Finding $a$ s.t the cone $\sqrt{x^{2}+y^{2}}=za$ divides the upper half of the unit ball into two parts with the same volume My friend gave me the following question:

For which value of the parameter $a$ does the cone
  $\sqrt{x^{2}+y^{2}}=za$ divides $$\{(x,y,z):\,
 x^{2}+y^{2}+z^{2}\leq1,z\geq0\}$$ into two parts with the same volume ?

I am having some difficulties with the question.
What I did:
First, a ball with radius $R$ have the volume $\frac{4\pi}{3}R^{3}$
hence the volume of the upper half of the unit ball is $\frac{\pi}{3}$.
Secondly: I found where does the cone intersect with the boundary
of the ball:
$$
\sqrt{x^{2}+y^{2}}=za
$$
hence
$$
z=\sqrt{\frac{x^{2}+y^{2}}{a^{2}}}
$$
and 
$$
x^{2}+y^{2}+z^{2}=1
$$
setting $z$ we get 
$$
x^{2}(\frac{a^{2}+1}{a^{2}})+y^{2}(\frac{a^{2}+1}{a^{2}})=1
$$
hence 
$$
x^{2}+y^{2}=\frac{a^{2}}{a^{2}+1}
$$
using the equation for the cone we get 
$$
z=\frac{1}{\sqrt{a^{2}+1}}
$$
I then did (and I am unsure about the boundaries) : $0<z<\frac{1}{\sqrt{a^{2}+1}},0<r<az$
and using the coordinates $x=r\cos(\theta),y=r\sin(\theta),z=z$ I
got that the volume that the cone enclose in the ball is 
$$
\int_{0}^{\frac{1}{\sqrt{a^{2}+1}}}dz\int_{0}^{az}dr\int_{0}^{2\pi} r  d\theta
$$
which evaluates to 
$$
\frac{\pi a^{2}}{3(a^{2}+1)^{\frac{3}{2}}}
$$
I then required this will be equal to e the volume of the upper half
of the unit ball :
$$
\frac{\pi a^{2}}{3(a^{2}+1)^{\frac{3}{2}}}=\frac{\pi}{3}
$$
and got 
$$
a^{2}=(a^{2}+1)^{\frac{3}{2}}
$$
which have no real solution, according to WA.
Can someone please help me understand where I am wrong and how to
solve this question ?

Check your bounds again.  I believe they should be 
$$\begin{align}0<&z<\frac{1}{\sqrt{a^2+1}}\\az<&r<\sqrt{1-z^2}\end{align}$$
Finishing the integral with these bounds should yield $a=\sqrt3$.
