How to find geometrical figures areas? Find the area of the region that lies inside the circle $r = 1$
and outside the cardioid $r=1-cos\alpha$
We know that area can not be negative value(at least basic calculus). 
I wonder where I made a mistake,I tried to show equations.
$$ 1=1-cos\alpha $$
$$\alpha = \pi/2,3\pi/2$$
\begin{equation}
\frac{1}{2}\int_{3\pi/2}^{\pi/2}f(x)^2d\alpha
\end{equation}
\begin{equation}
\frac{1}{2}(\int_{\pi/2}^{3\pi/2}1^2d\alpha-\int_{\pi/2}^{3\pi/2}(1-cos\alpha)^2d\alpha)  
\end{equation}
$$ \int_{3\pi/2}^{\pi/2}1^2d\alpha = $$
$$ \frac{3\pi}{2} -\frac{\pi}{2}=\pi $$
$$ \int_{\pi/2}^{3\pi/2}(1-cos\alpha)^2d\alpha = $$
$$ 9\pi/2 - 3\pi/2 + 2 - (-2) + 0 - 0 $$
$$ 2\frac{1}{2}(\pi - 3\pi - 4)$$
$$ Answer: (-2\pi-4) $$
symmterical no required multiple by 2.
 A: You're mixing up polar and rectangular coordinates, your $dx$ should be $d\alpha$. But also the subtraction should be the other way around.
Finally, fix the computations.
You want to find the of the semicircle and subtract the area of the cardioid for $-\pi/2\le\theta\le\pi/2$. So you need
\begin{align}
\int_{-\pi/2}^{\pi/2}\frac{1^2}{2}\,d\alpha-\int_{-\pi/2}^{\pi/2}\frac{(1-\cos\alpha)^2}{2}\,d\alpha
&=\int_{-\pi/2}^{\pi/2}\frac{1-(1-\cos\alpha)^2}{2}\,d\alpha\\
&=\int_{-\pi/2}^{\pi/2}\frac{2\cos\alpha-\cos^2\alpha}{2}\,d\alpha
\end{align}
which is essentially elementary; an antiderivative is
$$
\sin\alpha-\dfrac{1}{4}(\alpha+\sin\alpha\cos\alpha)
$$
so you get
$$
\Bigl(1-\frac{\pi}{8}\Bigr)-\Bigl(-1+\frac{\pi}{8}\Bigr)=2-\frac{\pi}{4}
$$
A: 
if you look at the picture you  are asked to calculate have half a circle + the 2 times the area of the cardioid from 0 to pi/2, not  what you did! your calculation is right you subtracted the big left part of the cardiode from half the circle , no wonder it is negative.
