I'm having some trouble with the setup for this one.
I tried $\int^{2\pi}_0\int^{\pi/2}_0\int^2_0\rho^2sin(\phi)d\rho d\phi d\theta$ , and got $\frac{16\pi}{3}$, confirmed in symbolab. but I seen that since I just have a z and not $z^2$ I can't get $\rho$ the way I was thinking. Some one else gave me $0<=\rho<=cos(\phi)$, and $\pi/4$ for $\phi$. So I set it up as
$2\pi \int _0^{\frac{\pi }{4}}\:\int _0^{cos\left(\theta \right)}\:p^2sin\left(\theta \right)dp\:d\theta $
And then checked again in symbolab before trying going through the eval. But that gave me $\pi/8$, where the answer should be $8\pi$.
$$ z=4-x^2-y^2\quad \Rightarrow \quad \rho\cos\phi = 4- \rho^2\sin^2\phi $$ Solving for $\rho \ge 0$ yields $$ \rho = \frac{-\cos\phi+\sqrt{1+15\sin^2\phi}}{2\sin^2\phi} $$ So the bounds are $$ 0 \le \theta \le 2\pi, \quad 0 \le \rho \le \frac{-\cos\phi+\sqrt{1+15\sin^2\phi}}{2\sin^2\phi}, \quad 0 \le \phi \le \frac{\pi}{2} $$ You can verify that $$ \int_0^{2\pi}\int_{0}^{\pi/2}\int_0^{\frac{-\cos\phi+\sqrt{1+15\sin^2\phi}}{2\sin^2\phi}} \rho^2\sin\phi\; d\rho d\phi d\theta = 8\pi $$
Note The final computation is not trivial. If you have the choice, you are better of with cylindrical coordinates.