# Let F = [x^2,0,z^2], and S the surface of the box abs(x) <= 1, abs(y) <= 3, 0 <= z <= 2. Evaluate the surface integral int int_S F*n dA by the divergence theorem.

Let $F=\left[{x}^{2},0,{z}^{2}\right]$, and S the surface of the box $|x|\le 1,|y|\le 3,0\le z\le 2$.
Evaluate the surface integral $\int {\int }_{S}F\cdot ndA$ by the divergence theorem.
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Step 1
The divergence of the function F is:
$\mathrm{\nabla }\cdot F=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$
$÷F=\frac{\partial }{\partial x}\left({x}^{2}\right)+\frac{\partial }{\partial y}\left(0\right)+\frac{\partial }{\partial z}\left({z}^{2}\right)$
$=2x+2z$
Step 2
By the divergence theorem convert the surface integral into a triple integral:
$\int {\int }_{S}Pdydz+Qdxdz+Rdxdy=\int {\int }_{V}\int \left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dxdydx$
$=\int {\int }_{V}\int \left(2x+2z\right)dxdydz$
Step 3
$\int {\int }_{V}\int \left(2x+2z\right)dxdydz={\int }_{-1}^{1}dx{\int }_{3}^{3}dy{\int }_{0}^{2}\left(2x+2z\right)dz$
$=2{\int }_{1}^{1}dx{\int }_{3}^{3}dy{\in }_{0}^{2}\left(x+z\right)dz$
$=2{\int }_{1}^{1}dx{\int }_{3}^{3}dy\left[\left(xz+\frac{{z}^{2}}{2}\right){\mid }_{0}^{2}\right]$
$=2{\int }_{1}^{1}dx{\int }_{3}^{3}\left(2x+2\right)dy$
$=4{\int }_{1}^{1}dx\left[\left(xy+y\right){\mid }_{-3}^{3}\right]$
$=4{\int }_{1}^{1}dx\left[\left(3x+3\right)-\left(-3x-3\right)\right]$