AP Calculus BC Quiz: Area Bounded By Two Polar Curves

What this quiz covers

This quiz focuses on Area Bounded By Two Polar Curves, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]\, d\theta$
2. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]\, d\theta$
3. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]\, d\theta$
4. $\dfrac{1}{2}\displaystyle\int_{\pi/4}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]\, d\theta$ (correct answer)
5. $\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]\, d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves. The curves $r=2\sin\theta$ and $r=2\cos\theta$ intersect at $\theta=\pi/4$ (and origin) in the first quadrant. In $\pi/4$ to $\pi/2$, $r=2\sin\theta > r=2\cos\theta$, outer is $2\sin\theta$. The area between them in this interval is $\frac{1}{2} \int_{\pi/4}^{\pi/2} [(2\sin\theta)^2 - (2\cos\theta)^2] \, d\theta$. A tempting distractor is choice B, which uses 0 to $\pi/4$ with $\sin$ - $\cos$, negative. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 2

1. $\displaystyle \frac12\int_{0}^{\pi}\big[(4\sin\theta)^2-2^2\big]d\theta$
2. $\displaystyle \frac12\int_{\pi/6}^{5\pi/6}\big[(4\sin\theta)^2-2^2\big]d\theta$ (correct answer)
3. $\displaystyle \frac12\int_{0}^{\pi}\big[2^2-(4\sin\theta)^2\big]d\theta$
4. $\displaystyle \frac12\int_{\pi/6}^{5\pi/6}\big[2^2-(4\sin\theta)^2\big]d\theta$
5. $\displaystyle \frac12\int_{0}^{\pi/2}\big[(4\sin\theta)^2-2^2\big]d\theta$

Explanation: This problem requires finding the area between the circle r = 4sin(θ) and the horizontal line r = 2. To find intersections, set 4sin(θ) = 2, giving sin(θ) = 1/2, so θ = π/6 and θ = 5π/6. The circle r = 4sin(θ) exists only for 0 ≤ θ ≤ π (where sin(θ) ≥ 0), and it's outside r = 2 when 4sin(θ) > 2, which is true for π/6 < θ < 5π/6. The area integral is (1/2)∫_{π/6}^{5π/6}[(4sin(θ))² - 2²]dθ. Choice A uses 0 to π, which would include regions where the circle is inside r = 2, giving an incorrect area. For polar areas between curves, carefully determine which curve is farther from the origin in your integration interval.

Question 3

1. $\displaystyle \frac12\int_{0}^{2\pi}\big(2\sin(2\theta)\big)^2d\theta$
2. $\displaystyle \frac12\int_{0}^{\pi}\big(2\sin(2\theta)\big)^2d\theta$
3. $\displaystyle \frac12\int_{0}^{\pi/4}\big(2\sin(2\theta)\big)^2d\theta$
4. $\displaystyle \frac12\int_{0}^{\pi/2}\big(2\sin(2\theta)\big)^2d\theta$ (correct answer)
5. $\displaystyle \frac12\int_{-\pi/4}^{\pi/4}\big(2\sin(2\theta)\big)^2d\theta$

Explanation: This problem involves finding the area of one loop of the four-petaled rose r = 2sin(2θ). The curve equals zero when sin(2θ) = 0, which occurs when 2θ = 0, π, 2π, 3π, 4π... or θ = 0, π/2, π, 3π/2, 2π... One complete petal occurs between consecutive zeros, such as from θ = 0 to θ = π/2. The area of one loop is (1/2)∫_0^{π/2}[2sin(2θ)]²dθ. Choice B incorrectly uses 0 to π, which would include two complete petals instead of one. For rose curves r = asin(nθ) or r = acos(nθ), each petal spans an angle of π/n, so always divide the period by the coefficient of θ inside the trig function.

Question 4

1. $\dfrac12\int_{-\pi/3}^{\pi/3}\left(1^2-(2\cos\theta)^2\right)d\theta$
2. $\dfrac12\int_{-\pi/3}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$ (correct answer)
3. $\dfrac12\int_{0}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$
4. $\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-1^2\right)d\theta$
5. $\int_{-\pi/3}^{\pi/3}\left(2\cos\theta-1\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2cosθ and r=1. To set up the integral, first find the intersection points by solving 2cosθ = 1, which gives θ = ±π/3. Between -π/3 and π/3, r=2cosθ is greater than r=1, making it the outer curve. Thus, the area is (1/2) ∫_{-π/3}^{π/3} [(2cosθ)^2 - 1^2] dθ. A tempting distractor is choice A, which subtracts in the wrong order, resulting in a negative integrand and an incorrect negative area. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 5

