Question

find the area of the region that is inside the cardioid r=4+4cos(theta) and outside the circle r=6

Answer 1

I would first recommend finding where the cardioid and circle intersect. For what \(\theta\) do you have \(4+4\cos\theta = 6\)?

Answer 2

+/- pi/3

Answer 3

Alright, first thing you want to note is that the region is symmetric about the polar axis; so it's much easier to find the area defined between \(0\leq \theta\leq \dfrac{\pi}{3}\) and double it.

The larger radial value over the interval \(0\leq \theta\leq \dfrac{\pi}{3}\) is \(r=4+4\cos\theta\) and the smaller radial value is \(r=6\). Hence the area enclosed by this region is given by
\[\begin{aligned}A &= 2\left(\dfrac{1}{2}\int_0^{\pi/3} (4+4\cos\theta)^2 - 6^2\,d\theta\right)\\ &= \int_0^{\pi/3} 16+32\cos\theta+16\cos^2\theta - 36\,d\theta\\ &= \int_0^{\pi/3}-20 + 32\cos\theta+16\cos^2\theta\,d\theta \\ &= \int_0^{\pi/3} -20 + 32\cos\theta + 8+8\cos(2\theta)\,d\theta \quad\left(\text{using }\cos^2\theta = \dfrac{1}{2}(1+\cos(2\theta))\right)\\ &= \int_0^{\pi/3}-12 + 32\cos\theta +8\cos(2\theta)\,d\theta \end{aligned}\]Can you take things from here? :-)

Answer 4

yes i can thank you very much!