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?


You are calculating the area in the following graph enter image description here

and your calculation is correct.

But the question asks for the area "inside" both graphs which can be seen below: enter image description here

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.

  • $\begingroup$ Could you please tell me what program can draw these graph? $\endgroup$ – user143993 Dec 18 '14 at 8:25
  • $\begingroup$ @ user143993: PolarPlot function of Mathematica. $\endgroup$ – Math-fun Dec 18 '14 at 8:31
  • $\begingroup$ Mathematica. Thanks a lot. $\endgroup$ – user143993 Dec 18 '14 at 10:01

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.