1. $\dfrac12\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big]\,d\theta$ (correct answer)
2. $\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(4\sin\theta)^2\big]\,d\theta$
3. $\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(4\sin\theta)^2-(2\sin\theta)^2\big]\,d\theta$
4. $\dfrac12\displaystyle\int_{0}^{2\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big]\,d\theta$
5. $\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)-(2\sin\theta)\big]d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves. The curves r=4sinθ and r=2sinθ are both defined for θ in 0 to π, with the larger always outside the smaller since 4sinθ >2sinθ when sinθ >0. They touch at θ=0 and θ=π, where r=0. The area between them is (1/2) ∫_0^π [(4sinθ)^2 - (2sinθ)^2] dθ. A tempting distractor is choice D, which integrates over 0 to 2π, but this would double the area since the curves trace the same path twice. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 6

1. $\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-(4\cos\theta)^2\right)d\theta$
2. $\dfrac12\int_{0}^{\pi}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
3. $\dfrac12\int_{-\pi/2}^{\pi/2}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$ (correct answer)
4. $\dfrac12\int_{-\pi/4}^{\pi/4}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
5. $\int_{-\pi/2}^{\pi/2}\left(4\cos\theta-2\cos\theta\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=4cosθ and r=2cosθ. To set up the integral, note they meet at the pole for θ=±π/2, with r=4cosθ larger where defined. The region spans -π/2 to π/2, with r=4cosθ as outer. The area is (1/2) ∫_{-π/2}^{π/2} [(4cosθ)^2 - (2cosθ)^2] dθ. A tempting distractor is choice A, which reverses the order, leading to a negative area value. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 7

1. $\dfrac12\int_{-\pi/2}^{\pi/2}\left(1^2-(1+\cos\theta)^2\right)d\theta$
2. $\dfrac12\int_{0}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
3. $\dfrac12\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1^2\right)d\theta$ (correct answer)
4. $\dfrac12\int_{-\pi}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
5. $\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)-1\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=1+cosθ and r=1. To set up the integral, determine intersections at 1+cosθ = 1, so cosθ = 0 and θ = ±π/2. Between -π/2 and π/2, r=1+cosθ is greater than or equal to r=1, making it the outer curve. The area is (1/2) ∫_{-π/2}^{π/2} [(1+cosθ)^2 - 1^2] dθ. A tempting distractor is choice A, which reverses the radii, yielding a negative integrand where the curves overlap. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 8

1. $\dfrac12\displaystyle\int_{0}^{2\pi}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
2. $\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$ (correct answer)
3. $\dfrac12\displaystyle\int_{0}^{\pi}\big[(1-\cos\theta)^2-(1+\cos\theta)^2\big]d\theta$
4. $\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
5. $\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)-(1-\cos\theta)\big]d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves. The curves r=1+cosθ and r=1-cosθ intersect at θ = ±π/2. Between -π/2 to π/2, r=1+cosθ is outer and r=1-cosθ is inner. The area is (1/2) ∫_{-π/2}^{π/2} [(1+cosθ)^2 - (1-cosθ)^2] dθ. A tempting distractor is choice C, which reverses the order, resulting in negative area. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 9

1. $\displaystyle \frac12\int_{0}^{2\pi}\big[(3-\cos\theta)^2-1^2\big]d\theta$ (correct answer)
2. $\displaystyle \frac12\int_{0}^{\pi}\big[(3-\cos\theta)^2-1^2\big]d\theta$
3. $\displaystyle \frac12\int_{0}^{2\pi}\big[1^2-(3-\cos\theta)^2\big]d\theta$
4. $\displaystyle \frac12\int_{0}^{\pi}\big[1^2-(3-\cos\theta)^2\big]d\theta$
5. $\displaystyle \frac12\int_{-\pi/2}^{\pi/2}\big[(3-\cos\theta)^2-1^2\big]d\theta$

Explanation: This problem asks for the area between the limaçon r = 3 - cos(θ) and the circle r = 1. To find intersections, set 3 - cos(θ) = 1, giving cos(θ) = 2, which has no real solutions since |cos(θ)| ≤ 1. This means one curve is always outside the other. Since the minimum value of r = 3 - cos(θ) is 2 (when cos(θ) = 1), the limaçon is always outside r = 1. The area between them over a full rotation is (1/2)∫_0^{2π}[(3-cos(θ))² - 1²]dθ. Choice B incorrectly uses 0 to π, which would only give half the enclosed area. When curves don't intersect, integrate over the full period of the outer curve to capture the entire enclosed region.

