Question

# Use the divergence theorem to evaluate int int_S F*NdS, where F(x,y,z)=y^2zi + y^3j + xzk and S is the boundary of the cube defined by -5 <= x,= 5, -5 <= y <= 5, and 0 <= z<= 10.

Use the divergence theorem to evaluate $$\displaystyle\int\int_{{S}}{F}\cdot{N}{d}{S}$$, where $$\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}$$ and S is the boundary of the cube defined by $$\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}$$.

2021-03-12

Step 1
The given field is $$\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}$$ and S is the boundary of the cube
$$\displaystyle-{5}\le{x}\le{5},-{5}\le{y}\le{5},{0}\le{z}\le{10}$$.
Step 2
By the Divergence theorem, we have $$\displaystyle\int\int_{{S}}{F}.{n}{d}{S}=\int\int_{{R}}\int\nabla\cdot{F}{d}{V}$$.
Here,
$$\displaystyle\nabla\cdot{F}={\left(\frac{\partial}{{\partial{x}}},\frac{\partial}{{\partial{y}}},\frac{\partial}{{\partial{z}}}\right)}\cdot{\left({y}^{{2}}{z},{y}^{{3}},{x}{z}\right)}$$
$$\displaystyle=\frac{\partial}{{\partial{x}}}{\left({y}^{{2}}{z}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}^{{3}}\right)}+\frac{\partial}{{\partial{z}}}{\left({x}{z}\right)}$$
$$\displaystyle{3}{y}^{{2}}+{x}$$
Hence,
$$\displaystyle\int\int_{{S}}{F}.{n}{d}{S}=\int\int_{{R}}\int\nabla\cdot{F}{d}{V}$$
$$\displaystyle={\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\int_{{{z}={0}}}^{{10}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{{\left({y}^{{3}}+{x}{y}\right)}_{{{y}=-{5}}}^{{5}}}{\left.{d}{x}\right.}$$
$$\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{\left({250}+{10}{x}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={10}{{\left({250}{x}+{5}{x}^{{2}}\right)}_{{-{5}}}^{{5}}}$$
$$=25000$$