Step 1

Divergence Theorem.

Let F be a vector field whose components have continuous first partial derivatives and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface's outward unit normal field n equals the triple integral of the divergence grad * F over the solid region D enclosed by the surface.

\(\displaystyle\int\int_{{S}}{F}\cdot{n}{d}\sigma=\int\int_{{D}}\int\nabla\cdot{F}{d}{V}\).

Step 2

Write the vector field.

\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}\)

Find the divergence of Equation(1).

\(\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{y}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{z}}}{\left({y}\right)}\)

\(\displaystyle={\left({y}^{{2}}\right)}+{\left({x}^{{2}}\right)}+{\left({0}\right)}\)

\(\displaystyle={x}^{{2}}+{y}^{{2}}\)

Step 3

Find the flux of the vector field.

Flux = \(\displaystyle\int\int_{{D}}\int\nabla{F}{d}{V}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{\int_{{-{{1}}}}^{{1}}}{r}^{{2}}{\left.{d}{z}\right.}{r}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{{\left[{z}\right]}_{{1}}^{{1}}}{d}{r}{\left.{d}{z}\right.}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{\left({1}+{1}\right)}{d}{r}{\left.{d}{z}\right.}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{2}{r}^{{3}}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{{2}{r}^{{4}}}}{{4}}\right]}_{{0}}^{{1}}}{d}{0}\)

Flux = \(\displaystyle{\int_{{0}}^{{{2}\pi}}}{\left[{\left(\frac{{{\left({1}\right)}^{{4}}}}{{2}}\right)}-{0}\right]}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}\frac{{1}}{{2}}{d}{0}\)

\(\displaystyle=\frac{{1}}{{2}}{{\left[{0}\right]}_{{0}}^{{{2}\pi}}}\)

\(\displaystyle=\frac{{1}}{{2}}{\left({2}\pi-{0}\right)}\)

\(\displaystyle=\pi\)

Step 4

Therefore, the outward flux of the vector field \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}\) by using the Divergence Theorem is \(\displaystyle\pi\).

Divergence Theorem.

Let F be a vector field whose components have continuous first partial derivatives and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface's outward unit normal field n equals the triple integral of the divergence grad * F over the solid region D enclosed by the surface.

\(\displaystyle\int\int_{{S}}{F}\cdot{n}{d}\sigma=\int\int_{{D}}\int\nabla\cdot{F}{d}{V}\).

Step 2

Write the vector field.

\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}\)

Find the divergence of Equation(1).

\(\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{y}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{z}}}{\left({y}\right)}\)

\(\displaystyle={\left({y}^{{2}}\right)}+{\left({x}^{{2}}\right)}+{\left({0}\right)}\)

\(\displaystyle={x}^{{2}}+{y}^{{2}}\)

Step 3

Find the flux of the vector field.

Flux = \(\displaystyle\int\int_{{D}}\int\nabla{F}{d}{V}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{\int_{{-{{1}}}}^{{1}}}{r}^{{2}}{\left.{d}{z}\right.}{r}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{{\left[{z}\right]}_{{1}}^{{1}}}{d}{r}{\left.{d}{z}\right.}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{3}}{\left({1}+{1}\right)}{d}{r}{\left.{d}{z}\right.}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{2}{r}^{{3}}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{{2}{r}^{{4}}}}{{4}}\right]}_{{0}}^{{1}}}{d}{0}\)

Flux = \(\displaystyle{\int_{{0}}^{{{2}\pi}}}{\left[{\left(\frac{{{\left({1}\right)}^{{4}}}}{{2}}\right)}-{0}\right]}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}\frac{{1}}{{2}}{d}{0}\)

\(\displaystyle=\frac{{1}}{{2}}{{\left[{0}\right]}_{{0}}^{{{2}\pi}}}\)

\(\displaystyle=\frac{{1}}{{2}}{\left({2}\pi-{0}\right)}\)

\(\displaystyle=\pi\)

Step 4

Therefore, the outward flux of the vector field \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}\) by using the Divergence Theorem is \(\displaystyle\pi\).