Integral $\int_{0}^{1} dx \int_{- \sqrt{x- x^2 }}^{\sqrt{x- x^2}}\sqrt{1- x^2 - y^2} dy $ I am having problems with integration.
I have a function $f(x, y) = \sqrt{1- x^2 - y^2}$
And I need to integrate it on a circle $(x- \frac{1}{2} )^2 + y^2 = \frac{1}{4}$ 
I couldn't get far with usual coordinates:
$$ 
\int_{0}^{1} dx \int_{- \sqrt{x-  x^2 }}^{\sqrt{x- x^2}}\sqrt{1- x^2 - y^2} dy
$$
So I tried with polar coordinates.
$$
x= r \cdot \cos\phi   \\\
y= r \cdot \sin \phi    \\\
\phi \in (-\frac{\pi}{2}, \frac{\pi}{2})   \\\
r \in (0, \cos\phi)    \\\
\int_{-\pi / 2}^{\pi / 2} d\phi \int_{- \sqrt{x-  x^2 }}^{\sqrt{x- x^2}} \sqrt{1- r^2} r dr  
$$ 
I tried with a new variable $u = 1- r^2$,
$du= -2r dr$.
Then I got. 
$$ 
\int_{-\frac{\pi}{ 2}}^{\frac{\pi}{2}}  \frac{-1}{3} ((1-\cos{\phi}^2)^{3/2}-1) d\phi =   
\\
\int_{-\pi / 2}^{\pi / 2}  \frac{-1}{3} (\sin{\phi}^3 - 1) d\phi =   
\\
\frac{-1}{3} \bigg( \int_{-\pi / 2}^{\pi / 2} \sin{\phi}^3 d\phi - \int_{-\pi / 2}^{\pi / 2} d\phi\bigg) =   \\
0 + \frac{-1}{3} (\pi / 2 + \pi /2)
$$
But it's not right. $\sin$ function is odd and on a symmetric interval its integral should be zero, right?

Near the end of your work, it should be $(1-\cos{\phi}^2)^{3/2}=|\sin{\phi}|$ (with the absolute value!) and therefore 
$$\frac{-1}{3} \bigg( \int_{-\pi / 2}^{\pi / 2} |\sin{\phi}|^3 d\phi - \int_{-\pi / 2}^{\pi / 2} d\phi\bigg)=-\frac{2}{3} \int_{0}^{\pi / 2} \sin^3{\phi}\, d\phi +\frac{\pi}{3}=-\frac{4}{9}+\frac{\pi}{3}.$$
