Calculus 2 Quiz: Area Of A Polar Region
4 questions · exam conditions
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Area Of A Polar RegionQuestion 1 of 4

What is the area of the region bounded by the polar curve r=θr = \theta for 0≤θ≤2π0 \leq \theta \leq 2\pi?

4π33\frac{4\pi^3}{3}
8π33\frac{8\pi^3}{3}
2π32\pi^3
4π34\pi^3
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Calculus 2 Quiz

Calculus 2 Quiz: Area Of A Polar Region

Practice Area Of A Polar Region in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Area Of A Polar Region, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

What is the area of the region bounded by the polar curve r=θr = \theta for 0≤θ≤2π0 \leq \theta \leq 2\pi?

  1. 4π33\frac{4\pi^3}{3} (correct answer)
  2. 8π33\frac{8\pi^3}{3}
  3. 2π32\pi^3
  4. 4π34\pi^3
Explanation: The spiral r=θr = \theta for 0≤θ≤2π0 \leq \theta \leq 2\pi forms an Archimedean spiral. The area is given by 12∫02πr2dθ=12∫02πθ2dθ=12[θ33]02π=12⋅(2π)33=12⋅8π33=4π33\frac{1}{2}\int_0^{2\pi} r^2 d\theta = \frac{1}{2}\int_0^{2\pi} \theta^2 d\theta = \frac{1}{2}[\frac{\theta^3}{3}]_0^{2\pi} = \frac{1}{2} \cdot \frac{(2\pi)^3}{3} = \frac{1}{2} \cdot \frac{8\pi^3}{3} = \frac{4\pi^3}{3}. Choice B omits the 12\frac{1}{2} factor in the polar area formula. Choice C uses an incorrect antiderivative. Choice D results from forgetting the 13\frac{1}{3} factor in the antiderivative of θ2\theta^2.

Question 2

The polar curve r=2cos⁡(3θ)r = 2\cos(3\theta) forms a three-petaled rose. What is the total area enclosed by all three petals?

  1. π\pi (correct answer)
  2. 3π2\frac{3\pi}{2}
  3. 2π2\pi
  4. 3π3\pi
Explanation: For r=2cos⁡(3θ)r = 2\cos(3\theta), the curve traces out 3 complete petals as θ\theta goes from 0 to π\pi. Each petal has the same area due to symmetry. One petal is traced from θ=0\theta = 0 to θ=π3\theta = \frac{\pi}{3}. The area of one petal is 12∫0π/3(2cos⁡(3θ))2dθ=12∫0π/34cos⁡2(3θ)dθ=2∫0π/31+cos⁡(6θ)2dθ=∫0π/3[1+cos⁡(6θ)]dθ=[θ+sin⁡(6θ)6]0π/3=π3+sin⁡(2π)6−0−0=π3\frac{1}{2}\int_0^{\pi/3}(2\cos(3\theta))^2 d\theta = \frac{1}{2}\int_0^{\pi/3}4\cos^2(3\theta) d\theta = 2\int_0^{\pi/3}\frac{1+\cos(6\theta)}{2} d\theta = \int_0^{\pi/3}[1+\cos(6\theta)] d\theta = [\theta + \frac{\sin(6\theta)}{6}]_0^{\pi/3} = \frac{\pi}{3} + \frac{\sin(2\pi)}{6} - 0 - 0 = \frac{\pi}{3}. Therefore, the total area is 3×π3=π3 \times \frac{\pi}{3} = \pi. Choice B would be if we miscounted petals or made an error in integration. Choice C would be if we incorrectly used the wrong bounds. Choice D would be if we thought each petal had area π\pi.

Question 3

Consider the polar curve r=2−2cos⁡θr = 2 - 2\cos\theta. This curve is traced completely as θ\theta varies from 00 to 2π2\pi. What is the area enclosed by this curve?

  1. 4π4\pi
  2. 6π6\pi (correct answer)
  3. 8π8\pi
  4. 12π12\pi
Explanation: The curve r=2−2cos⁡θr = 2 - 2\cos\theta is a cardioid. The area is 12∫02πr2dθ=12∫02π(2−2cos⁡θ)2dθ=12∫02π[4−8cos⁡θ+4cos⁡2θ]dθ\frac{1}{2}\int_0^{2\pi} r^2 d\theta = \frac{1}{2}\int_0^{2\pi} (2 - 2\cos\theta)^2 d\theta = \frac{1}{2}\int_0^{2\pi} [4 - 8\cos\theta + 4\cos^2\theta] d\theta. Using cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1+\cos(2\theta)}{2}: =12∫02π[4−8cos⁡θ+2(1+cos⁡(2θ))]dθ=12∫02π[6−8cos⁡θ+2cos⁡(2θ)]dθ=12[6θ−8sin⁡θ+sin⁡(2θ)]02π=12[12π−0+0−0+0−0]=6π= \frac{1}{2}\int_0^{2\pi} [4 - 8\cos\theta + 2(1+\cos(2\theta))] d\theta = \frac{1}{2}\int_0^{2\pi} [6 - 8\cos\theta + 2\cos(2\theta)] d\theta = \frac{1}{2}[6\theta - 8\sin\theta + \sin(2\theta)]_0^{2\pi} = \frac{1}{2}[12\pi - 0 + 0 - 0 + 0 - 0] = 6\pi. Choice A would result from using the wrong coefficient. Choice C would come from omitting the 12\frac{1}{2} factor. Choice D would result from computational errors in the integration.

Question 4

The area enclosed by one loop of the polar curve r=3sin⁡(2θ)r = 3\sin(2\theta) is:

  1. 9π8\frac{9\pi}{8} (correct answer)
  2. 9π4\frac{9\pi}{4}
  3. 94\frac{9}{4}
  4. 9π2\frac{9\pi}{2}
Explanation: The rose curve r=3sin⁡(2θ)r = 3\sin(2\theta) has 4 petals. One loop occurs from θ=0\theta = 0 to θ=π2\theta = \frac{\pi}{2}. The area of one loop is 12∫0π/2(3sin⁡(2θ))2dθ=12∫0π/29sin⁡2(2θ)dθ=92∫0π/21−cos⁡(4θ)2dθ=94∫0π/2[1−cos⁡(4θ)]dθ=94[θ−sin⁡(4θ)4]0π/2=94[π2−sin⁡(2π)4−0+0]=94⋅π2=9π8\frac{1}{2}\int_0^{\pi/2}(3\sin(2\theta))^2 d\theta = \frac{1}{2}\int_0^{\pi/2}9\sin^2(2\theta) d\theta = \frac{9}{2}\int_0^{\pi/2}\frac{1-\cos(4\theta)}{2} d\theta = \frac{9}{4}\int_0^{\pi/2}[1-\cos(4\theta)] d\theta = \frac{9}{4}[\theta - \frac{\sin(4\theta)}{4}]_0^{\pi/2} = \frac{9}{4}[\frac{\pi}{2} - \frac{\sin(2\pi)}{4} - 0 + 0] = \frac{9}{4} \cdot \frac{\pi}{2} = \frac{9\pi}{8}. Choice B gives the total area of all 4 petals. Choice C omits the π\pi factor. Choice D is twice the total area.