Find the area of the region that lies inside the circle $r = 1$ and outside the cardioid $r = 1-cos(\theta)$
I drew the graph and set it up like this:
$$ \int_0^\pi \frac{1}{2} [ (1)^2 – (1-cos(\theta))^2 ] d\theta $$
Squaring this out, we get:
$$ \frac{1}{2} \int_0^\pi 1 – 1 + cos^2(\theta) – 2 cos(\theta) $$
Ones cancel leaving us with (replace $cos^2(\theta)$ with $\frac{1+cos(2\theta)}{2}$:
$$ \frac{1}{2} \int_0^\pi \frac{1+cos(2\theta)}{2} – 2 cos(\theta) $$
This gives me
$$ \frac{1}{2} [\frac{\theta}{2} + \frac{sin(2\theta)}{4} – 2sin(\theta)] |_0^\pi$$
Ultimately giving me
$$ \frac{\theta}{4} + \frac{sin(2\theta)}{8} – sin(\theta)] |_0^\pi$$
Is this correct?
Best Answer
There are some errors both in the setup and in the algebra. For the setup, the limits of integration are wrong: the region inside the circle and outside the cardioid lies in the first and fourth quadrants, so the correct range of angles is from $-\frac{\pi}2$ to $\frac{\pi}2$. However, the region is symmetric about the $x$-axis, so we might as well just calculate the area of the half of the region that’s in the first quadrant and double the result to get the area of the whole region. If we do, we get
$$\begin{align*} A&=2\int_0^{\pi/2}\frac12\Big(1^2-(1-\cos\theta)^2\Big)d\theta\\\\ &=\int_0^{\pi/2}\Big(1-\big(1-2\cos\theta+\cos^2\theta\big)\Big)d\theta\\\\ &=\int_0^{\pi/2}\left(2\cos\theta-\cos^2\theta\right)d\theta\;. \end{align*}$$
If you compare that last integrand with yours, you’ll see that you made a sign error in clearing the inner parentheses. Your use of the half angle formula is correct, as are your antiderivatives. The only other problem that I see is a small notational one: in two of your three integrals you’re missing the $d\theta$ that I consider a required part of the notation.