Considere o campo vetorial
$$\mathbf{F}(x,y,z)=(\frac{x^3}{3}+x{y}^2)\mathbf{i}+(\frac{y^3}{3}+y{x}^2)\mathbf{j}+z(x^2+y^2)\mathbf{k}$$
e seja $S$ a porção do paraboloide $z=x^2+y^2$ no semiespaço $y\ge 0$ e abaixo do plano $z=1$ com normal $\mathbf{n}$ tal que $\mathbf{n}\cdot \mathbf{k}>0$.
Calcule ${\iint }_S\mathbf{F}\cdot \mathbf{n}dS=\iiint \int \mathrm{div}Fdv$

Step 1 ：Consider the vector field $\mathbf{F}(x,y,z)=(\frac{x^3}{3}+x{y}^2)\mathbf{i}+(\frac{y^3}{3}+y{x}^2)\mathbf{j}+z(x^2+y^2)\mathbf{k}$ and the surface S, which is the portion of the paraboloid $z=x^2+y^2$ in the half-space $y\ge 0$ 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: $\mathrm{div}\mathbf{F}=3x^2+3y^2$.

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

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: $\iiint \mathrm{div}{\mathbf{F}}_{cylindrical}dV=\frac{\pi }{4}$.

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