Anish Buchanan
2021-01-31
Answered

Use the Divergence Theorem to calculate the surface integral $\int {\int}_{S}F\xb7dS$ , that is, calculate the flux of F across S.

$F(x,y,z)=({x}^{3}+{y}^{3})i+({y}^{3}+{z}^{3})j+({z}^{3}+{x}^{3})k$ , S is the sphere with center the origin and radius 2.

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Bentley Leach

Answered 2021-02-01
Author has **109** answers

Let E be a simplesolid region and S is the boundary surface of E with positive orientation and Let $\overrightarrow{F}$ be a vector field whose components have continous first orderpartial derivatives. then

$\int {\int}_{S}\overrightarrow{F}\cdot d\overrightarrow{S}=\int {\int}_{E}\int \xf7\overrightarrow{F}dV$

Step 2

Apply the above theorem as follows.

Given that$F(x,y,z)=({x}^{3}+{y}^{3})\overrightarrow{i}+({y}^{3}+{z}^{3})\overrightarrow{j}+({z}^{3}+{x}^{3})\overrightarrow{k}$

div$\overrightarrow{F}=\mathrm{\nabla}\overrightarrow{F}=3{x}^{2}+3{y}^{2}+3{z}^{2}$

$\int {\int}_{S}\overrightarrow{F}\cdot d\overrightarrow{S}=\int {\int}_{E}\int idv\overrightarrow{F}dV$

$=\int {\int}_{E}\int (3{x}^{2}+3{y}^{2}+3{z}^{2})dV$

$=3\int {\int}_{E}\int ({x}^{2}+{y}^{2}+{z}^{2})dV$

$=36\int {\int}_{E}\int \left(2\right)dV\u27e8{}^{\prime}{x}^{2}+{y}^{2}+{z}^{2}={2}^{2})$

$=72\int {\int}_{E}\int 1dV$

$=72(\frac{4\pi}{3}\times {2}^{3})$ ($\because$ volume of the sphere is $\frac{4\pi}{3}{r}^{3})$

$=768\pi$

Step 2

Apply the above theorem as follows.

Given that

div

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