Finding the area inside the polar curve $r=1+\cos\theta$ and outside the polar curve $r=2\cos\theta$ 
Find the area of the region lying inside the polar curve $r=1+\cos\theta$ and outside the polar curve $r=2\cos\theta$.

Let
$$A_1 = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2d\theta = \frac{3\pi}{2}$$ and $$A_2=\pi(1)^2 = \pi$$ because the radius of the curve $r=2\cos\theta$ is $1$. $$A_1 - A_2 = \frac{\pi}{2}$$ which is the final answer that I got.
However, when I tried to solve the problem on Wolfram Alpha with the input $$\frac{1}{2}\int_0^{2\pi}((1+\cos\theta)^2-(2\cos\theta)^2)d\theta$$ it gave me $-\frac{\pi}{2}$ even though the curve $r=1+\cos\theta$ looks bigger than $r=2\cos\theta$.
How do I reconcile these two answers? Any help would be appreciated!
 A: Note that to find the area of $r=2\cos\theta$, you should compute $$2\int_{0}^\pi \cos^2\theta \,d\theta$$
instead of 
$$2\int_{0}^{2\pi} \cos^2\theta \,d\theta$$
The issue  is that when you plugged in your expression into Wolfram Alpha, it subtracted the area of $r=2\cos\theta$ twice, yielding $\frac{3\pi}{2}-2\pi$.
A: I see where your problem lies, in the integral around $2\cos t$ you integrate from $0$ to $2\pi$ but this goes around the circle twice, double counting the area, the right bounds should be from $0$ to $\pi$.
A: You need to calculate one integral from $0$ to $2\pi$ (or from $-\pi$ to $\pi$) and the other one from $-\pi/2$ to $\pi/2$.
The second part of your last integral $$\frac{1}{2}\color{red}{\int_0^{2\pi}}((1+\cos\theta)^2-\color{red}{(2\cos\theta)^2})\color{red}{d\theta}$$
sweeps the circle twice, hence its value is $2\pi$ instead of expected $\pi$.
As a result you effectively make a subtraction like $$\dfrac{3\pi}2-\color{red}{2}\pi = -\dfrac\pi 2$$ instead of supposed $$\dfrac{3\pi}2-\pi = \dfrac\pi 2$$
Compare the graph by Wolfram Alpha:

Source: https://www.wolframalpha.com/input?i=polar+r%3D1%2B%5Ccos+%CE%B8+%2C+polar+r%3D2%5Ccos+%CE%B8
