Answer
$\phi = \frac{4}{3}\pi $
Work Step by Step
Let \(\vec{F}\) be the vector field: \[ \vec{F} = 2x \hat{i} - yz \hat{j} + z^2 \hat{k} \] and the surface \(\sigma\) given by the paraboloid \(z = x^2 + y^2\) capped by the disk \(x^2 + y^2 \leq 1\) in the plane \(z = 1\). The projection of this surface on the xy plane is the circle \(\mathcal{R}\): \[ \mathcal{R}: \{ (r, \theta) \mid 0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq 1 \} \] where we have used polar coordinates. Using the Divergence Theorem: \[ \begin{align*} \phi &= \iint_{\sigma} \vec{F} \cdot \hat{n} \,dS \\ &= \iiint_{V} \nabla \cdot \vec{F} \,dV \end{align*} \] where \[ \nabla \cdot \vec{F} = 2 - z + 2z = 2 + z \] and \[ \nabla \cdot \vec{F} = 2 - z + 2z = 2 + z \] Since \(x^2 + y^2 = r^2\) and \(r^2 \leq z \leq 1\): \[ \begin{align*} \phi &= \iiint_{V} \nabla \cdot \vec{F} \,dV \\ &= \int_0^{2\pi} \int_0^1 \int_{r^2}^1 (2 + z) \,rdzdrd\theta \\ &= \int_0^{2\pi} \int_0^1 \left( -\frac{r^5}{2} - 2r^3 + \frac{5r}{2} \right) \,drd\theta \\ &= \frac{4}{3}\pi \end{align*} \] Finally, \[ \text{flux} = \phi = \frac{4}{3}\pi \] So, the flux is \(\frac{4}{3}\pi\).
