Use the Divergence Theorem to find the flux of F = xy^2i + x^2yj + yk outward through the surface of the region enclosed by the cylinder x^2 + y^2 = 1 and the planes z = 1 and z =-1.

Use the Divergence Theorem to find the flux of F = xy^2i + x^2yj + yk outward through the surface of the region enclosed by the cylinder x^2 + y^2 = 1 and the planes z = 1 and z =-1.

Question
Use the Divergence Theorem to find the flux of \(\displaystyle{F}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}\) outward through the surface of the region enclosed by the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={1}\) and the planes z = 1 and z =-1.

Answers (1)

2020-12-30
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 \(\displaystyle\pi\).
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