Question

Use the Divergence Theorem to calculate the surface integral \(\displaystyle\int\int_{{S}}{F}·{d}{S}\), that is, calculate the flux of F across S.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}\), S is the sphere with center the origin and radius 2.

Answer 1

Let E be a simplesolid region and S is the boundary surface of E with positive orientation and Let \(\displaystyle\vec{{{F}}}\) be a vector field whose components have continous first orderpartial derivatives. then
\(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{d}\vec{{{S}}}=\int\int_{{E}}\int\div\vec{{{F}}}{d}{V}\)
Step 2
Apply the above theorem as follows.
Given that \(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}\vec{{{i}}}+{\left({y}^{{3}}+{z}^{{3}}\right)}\vec{{{j}}}+{\left({z}^{{3}}+{x}^{{3}}\right)}\vec{{{k}}}\)
div \(\displaystyle\vec{{{F}}}=\nabla\vec{{{F}}}={3}{x}^{{2}}+{3}{y}^{{2}}+{3}{z}^{{2}}\)
\(\displaystyle\int\int_{{S}}\vec{{{F}}}\cdot{d}\vec{{{S}}}=\int\int_{{E}}\int{i}{d}{v}\vec{{{F}}}{d}{V}\)
\(\displaystyle=\int\int_{{E}}\int{\left({3}{x}^{{2}}+{3}{y}^{{2}}+{3}{z}^{{2}}\right)}{d}{V}\)
\(\displaystyle={3}\int\int_{{E}}\int{\left({x}^{{2}}+{y}^{{2}}+{z}^{{2}}\right)}{d}{V}\)
\(\displaystyle={36}\int\int_{{E}}\int{\left({2}\right)}{d}{V}{\left\langle'{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={2}^{{2}}\right)}\)
\(\displaystyle={72}\int\int_{{E}}\int{1}{d}{V}\)
\(\displaystyle={72}{\left(\frac{{{4}\pi}}{{3}}\times{2}^{{3}}\right)}\) (\(\displaystyle\because\) volume of the sphere is \(\displaystyle\frac{{{4}\pi}}{{3}}{r}^{{3}}{)}\)
\(\displaystyle={768}\pi\)