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$r = 2\cos(\theta)$ and $ r = 1$

I went ahead and tried it and my answer was just $2\pi$. I was wondering if someone could check if I got it right, and if I didn't, tell me what I did wrong?

Integral:

$\int_{0}^{2\pi} 2{\cos(\theta)}^{2} d\theta$

I use the power reduction rule

simplified down to $ 1 + cos(2\theta) d\theta$

$\int_{0}^{2\pi}( \theta +\frac{\sin{2\theta}}{2}) d\theta$

plugging in $2\pi$ and substracting $0$ when its plugged in, I ended up with my answer of $2\pi$.

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  • $\begingroup$ No, you don't have it right. Have you drawn a picture. Can you see why it very clearly must be less than $\pi$? $\endgroup$
    – Doug M
    Jul 9 '16 at 0:32
  • $\begingroup$ Its a half circle with a diameter of 2pi. So its area is probably pi? $\endgroup$
    – dmscs
    Jul 9 '16 at 1:21
  • $\begingroup$ $r = 1$ is a circle of radius 1 and has area $\pi. r = 2\cos\theta$ is also a circle of radius 1. The area overlapped by both circles must be less than the area of each circle. $\endgroup$
    – Doug M
    Jul 9 '16 at 1:46
  • $\begingroup$ So what's wrong with the integral then? Because the one I am using gives me 2pi as the answer, symbolab agrees with me. So the integral must be wrong. Here's a screenshot: i.gyazo.com/8078966cfadf2639df404afc1e041b1a.png $\endgroup$
    – dmscs
    Jul 9 '16 at 1:49
  • $\begingroup$ Sketch the curve, and tell me if 0 to 2pi are the correct limits of integration? $\endgroup$
    – Doug M
    Jul 9 '16 at 1:56
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Here is a picture of your region of integration.enter image description here

the area = $2 (\frac 12 \int_0^{\pi/3} d\theta + \frac 12 \int_{\pi/3}^{\pi/2} (2cos\theta)^2 d\theta)$

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  • $\begingroup$ Why do you multiply it all by 2? And what is the reasoning for ∫ from 0 to pi/3 dρ, $\endgroup$
    – dmscs
    Jul 9 '16 at 2:31
  • $\begingroup$ The two curves intersect at $\pi/3$ and $-\pi/3$ To get the area between the curves we need the area inside one curve, and then the area inside the other curve. Why is it doubled. I thought it was simpler to look at the area above the x axis and double it, then have 3 regions to integrate. $\endgroup$
    – Doug M
    Jul 9 '16 at 2:46
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$$f(\theta)=1$$ $$g(\theta)=2\cos(\theta)$$

We are trying to compute the following:

$$2\int_{0}^{\frac{\pi}{3}}\frac{1}{2}\left(f(\theta)\right)^2d\theta + 2\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{1}{2}\left(g(\theta)\right)^2d\theta$$

$$2\int_{0}^{\frac{\pi}{3}}\frac{1}{2}d\theta + 2\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}2\cos^2(\theta)d\theta$$ $$\frac{\pi}{3}+\sin(\pi)+\pi-{\sin\left(\frac{2\pi}{3}\right)-\frac{2\pi}{3}}$$ $$\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$$

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