alesterp

2021-02-20

Set up the integral for the divergence theorem both ways. Then find the flux.
$F\left(x,y,z\right)=3x\stackrel{^}{i}+xy\stackrel{^}{j}+2xz\stackrel{^}{k}$
E is the cube bounded by the planes .

Macsen Nixon

Step 1
Considering the given dunction is
$f\left(x,y,z\right)=3x\stackrel{^}{i}+xy\stackrel{^}{j}+2xz\stackrel{^}{k}$.
Where E is the cube bounded by the planes
Find the flux.
Step 2
Consider
$F\left(x,y,z\right)=3x\stackrel{^}{i}+xy\stackrel{^}{j}+2xz\stackrel{^}{k}$.
Find the div(f), then
grad $F=3+x+2x$
$=3+3x$
By using the divergence theorem, then
$Flux=\int \int \int \mathrm{\nabla }Fdv$
$={\int }_{0}^{3}{\int }_{0}^{3}{\int }_{0}^{3}\left(3+3x\right)dzdydx$
$={\int }_{0}^{3}\left(3+3x\right){\int }_{0}^{3}{\left(z\right)}_{0}^{3}dydx$
$=3{\int }_{0}^{3}\left(3+3x\right){\int }_{0}^{3}dydx$
$=3{\int }_{0}^{3}\left(3+3x\right){\left(y\right)}_{0}^{3}dx$
$=9{\int }_{0}^{3}\left(3+3x\right)dx$
$=9{\left(3x+\frac{3{x}^{2}}{2}\right)}_{0}^{3}$
$=\frac{405}{2}$
Hence, total flux is $\frac{405}{2}$.

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