Question

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 \(\displaystyle{F}={\left[{x}^{{2}},{0},{z}^{{2}}\right]}\), and S the surface of the box \(\displaystyle{\left|{{x}}\right|}\le{1},{\left|{{y}}\right|}\le{3},{0}\le{z}\le{2}\).
Evaluate the surface integral \(\displaystyle\int\int_{{S}}{F}\cdot{n}{d}{A}\) by the divergence theorem.

Answers (1)

2021-02-22

Step 1
The divergence of the function F is:
\(\displaystyle\nabla\cdot{F}=\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}+\frac{{\partial{R}}}{{\partial{z}}}\)
\(\displaystyle\div{F}=\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({0}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}^{{2}}\right)}\)
\(=2x+2z\)
Step 2
By the divergence theorem convert the surface integral into a triple integral:
\(\displaystyle\int\int_{{S}}{P}{\left.{d}{y}\right.}{\left.{d}{z}\right.}+{Q}{\left.{d}{x}\right.}{\left.{d}{z}\right.}+{R}{\left.{d}{x}\right.}{\left.{d}{y}\right.}=\int\int_{{V}}\int{\left(\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}+\frac{{\partial{R}}}{{\partial{z}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle=\int\int_{{V}}\int{\left({2}{x}+{2}{z}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
Step 3
\(\displaystyle\int\int_{{V}}\int{\left({2}{x}+{2}{z}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}={\int_{{-{{1}}}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\int_{{0}}^{{2}}}{\left({2}{x}+{2}{z}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\in_{{0}}^{{2}}}{\left({x}+{z}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\left[{\left({x}{z}+\frac{{z}^{{2}}}{{2}}\right)}{{\mid}_{{0}}^{{2}}}\right]}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left({2}{x}+{2}\right)}{\left.{d}{y}\right.}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({x}{y}+{y}\right)}{{\mid}_{{-{{3}}}}^{{3}}}\right]}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({3}{x}+{3}\right)}-{\left(-{3}{x}-{3}\right)}\right]}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({6}{x}+{6}\right)}\right]}\)
\(\displaystyle={24}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}\)
\(\displaystyle={24}{{\left[{x}\right]}_{{-{{1}}}}^{{1}}}\)
\(=48\)

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