Answer
$\Phi = \frac{192}{5}\pi$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = (x^3 - e^y)\mathbf{i} + (y^3 + \sin z)\mathbf{j} + (z^3 - xy)\mathbf{k}\) across the hemisphere \(z = 4 - x^2 - y^2\) is given by \[ \Phi = \iint \mathbf{F} \cdot \mathbf{n} \, dS = \iiint \nabla \cdot \mathbf{F} \, dV \] where \(\mathbf{G}\) is the solid bounded by the hemisphere. Using spherical coordinates, \[ G: \left\{ (r, 0, \phi) \, \mid \, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \frac{\pi}{2}, 0 \leq r \leq 2 \right\} \] and taking into account that \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^3 - e^y) + \frac{\partial}{\partial y}(y^3 + \sin z) + \frac{\partial}{\partial z}(z^3 - xy) = 3(x^2 + y^2 + z^2) = 3r^2 \] the flux is given by \[ \Phi = \int_0^{2\pi} \int_0^{\pi/2} \int_0^2 3r^2 (r^2 \sin \phi) \, dr \, d\phi \, d\theta = 3\int_0^{2\pi} \int_0^{\pi/2} \sin \phi \, d\phi \int_0^2 r^4 \, dr = 3(2\pi - 0)(1 - 0)\left(\frac{2^5}{5} - 0\right) = \frac{192}{5}\pi \] Finally, \[ \text{flux} = \Phi = \frac{192}{5}\pi \]
