How to find the volume of a part of sphere from $z=0.5r$ in spherical coordinates? 
Given sphere  $x^2+y^2+z^2=a^2$ find the volume above $z=0.5a, a>0$. The solution must use spherical coordinates.

It looks like the radius is not constant as it depends on the angle $\phi$ so $0.5a\le r\le\frac{0.5a}{\cos\phi}$
I think regarding the ranges:
$$
0\le\theta\le2\pi\\
0\le\phi\le\pi/3\quad\text{because }\cos\phi=\frac{0.5a}{a}=1/2
$$
Because we're using spherical coordinates we also need to multiply by the Jacobian:
$$
\iiint r^2\sin\phi
$$
Finally the computation:
$$
\iint\sin\phi\bigg[\frac{r^3}{3}\bigg]_{0.5a}^{\frac{0.5a}{\cos\phi}}=\iint\frac{a^3\sin\phi}{24\cos^3\phi}-\iint \frac{a^3\sin\phi}{24}
$$
We can solve the integrals separately:
$$
\iint\frac{a^3\sin\phi}{24\cos^3\phi}\quad\text{using substitution}\quad u=\sin\phi\\
\iint\frac{a^3\sin\phi}{24\cos^3\phi}=\iint-\frac{a^3}{24u^3}=\frac{a^3}{96}\int\bigg[\frac{1}{\cos^4\phi}\bigg]_0^{\pi/3}=\frac{a^3}{96}\int 15=\frac{15a^3}{96}\cdot 2\pi
$$
I'm really not sure about the ranges I chose and the result seems not correct to me.

Using spherical coordinates, the radius $r$ should be $\frac{a}{2 \cos \phi}<r<a$ and indeed the angle $0<\phi<\pi/3$. So you get 
$$
2 \pi \int_0^{\pi/3}\sin\phi\bigg[\frac{r^3}{3}\bigg]_{\frac{a}{2 \cos \phi}}^{a} \rm d \phi 
= 2 \pi \int_0^{\pi/3}\frac{a^3\sin\phi}{3} \rm d \phi -2 \pi \int_0^{\pi/3} \frac{a^3\sin\phi}{24 \cos^3 \phi} \rm d \phi \\
= 2 \pi \frac{a^3}{3} \bigg[ 1 - \cos\pi/3 -\frac{1}{16} ( \cos^{-2} \phi)_0^{\pi/3}  \bigg] \\
= 2 \pi \frac{a^3}{3} \bigg[ 8/16 -\frac{1}{16} (4-1 )\bigg]\\
= 2 \pi \frac{a^3}{3} \frac{5}{16} = \pi {a^3} \frac{5}{24}
$$
Compare this to the standard result (which uses cylindrical coordinates) which is (see my comment above):  
$$
V = \dfrac{\pi (a/2)^2}{3} (3a -a/2 ) = \dfrac{5 \pi a^3}{24} 
$$
Here you go!
