Calculus 2 Quiz: Area Between Two Polar Curves
2 questions · exam conditions
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Area Between Two Polar CurvesQuestion 1 of 2

The area enclosed by one loop of r2=4cos⁡(2θ)r^2 = 4\cos(2\theta) is:

11
22
π\pi
44
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Calculus 2 Quiz

Calculus 2 Quiz: Area Between Two Polar Curves

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

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

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Question 1

The area enclosed by one loop of r2=4cos⁡(2θ)r^2 = 4\cos(2\theta) is:

  1. 11
  2. 22 (correct answer)
  3. π\pi
  4. 44
Explanation: The lemniscate r2=4cos⁡(2θ)r^2 = 4\cos(2\theta) exists when cos⁡(2θ)≥0\cos(2\theta) \geq 0, which occurs for −π/4≤θ≤π/4-\pi/4 \leq \theta \leq \pi/4 and 3π/4≤θ≤5π/43\pi/4 \leq \theta \leq 5\pi/4. One loop corresponds to −π/4≤θ≤π/4-\pi/4 \leq \theta \leq \pi/4 where r=2cos⁡(2θ)r = 2\sqrt{\cos(2\theta)}. The area of one loop is 12∫−π/4π/4r2dθ=12∫−π/4π/44cos⁡(2θ)dθ=2∫−π/4π/4cos⁡(2θ)dθ=2[12sin⁡(2θ)]−π/4π/4=[sin⁡(2θ)]−π/4π/4=sin⁡(π/2)−sin⁡(−π/2)=1−(−1)=2\frac{1}{2}\int_{-\pi/4}^{\pi/4}r^2 d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}4\cos(2\theta)d\theta = 2\int_{-\pi/4}^{\pi/4}\cos(2\theta)d\theta = 2[\frac{1}{2}\sin(2\theta)]_{-\pi/4}^{\pi/4} = [\sin(2\theta)]_{-\pi/4}^{\pi/4} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2.

Question 2

Find the area of the region inside r=3r = 3 but outside r=3−3cos⁡θr = 3 - 3\cos\theta.

  1. 9π2+274\frac{9\pi}{2} + \frac{27}{4}
  2. 9π2−274\frac{9\pi}{2} - \frac{27}{4} (correct answer)
  3. 9π4+278\frac{9\pi}{4} + \frac{27}{8}
  4. 9π4−278\frac{9\pi}{4} - \frac{27}{8}
Explanation: The cardioid r=3−3cos⁡θ=3(1−cos⁡θ)r = 3 - 3\cos\theta = 3(1 - \cos\theta) intersects the circle r=3r = 3 when 3=3−3cos⁡θ3 = 3 - 3\cos\theta, giving cos⁡θ=0\cos\theta = 0, so θ=π/2,3π/2\theta = \pi/2, 3\pi/2. The area inside the circle but outside the cardioid is 12∫π/23π/2[32−(3−3cos⁡θ)2]dθ=12∫π/23π/2[9−9(1−cos⁡θ)2]dθ=92∫π/23π/2[1−(1−2cos⁡θ+cos⁡2θ)]dθ=92∫π/23π/2(2cos⁡θ−cos⁡2θ)dθ\frac{1}{2}\int_{\pi/2}^{3\pi/2}[3^2 - (3-3\cos\theta)^2]d\theta = \frac{1}{2}\int_{\pi/2}^{3\pi/2}[9 - 9(1-\cos\theta)^2]d\theta = \frac{9}{2}\int_{\pi/2}^{3\pi/2}[1 - (1-2\cos\theta+\cos^2\theta)]d\theta = \frac{9}{2}\int_{\pi/2}^{3\pi/2}(2\cos\theta - \cos^2\theta)d\theta. Evaluating this integral gives 9π2−274\frac{9\pi}{2} - \frac{27}{4}.