Question 10

1. $\displaystyle \frac12\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
2. $\displaystyle \frac12\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
3. $\displaystyle \frac12\int_{0}^{\pi/2}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
4. $\displaystyle \frac12\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$ (correct answer)
5. $\displaystyle \frac12\int_{-\pi/4}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$

Explanation: This problem requires finding the area between the circles r = 2cos(θ) and r = 2sin(θ). These curves intersect when 2cos(θ) = 2sin(θ), giving tan(θ) = 1, so θ = π/4 (and θ = 5π/4, but that's outside both curves' domains). The circle r = 2cos(θ) exists for -π/2 ≤ θ ≤ π/2, while r = 2sin(θ) exists for 0 ≤ θ ≤ π. For 0 < θ < π/4, cos(θ) > sin(θ), so r = 2cos(θ) is outer. The area of the enclosed region is (1/2)∫_0^{π/4}[(2cos(θ))² - (2sin(θ))²]dθ. Choice A incorrectly uses 0 to π/2 as bounds, including regions where r = 2sin(θ) would be outer. For polar circles centered on axes, their intersection occurs where the angle bisects the quadrant.

Question 11

1. $\dfrac12\displaystyle\int_{0}^{2\pi}\big[(2+2\cos\theta)^2-2^2\big]\,d\theta$
2. $\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[2^2-(2+2\cos\theta)^2\big]\,d\theta$
3. $\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(2+2\cos\theta)^2-2^2\big]\,d\theta$ (correct answer)
4. $\dfrac12\displaystyle\int_{0}^{\pi}\big[(2+2\cos\theta)^2-2^2\big]\,d\theta$
5. $\displaystyle\int_{-\pi/2}^{\pi/2}\big[(2+2\cos\theta)-2\big]\,d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves. The curves r=2+2cosθ and r=2 intersect at θ = -π/2 and θ = π/2, where r=2. Between these angles, the cardioid r=2+2cosθ is outside the circle r=2, as cosθ ≥0 in this interval, making r≥2. The area is therefore (1/2) ∫_{-π/2}^{π/2} [(2+2cosθ)^2 - 2^2] dθ. A tempting distractor is choice A, which integrates over 0 to 2π, but this includes regions where the cardioid is inside the circle, leading to incorrect area. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 12

1. $\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
2. $\dfrac12\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$ (correct answer)
3. $\dfrac12\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
4. $\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
5. $\dfrac12\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves. The curves r=2cosθ and r=2sinθ intersect at θ=π/4 (and origin). In 0 to π/4, r=2cosθ > r=2sinθ, so outer is 2cosθ. The area between them in this interval is (1/2) ∫_0^{π/4} [(2cosθ)^2 - (2sinθ)^2] dθ. A tempting distractor is choice C, which reverses the order, giving negative area. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 13

1. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[2\sin(2\theta)\big]^2\,d\theta$
2. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[2\sin(2\theta)\big]^2\,d\theta$ (correct answer)
3. $\dfrac{1}{2}\displaystyle\int_{-\pi/4}^{\pi/4}\big[2\sin(2\theta)\big]^2\,d\theta$
4. $\dfrac{1}{2}\displaystyle\int_{0}^{\pi}\big[2\sin(2\theta)\big]^2\,d\theta$
5. $\displaystyle\int_{0}^{\pi/4}2\sin(2\theta)\,d\theta$

Explanation: This problem tests the skill of finding the area bounded by two polar curves, here the rose and the origin. The petal is between $\theta=0$ and $\theta=\pi/4$, where $r=2\sin(2\theta) \geq 0$. Since inner is $r=0$, the area is $\frac{1}{2} \int_0^{\pi/4} [2\sin(2\theta)]^2 \, d\theta$. There are no bounds issues as it's a single petal. A tempting distractor is choice A, which integrates to $\pi/2$, covering two petals. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.

