4
$\begingroup$

I'm trying to find the area in the curve $r^2=2\cos \theta$ and out of $r=2(1-\cos \theta)$

The intersections are at $\theta=\frac{\pi}{3}$ and $\theta=\frac{-\pi}{3}$, then, the integral to find the area is:

$$A=\frac{1}{2} \int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} (\sqrt{2 \cos{\theta}})^2-(2-2\cos{\theta})^2 d\theta=9\sqrt{3}-4\pi$$

Using the result that the area of ​​a region in polar coordinates is given by:

$$\frac{1}{2} \int_{\theta_1}^{\theta_2} (f(\theta))^2 d\theta$$

Is this correct?

Thanks for your help.

$\endgroup$
1
$\begingroup$

I agree with your integral setup for $A$, but I think you may have lost the $\frac{1}{2}$, since I get half your answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.