Find the area of the region that is enclosed by:

Question

$r = 2\cos(\theta)$ 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 $2\pi$.

Answer 1

Here is a picture of your region of integration.

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

Answer 2

$$f(\theta)=1$$ $$g(\theta)=2\cos(\theta)$$

We are trying to compute the following:

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

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