Answer
$ \Phi = 180\pi$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = x^3\mathbf{i} + y^3\mathbf{j} + z^3\mathbf{k}\) across the surface of the cylindrical solid bounded by the cylinder \(x^2 + y^2 = 4\) and the planes \(z = 0\), \(z = 4\) is given by \[ \Phi = \iint \mathbf{F} \cdot \mathbf{n} \, dS = \iiint \nabla \cdot \mathbf{F} \, dV \] where \[ G: \left\{ (x, y, z) \, \mid \, -2 \leq x \leq 2, -4 - x^2 \leq y \leq 4 - x^2, 0 \leq z \leq 3 \right\} \] Using cylindrical coordinates, this can be expressed as \[ G: \left\{ (r, \theta, z) \, \mid \, 0 \leq \theta \leq 2\pi, 0 \leq r \leq 2, 0 \leq z \leq 3 \right\} \] and taking into account that \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(z^3) = 3(x^2 + y^2 + z^2) = 3(r^2 + z^2) \] the flux result is \[ \Phi = \int_0^{2\pi} \int_0^2 \int_0^3 3(r^2 + z^2) \, r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^2 (9r^3 + 27r) \, dr \, d\theta = \int_0^{2\pi} \left(\frac{4}{9}r^4 + \frac{2}{27}r^2\right)\Bigg|_0^2 \, d\theta = (2\pi - 0)(90) = 180\pi \] Finally, \[ \text{flux} = \Phi = 180\pi \]
