Pretty simple polar integration question that I've been having trouble with…
The question says it all. I identified the limits of integration by setting $1 – \cos(\theta) = \cos(\theta)$ so that $\cos(\theta) = \frac{1}{2}$ and $\theta = \pm \frac{\pi}{3}$.
I've tried the integral
$$ \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) – (1 – \cos(\theta))^2 = \sqrt{3} – \frac{\pi}{3}$$
Back of the book says the answer is $\frac{7\pi}{12} – \sqrt{3}$.
I think the polar curves are messing me up. Am I interpreting $r = \cos(\theta)$ wrong or something? Do I need to change the bounds of integration?
Best Answer
You are calculating the area in the following graph
and your calculation is correct.
But the question asks for the area "inside" both graphs which can be seen below:
This area is then $$A=\int_0^{\pi/3}(1-\cos x)^2dx+\int_{\pi/3}^{\pi/2}\cos^2xdx$$ which is what the back of your book says.