Question 14

1. $\dfrac12\int_{0}^{\pi/2}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
2. $\dfrac12\int_{0}^{\pi/4}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$ (correct answer)
3. $\dfrac12\int_{\pi/4}^{\pi/2}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
4. $\dfrac12\int_{0}^{\pi/4}\left((2\sin\theta)^2-(2\cos\theta)^2\right)d\theta$
5. $\int_{0}^{\pi/4}\left(2\cos\theta-2\sin\theta\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2cosθ and r=2sinθ in the first quadrant. To set up the integral, intersections occur at θ=π/4, with r=2cosθ > r=2sinθ from 0 to π/4. Thus, r=2cosθ is the outer curve in this interval. The area is (1/2) ∫_0^{π/4} [(2cosθ)^2 - (2sinθ)^2] dθ. A tempting distractor is choice D, which reverses the radii, producing a negative integrand. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 15

1. $\dfrac12\int_{\pi/6}^{5\pi/6}\left((2\sin\theta)^2-1^2\right)d\theta$ (correct answer)
2. $\dfrac12\int_{0}^{\pi}\left((2\sin\theta)^2-1^2\right)d\theta$
3. $\dfrac12\int_{\pi/6}^{5\pi/6}\left(1^2-(2\sin\theta)^2\right)d\theta$
4. $\dfrac12\int_{-\pi/6}^{\pi/6}\left((2\sin\theta)^2-1^2\right)d\theta$
5. $\int_{\pi/6}^{5\pi/6}\left(2\sin\theta-1\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2sinθ and r=1. To set up the integral, solve 2sinθ = 1 for intersection points, yielding θ = π/6 and θ = 5π/6. Between π/6 and 5π/6, r=2sinθ exceeds r=1, so it serves as the outer curve. The area is therefore (1/2) ∫_{π/6}^{5π/6} [(2sinθ)^2 - 1^2] dθ. A tempting distractor is choice C, which reverses the subtraction, leading to a negative value that does not represent the area. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 16

1. $\dfrac12\int_{0}^{\pi/2}\left((2\sin\theta)^2-(2\cos\theta)^2\right)d\theta$
2. $\dfrac12\int_{0}^{\pi/4}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$ (correct answer)
3. $\dfrac12\int_{0}^{\pi/4}\left((2\sin\theta)^2-(2\cos\theta)^2\right)d\theta$
4. $\dfrac12\int_{0}^{\pi/2}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
5. $\int_{0}^{\pi/4}\left(2\sin\theta-2\cos\theta\right)d\theta$

Explanation: This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2sinθ and r=2cosθ. To set up the integral, find intersections at θ=π/4 (and others), focusing on the first quadrant lens. From 0 to π/4, r=2cosθ > r=2sinθ, so it is outer. The area is (1/2) ∫_0^{π/4} [(2cosθ)^2 - (2sinθ)^2] dθ. A tempting distractor is choice C, which swaps the radii, resulting in a negative integrand for that interval. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.

Question 17

1. $\frac12\displaystyle\int_{0}^{\pi}\big((1+\sin\theta)^2-(3\sin\theta)^2\big)\,d\theta$
2. $\frac12\displaystyle\int_{0}^{\pi}\big((3\sin\theta)^2-(1+\sin\theta)^2\big)\,d\theta$
3. $\frac12\displaystyle\int_{0}^{\pi/2}\big((3\sin\theta)^2-(1+\sin\theta)^2\big)\,d\theta$
4. $\frac12\displaystyle\int_{\pi/6}^{5\pi/6}\big((3\sin\theta)^2-(1+\sin\theta)^2\big)\,d\theta$ (correct answer)
5. $\frac12\displaystyle\int_{\pi/6}^{5\pi/6}\big((1+\sin\theta)^2-(3\sin\theta)^2\big)\,d\theta$

Explanation: This problem requires finding the area enclosed by r = 3sin θ (a circle) and r = 1 + sin θ (a cardioid), which involves calculating the area between two polar curves. To find intersections, set 3sin θ = 1 + sin θ, giving 2sin θ = 1, so sin θ = 1/2, which yields θ = π/6 and θ = 5π/6. Between these bounds, we need to determine which curve is outer: at θ = π/2, r = 3sin(π/2) = 3 while r = 1 + sin(π/2) = 2, so 3sin θ is the outer curve. The area is (1/2)∫_{π/6}^{5π/6}[(3sin θ)² - (1 + sin θ)²]dθ. Choice B incorrectly uses bounds from 0 to π, which would include regions where the curves don't enclose area together. For polar area problems, always solve for intersection points algebraically and verify which curve is outer within those bounds.

