# Help calculating area inside circle and outside cardioid

I've calculated the area inside the circle $r=3a\cos(\theta)$ and outside the cardioid $r=a(1+\cos(\theta))$ and I got two answers: $a^2\pi + a^2\frac{\sqrt{3}}{2}$ and the answer: $a^2 \pi$. Can someone please help? Because I'm not quite sure what the correct answer is.

If $$r_1(\theta) = 3a \cos \theta$$ is the equation of the circle, and $$r_2\theta) = a(1+\cos \theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(\theta) = r_2(\theta)$, or $3a \cos \theta = a(1+\cos \theta)$, or $$2 \cos \theta = 1.$$ This gives us the intersection points $$\theta = \pm \frac{\pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $\theta \in [-\pi/3, \pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = \int_{\theta = -\pi/3}^{\pi/3} \frac{1}{2}\left(r_1^2(\theta) - r_2^2(\theta)\right) \, d\theta.$$ Note this is not the same as $$\int_{\theta = -\pi/3}^{\pi/2} \frac{1}{2}\left(r_1(\theta) - r_2(\theta)\right)^2 \, d\theta.$$ The second integral is incorrect for the area enclosed.
• @user332320 No, it is not correct. $$r_1^2(\theta) - r_2^2(\theta) = 9a^2 \cos^2 \theta - a^2 (1+\cos\theta)^2 = a^2 (8 \cos^2 \theta - 2 \cos \theta - 1).$$ The antiderivative is $$a^2 (2 \sin 2\theta - 2 \sin \theta + 3\theta).$$ Evaluating that at $\theta = \pi/3$ gives $a^2 \pi$. Commented Apr 18, 2016 at 18:33