Find the area lying inside the cardoid $r = a(1 + cos \theta)$ and outside the circle $r = a$.
Area inside the Cardoid and outside the Circle
integration
Related Question
- [Math] How to find the area inside the larger loop and outside the smaller loop of the limacon $r=\frac{1}{2} +\cos \theta$
- [Math] the area of the region that is outside the curve $r=1 – \cos\theta$ but inside the curve $r=1$
- [Math] Find area lies inside cardioid $r = 1 + \cos θ$ and outside circle $r = \cos θ$
- Find the area inside the curve $r^2=2\cos(5\theta)$ and outside the unit circle.
- Find the area outside $r = 3\cos\Theta$ and inside $r = 1 + \cos\Theta$
Best Answer
It looks like a fairly straightforward computation. Where do you have difficulty?
The first thing I would do is determine the points where the two curves intersect to get the limits of integration on $\theta$. That turns out to be simple: $r= a(1+ cos(\theta)= a$ so $1+ cos(\theta)= 1$, $cos(\theta)= 0$, $\theta= -\frac{\pi}{2}$ and $\theta= \frac{\pi}{2}$. The "differential of area" in polar coordinates is $rdrd\theta$ so the integral is $\int_{-\pi/2}^{\pi/2} \int_a^{a(1+ cos(\theta))} rdrd\theta= \int_{-\pi/2}^{\pi/2} \left[\frac{1}{2}r^2\right]_{a}^{a(1+ cos\theta))}d\theta$ $= \frac{1}{2}a^2\int_{-\pi/2}^{\pi/2} (1+ cos(\theta))^2- 1)dr= \frac{1}{2}a^2\int_{-\pi/2}^{\pi/2}2cos(\theta)+ cos^2(\theta) d\theta$.
Surely you can finish that yourself?