Answer
$\Phi = \frac{\pi}{2}$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}\) across the surface of the solid bounded by \(x^2 + y^2 = z\) and the plane \(z = 1\) is given by \[ \Phi = \iint \mathbf{F} \cdot \mathbf{n} \, dS = \iiint \nabla \cdot \mathbf{F} \, dV \] where in cylindrical coordinates, \[ G: \left\{ (r, \theta, z) \, \mid \, 0 \leq \theta \leq 2\pi, 0 \leq r \leq 1, r \leq z \leq 1 \right\} \] and \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2(x + y + z) \] Then, the flux is given by \[ \Phi = \int_0^{2\pi} \int_0^1 \int_{r}^1 2r(r\cos\theta + r\sin\theta + z) \, dz \, dr \, d\theta \] \[ = \int_0^{2\pi} \int_0^1 \left[2r(1 - r)(r\cos\theta + r\sin\theta) + \frac{1}{2}(1 - r^2)\right] \, dr \, d\theta \] \[ = \int_0^{2\pi} \left[r^2 - r^4 + \frac{r^3}{3}\cos\theta - \frac{r^4}{2}\cos\theta + \frac{r^3}{3}\sin\theta - \frac{r^4}{2}\sin\theta\right]_0^1 \, d\theta \] \[ = \int_0^{2\pi} \left[\frac{1}{4} + \frac{1}{6}\cos\theta + \frac{1}{6}\sin\theta\right] \, d\theta \] \[ = \frac{1}{4}(2\pi - 0) = \frac{\pi}{2} \] Finally, \[ \text{flux} = \Phi = \frac{\pi}{2} \]
