Question

POLAR..... Find area inside the circle r(\(\theta\))=2,
but outside the cardiod r(\(\theta\))=2-Cos(\(\theta\)).

Answer 1

https://www.desmos.com/calculator/azsqobis1k

Answer 2

I need someone to start me up, please.

Answer 3

@CallMeKiki

Answer 4

Sorry. I can't remember how to do this.

Answer 5

@wolf1728 can you help?

Answer 6

We know these are the intersection points because
\[2=2-2\cos\theta~~\implies~~\cos\theta=0\]
Notice that \(\dfrac{3\pi}{2}\) is an angle equivalent to \(-\dfrac{\pi}{2}\), so I'll use this instead. (If you actually meant the second curve to be \(2-\cos\theta\), like you wrote in your original question, the intersection points are the same.)

Now we set up the integral. I'll use the more general double integral first:
\[A=\iint_D dA=\int_{-\pi/2}^{\pi/2}\int_{2-2\cos\theta}^{2}r\,dr\,d\theta\]
Integrating with respect to \(r\) gives you the formula you're probably more familiar with:
\[\begin{align*}A&=\int_{-\pi/2}^{\pi/2}\left(\frac{1}{2}\left[r^2\right]_{2-2\cos\theta}^{2}\right)\,d\theta\\\\
&=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(2^2-(2-2\cos\theta)^2\right)\,d\theta
\end{align*}\]
Notice that due to the symmetry of both curves, we can rewrite as
\[A=2\times\frac{1}{2}\int_{0}^{\pi/2}\left(2^2-(2-2\cos\theta)^2\right)\,d\theta\]