Set up the integral for the divergence theorem both ways. Then find the flux. F(x,y,z) = 3x hat(i) + xy hat(j) + 2xz hat(k) E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.

Set up the integral for the divergence theorem both ways. Then find the flux. F(x,y,z) = 3x hat(i) + xy hat(j) + 2xz hat(k) E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.

Question
Set up the integral for the divergence theorem both ways. Then find the flux.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}\)
E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.

Answers (1)

2021-02-21
Step 1
Considering the given dunction is
\(\displaystyle{f{{\left({x},{y},{z}\right)}}}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}\).
Where E is the cube bounded by the planes x =0, x = 3, y = 0, y = 3 and z = 0, z =3.
Find the flux.
Step 2
Consider
\(\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}\).
Find the div(f), then
grad F = 3+x+2x
=3+3x
By using the divergence theorem, then
\(\displaystyle{F}{l}{u}{x}=\int\int\int\nabla{F}{d}{v}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\int_{{0}}^{{3}}}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\int_{{0}}^{{3}}}{{\left({z}\right)}_{{0}}^{{3}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{{\left({y}\right)}_{{0}}^{{3}}}{\left.{d}{x}\right.}\)
\(\displaystyle={9}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={9}{{\left({3}{x}+\frac{{{3}{x}^{{2}}}}{{2}}\right)}_{{0}}^{{3}}}\)
\(\displaystyle=\frac{{405}}{{2}}\)
Hence, total flux is \(\displaystyle\frac{{405}}{{2}}\).
0

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