Only Gauss can help us here

Nice to see some vector calc problems being posted. I will use the divergence theorem.

$\int \int \int_V (\nabla \cdot \vec{F}) \, dV = \int \int_S (\vec{F} \cdot \vec{n}) \, dS$

The above means that the volume integral of the vector field divergence over the interior of a closed surface is equal to the flux of the vector field over the surface. But for this, a closed surface is required, so I will add a disk of radius $2$ in the $xy$ plane to the top of the sphere, and the two of them together form a closed surface.

To evaluate the left side of the integral equation, calculate the vector field divergence:

$\nabla \cdot \vec{F} = \Big( \frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, \frac{\partial}{\partial{z}} \Big) \cdot \Big( z^2 x, \frac{y^3}{3} + \sin z, x^2 z + y^2 \Big) = z^2 + y^2 + x^2 = r^2$

The triple integral of the divergence is then:

$\int \int \int_V r^2 \, dV = \int_0^{\pi/2} \int_0^{2 \pi} \int_0^2 r^4 \sin \phi \, dr \, d \theta \, d \phi = \frac{64}{5} \pi = \Psi_{total}$

The above result is the total flux over the closed surface. Now find the flux over the bottom surface (the disk). On the disk, we have:

$\vec{F} = \Big( 0, \frac{y^3}{3} , y^2 \Big) \\ \vec{n} = (0, 0, -1) \\ \vec{F} \cdot \vec{n} = -y^2$

The flux over the disk is:

$\int \int_{disk} -y^2 \, dS = \int \int_{disk} -r^2 \sin^2 \theta \, (r \, dr \, d\theta) = \int_0^{2 \pi} \int_0^2 -r^3 \sin^2 \theta \, dr \, d\theta = - 4 \pi$

Now equate the two fluxes to the total flux:

$\Psi_{total} = \Psi_{disk} + \Psi_{half sphere} \\ \frac{64}{5} \pi = - 4 \pi + \Psi_{half sphere} \\ \frac{64}{5} \pi = - \frac{20}{5} \pi + \Psi_{half sphere} \\ \Psi_{half sphere} = \frac{84}{5} \pi$