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!