the region crosses the

x

-axis

D.Method A cannot represent regions symmetric about the

x

-axis

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The region is a circular sector whose boundaries are naturally expressed as $r=3$ and constant-angle rays. Polar coordinates describe such geometry directly with simple limits, while Cartesian conversion introduces line equations and potentially more complicated bounds. Although both methods can work, the polar setup is clearly more efficient here.

Q12. Consider the region $0\le r\le2\cos\theta$ for $-\pi/2\le\theta\le\pi/2$ . A student says its area must be $\pi(2)^2$ because the maximum value of $r$ is 2. Which reasoning correctly resolves the claim?

A.The region is a full disk of radius 2, so the student's answer is correct

B.The radius changes with angle, so the region is not a full disk; its polar area integral must account for

r=2\cos\theta

✅

C.The maximum radius is irrelevant because polar regions cannot have curved boundaries

D.The area is zero because the angular interval is symmetric

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Although $r$ reaches 2 when $\theta=0$ , it decreases as $|\theta|$ increases and becomes zero at the endpoints. Thus the boundary is not a circle $r=2$ . The region is a circle of a different center and radius when converted to Cartesian form, so using $\pi(2)^2$ overestimates the area.

Q13. Let $R$ be defined by $0\le r\le4\cos\theta$ for $-\pi/2\le\theta\le\pi/2$ . A second region $S$ is defined by $0\le r\le4\sin\theta$ for $0\le\theta\le\pi$ . Which relationship between their areas is correct?

A.The area of

R

is twice the area of

S

B.The area of

S

is twice the area of

R

C.Their areas are equal because the regions are rotations of one another ✅

D.Their areas cannot be compared because their angular intervals differ

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The first region has a boundary $r=4\cos\theta$ , while the second has $r=4\sin\theta$ . These describe congruent circles of radius 2 centered on the positive $x$ -axis and positive $y$ -axis, respectively. A rotation maps one region onto the other, so their areas are equal.

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