This is the exercise:
Let $B$ be the region of $\mathbb{R}^3$ $$B = {(x,y,z) | x^2+y^2+z^2 \le 1,\ z^2\geq x^2+y^2}$$ and and be the field $$\mathbf{F}(x,y,z) = (x,y,z)$$
My guess was since we are dealing with a hyperbole (top part) and a sphere, use spherical coordinates for the triple integral, and for the other part of the theorem use spherical for the sphere and cylindric for the upper cone.
the triple integral, I integrate 3, which is the result of the divergence of F with $0\leq \theta\leq2\pi$;$0\leq\phi\leq1\pi/4$ and $0\leq\rho\leq1$ but it didn't work, results are weird and for the other part of the theorem the sum of the double integrals did not give me as the triple integral.
Sorry about my english, is so bad.
The divergence theorem in spherical coordinates gives $$\begin{align} \iiint_D \text{div}(\mathbf{F})dxdydz&=\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4}\int_{\rho=0}^1 3\rho^2\sin(\phi) d\rho d\phi d\theta\\ &=2\pi\cdot 3[\rho^3/3]_0^1[-\cos(\phi)]_{\phi=0}^{\pi/4}=2\pi\left(1-\frac{1}{\sqrt{2}}\right). \end{align}$$ On the other hand, the flux through the conic surface is zero because $\mathbf{n}$ is orthogonal to $\mathbf{F}$ whereas the flux through the spherical cap is $$\iint_S \mathbf{F}\cdot d\mathbf{S}=|S|=\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4}\sin(\phi) d\phi d\theta=2\pi\left(1-\frac{1}{\sqrt{2}}\right).$$ because $\mathbf{n}=\mathbf{F}$ (the sphere has radius 1).