The problem: Consider the circle $r = (\sqrt{2} – 1)\cos(\theta)$ and the cardioid $r = 1 – \cos(\theta)$. Find the area $A$ of the region $R$ common to the inside of the circle and the inside of the cardioid.
I am confused about solving this. Can someone in a not too technical/difficult way explain how this would be solved?
Thank you
Best Answer
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(\sqrt2-1)\cos\theta=1-\cos\theta$ which holds when $\theta=\pm\pi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $\pm45$ degree angles away from the origin. Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $\theta\in[0,\pi/4]$ together with the part of the circle in the interval $\theta\in[\pi/4,\pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$ A=\frac12\int_0^{\pi/4}(1-\cos \theta)^2\,d\theta+\frac12\int_{\pi/4}^{\pi/2}(\sqrt2-1)^2\cos^2\theta\,d\theta. $$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $\theta\in[-\pi/4,\pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$ \frac12\int_{-\pi/4}^{\pi/4}\left((\sqrt2-1)^2\cos^2\theta-(1-\cos\theta)^2\right)\,d\theta$$ instead. $$
Moral: Do not forget to draw the picture.