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 \(\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}\).

Answers (1)

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,
\(\displaystyle\int\int_{{S}}{F}.{n}{d}{S}=\int\int_{{R}}\int\nabla\cdot{F}{d}{V}\)
\(\displaystyle={\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\int_{{{z}={0}}}^{{10}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{\int_{{{y}=-{5}}}^{{5}}}{\left({3}{y}^{{2}}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{{\left({y}^{{3}}+{x}{y}\right)}_{{{y}=-{5}}}^{{5}}}{\left.{d}{x}\right.}\)
\(\displaystyle={10}{\int_{{{x}=-{5}}}^{{5}}}{\left({250}+{10}{x}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={10}{{\left({250}{x}+{5}{x}^{{2}}\right)}_{{-{5}}}^{{5}}}\)
\(=25000\)

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