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Area of a Polar Region - Calculus 2

  1. 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?
    Answer: 6π6\pi 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.
  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?
    Answer: π\pi 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.
  3. What is the area of the region bounded by the polar curve r=θr = \theta for 0≤θ≤2π0 \leq \theta \leq 2\pi?
    Answer: 4π33\frac{4\pi^3}{3} 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.
  4. The area enclosed by one loop of the polar curve r=3sin⁡(2θ)r = 3\sin(2\theta) is:
    Answer: 9π8\frac{9\pi}{8} 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.

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