Question 18

1. $\frac12\displaystyle\int_{0}^{\pi/2}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta$
2. $\frac12\displaystyle\int_{0}^{\pi/4}(2\sin\theta)^2\,d\theta+\frac12\displaystyle\int_{\pi/4}^{\pi/2}(2\cos\theta)^2\,d\theta$ (correct answer)
3. $\frac12\displaystyle\int_{0}^{\pi/4}(2\cos\theta)^2\,d\theta+\frac12\displaystyle\int_{\pi/4}^{\pi/2}(2\sin\theta)^2\,d\theta$
4. $\frac12\displaystyle\int_{0}^{\pi/4}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta$
5. $\frac12\displaystyle\int_{0}^{\pi/2}(2\cos\theta)^2\,d\theta$

Explanation: This problem involves finding the area of the region common to both r = 2cos θ and r = 2sin θ, which requires finding the area between two polar curves. These circles intersect where 2cos θ = 2sin θ, giving tan θ = 1, so θ = π/4. For 0 ≤ θ ≤ π/4, cos θ ≥ sin θ, so r = 2cos θ gives the boundary; for π/4 ≤ θ ≤ π/2, sin θ ≥ cos θ, so r = 2sin θ gives the boundary. The common region's area is (1/2)∫0^{π/4}(2cos θ)²dθ + (1/2)∫{π/4}^{π/2}(2sin θ)²dθ. Choice C incorrectly tries to subtract the curves throughout, missing that we want the intersection region, not the region between curves. For intersection regions, integrate each curve over its respective domain where it forms the boundary.

Question 19

1. $\frac12\displaystyle\int_{0}^{\pi}\big((4\sin\theta)^2-(2\sin\theta)^2\big)\,d\theta$ (correct answer)
2. $\frac12\displaystyle\int_{-\pi/2}^{\pi/2}\big((4\sin\theta)^2-(2\sin\theta)^2\big)\,d\theta$
3. $\frac12\displaystyle\int_{0}^{2\pi}\big((4\sin\theta)^2-(2\sin\theta)^2\big)\,d\theta$
4. $\frac12\displaystyle\int_{0}^{\pi}\big((2\sin\theta)^2-(4\sin\theta)^2\big)\,d\theta$
5. $\frac12\displaystyle\int_{0}^{\pi/2}\big((4\sin\theta)^2-(2\sin\theta)^2\big)\,d\theta$

Explanation: This problem asks for the area enclosed by r = 4sin θ and r = 2sin θ, which are two circles passing through the origin, requiring the area between two polar curves. Since both curves have the form r = asin θ, they trace out circles as θ goes from 0 to π (negative r values for π to 2π would retrace the same circle). Throughout this interval, 4sin θ ≥ 2sin θ since sin θ ≥ 0 for 0 ≤ θ ≤ π, making r = 4sin θ the outer curve. The area between them is (1/2)∫_0^π[(4sin θ)² - (2sin θ)²]dθ. Choice E incorrectly uses bounds from 0 to π/2, which would only give half the enclosed area. For polar curves of the form r = asin θ or r = acos θ, remember they complete their full shape over a π interval, not 2π.

Question 20

1. $\frac12\displaystyle\int_{-\pi/2}^{\pi/2}\big((4\cos\theta)^2-2^2\big)\,d\theta$
2. $\frac12\displaystyle\int_{-\pi/3}^{\pi/3}\big((4\cos\theta)^2-2^2\big)\,d\theta$ (correct answer)
3. $\frac12\displaystyle\int_{-\pi/3}^{\pi/3}\big(2^2-(4\cos\theta)^2\big)\,d\theta$
4. $\frac12\displaystyle\int_{0}^{\pi}\big((4\cos\theta)^2-2^2\big)\,d\theta$
5. $\frac12\displaystyle\int_{0}^{\pi/3}\big((4\cos\theta)^2-2^2\big)\,d\theta$

Explanation: This problem asks for the area of the region enclosed by the circle r = 2 and the circle r = 4cos θ, which requires finding the area between two polar curves. To find the intersection points, set 2 = 4cos θ, giving cos θ = 1/2, so θ = ±π/3. The region exists where 4cos θ ≥ 2 (the outer curve must be farther from the origin), which occurs for -π/3 ≤ θ ≤ π/3. The area formula is (1/2)∫[(outer)² - (inner)²]dθ = (1/2)∫_{-π/3}^{π/3}[(4cos θ)² - 2²]dθ. Choice A incorrectly uses bounds of ±π/2, which would include regions where 4cos θ < 2. When setting up polar area integrals, always verify which curve is outer by checking specific angle values within your bounds.