Considere o campo vetorial \[ \mathbf{F}(x, y, z)=\left(\frac{x^{3}}{3}+x y^{2}\right) \mathbf{i}+\left(\frac{y^{3}}{3}+y x^{2}\right) \mathbf{j}+z\left(x^{2}+y^{2}\right) \mathbf{k} \] e seja $S$ a porção do paraboloide $z=x^{2}+y^{2}$ no semiespaço $y \geq 0$ e abaixo do plano $z=1$ com normal $\mathbf{n}$ tal que $\mathbf{n} \cdot \mathbf{k}>0$. Calcule $\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=\iiint \int \operatorname{div} F d v$

Step 1 ：Consider the vector field \(\mathbf{F}(x, y, z)=\left(\frac{x^{3}}{3}+x y^{2}\right) \mathbf{i}+\left(\frac{y^{3}}{3}+y x^{2}\right) \mathbf{j}+z\left(x^{2}+y^{2}\right) \mathbf{k}\) and the surface S, which is the portion of the paraboloid \(z=x^{2}+y^{2}\) in the half-space \(y \geq 0\) and below the plane \(z=1\) with normal \(\mathbf{n}\) such that \(\mathbf{n} \cdot \mathbf{k}>0\).

Step 2 ：Calculate the divergence of F: \(\operatorname{div} \mathbf{F} = 3x^2 + 3y^2\).

Step 3 ：Convert the divergence to cylindrical coordinates: \(\operatorname{div} \mathbf{F}_{cylindrical} = r(3r^2\sin^2(\theta) + 3r^2\cos^2(\theta))\).

Step 4 ：Set up the triple integral using cylindrical coordinates with the limits of integration for the volume enclosed by S: \(r: 0 \to 1\), \(\theta: 0 \to \pi\), and \(z: r^2 \to 1\).

Step 5 ：Calculate the triple integral: \(\iiint \operatorname{div} \mathbf{F}_{cylindrical} \, dV = \frac{\pi}{4}\).

Step 6 ：\(\boxed{\frac{\pi}{4}}\) is the surface integral of the given vector field F over the surface S.