Question

Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S. F(x, y, z) = 3xy^2i + xe^zj + z^3k, S is the surface of the solid bounded by the cylinder y^2 + z^2 = 9 and the planes x = −3 and x = 1.

Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}\),
S is the surface of the solid bounded by the cylinder \(\displaystyle{y}^{{2}}+{z}^{{2}}={9}\) and the planes x = −3 and x = 1.

Expert Answers (1)

2020-12-29
Step 1
Given \(\displaystyle\vec{{{F}}}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}\hat{{{i}}}+{x}{e}^{{z}}\hat{{{j}}}+{z}^{{3}}\hat{{{k}}}\)
We have to evaluate the surface integral \(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{N}{d}{S}\) where, S is the surface of the solid bounded by the cylinder \(\displaystyle{y}^{{2}}+{z}^{{2}}={9}\) and the plane x=−3 and x=1
The projection of the cylinder is on the yz−plane is a circle of radius 3
Use polar coordinate
\(\displaystyle{y}={r}{\cos{{0}}}{\quad\text{and}\quad}{z}={r}{\sin{{0}}}{\quad\text{and}\quad}{\left.{d}{y}\right.}{\left.{d}{z}\right.}={r}{d}{r}{d}{0}\)
And the solid E can be described as
\(\displaystyle{E}={\left\lbrace{\left({x},{t},{0}\right)}:-{3}\le{x}\le{1},{0}\le{r}\le{3},{0}\le{0}\le{2}\pi\right\rbrace}\)
Step 2
Use theorem \(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{N}{d}{S}=\int\int\int_{{E}}\div\vec{{{F}}}{d}{V}\)
Now,
div \(\displaystyle\vec{{{F}}}={\left(\frac{\partial}{{\partial{x}}}\hat{{{i}}}+\frac{\partial}{{\partial{y}}}\hat{{{j}}}+\frac{\partial}{{\partial{z}}}\hat{{{k}}}\right)}{\left({3}{x}{y}^{{2}}\hat{{{i}}}+{x}{e}^{{z}}\hat{{{j}}}+{z}^{{3}}\hat{{{k}}}\right)}\)
\(\displaystyle\frac{\partial}{{\partial{x}}}{\left({3}{x}{y}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}{e}^{{z}}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}^{{3}}\right)}\)
\(\displaystyle={3}{y}^{{2}}+{0}+{3}{z}^{{2}}\)
\(\displaystyle={3}{y}^{{2}}+{3}{z}^{{2}}\)
\(\displaystyle={3}{\left({y}^{{2}}+{z}^{{2}}\right)}\)
\(\displaystyle={3}{r}^{{2}}\)
Hence, div \(\displaystyle\vec{{{F}}}={3}{r}^{{2}}\)
Step 3
Hence,
\(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{N}{d}{S}={\int_{{-{{3}}}}^{{1}}}{\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{3}}}{3}{r}^{{2}}\cdot{r}{d}{r}{d}{0}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{-{{3}}}}^{{1}}}{\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{3}}}{r}^{{3}}{d}{r}{d}{0}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{-{{3}}}}^{{1}}}{\int_{{0}}^{{{2}\pi}}}{\left[{\int_{{0}}^{{3}}}{r}^{{3}}{d}{r}\right]}{d}{0}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{-{{3}}}}^{{1}}}{\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{r}^{{4}}}{{4}}\right]}_{{0}}^{{3}}}{d}{0}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{-{{3}}}}^{{1}}}{\int_{{0}}^{{{2}\pi}}}{\left[\frac{{{\left({3}\right)}^{{4}}}}{{4}}-{0}\right]}{d}{0}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{243}}{{4}}{\int_{{-{{3}}}}^{{1}}}{\left[{\int_{{0}}^{{{2}\pi}}}{d}{0}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{243}}{{4}}{\int_{{3}}^{{1}}}{{\left[{0}\right]}_{{0}}^{{{2}\pi}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{243}}{{4}}{\int_{{3}}^{{1}}}{\left[{2}\pi-{0}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{243}}{{3}}\cdot{2}\pi{\int_{{3}}^{{1}}}{\left.{d}{x}\right.}\)
\(\displaystyle\frac{{243}}{{2}}\pi{{\left[{x}\right]}_{{3}}^{{1}}}\)
\(\displaystyle=\frac{{243}}{{2}}\pi{\left[{1}-{\left(-{3}\right)}\right]}\)
\(\displaystyle=\frac{{243}}{{2}}\pi{\left[{4}\right]}\)
\(\displaystyle={486}\pi\)
Hence, \(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{N}{d}{S}={486}\pi\)
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