Using the Divergence Theorem, evaluate int int_S F.NdS, where F(x,y,z)=(z^3i-x^3j+y^3k) and S is the sphere x^2+y^2+z^2=a^2, with outward unit normal vector N.

Question
Using the Divergence Theorem, evaluate \(\displaystyle\int\int_{{S}}{F}.{N}{d}{S}\), where \(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({z}^{{3}}{i}-{x}^{{3}}{j}+{y}^{{3}}{k}\right)}\) and S is the sphere \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={a}^{{2}}\), with outward unit normal vector N.

Answers (1)

2021-01-20
Step 1
Divergence theorem relates surface integrals and volume integrals. By using the Gauss divergence theorem we can evaluate this surface integral.
The Gauss divergence theorem formula can be stated as follows:
\(\displaystyle\int\int_{{S}}{F}.{N}{d}{S}=\int\int\int_{{V}}\div{F}.{d}{V}\)
\(\displaystyle\div{F}={\left(\partial\frac{{{z}^{{3}}{i}}}{{\partial{x}}}\right)}+{\left(\partial\frac{{-{x}^{{3}}{j}}}{{\partial{y}}}\right)}+{\left(\partial\frac{{{y}^{{3}}{k}}}{{\partial{z}}}\right)}\)
Step 2
div F = 0, since the partial derivative of \(\displaystyle{z}^{{3}}\) with respect to x is zero and partial derivative of \(\displaystyle-{x}^{{3}}\) with respect to y is zero and the partial derivative of \(\displaystyle{y}^{{3}}\) with respect to z is zero.
Then \(\displaystyle\int\int\int_{{V}}{0}.{d}{V}={0}\)
Since \(\displaystyle\int\int_{{S}}{F}.{N}{d}{S}=\int\int\int_{{V}}\div{F}.{d}{V}\)
Then by Gauss Divergence theorem, we can say \(\displaystyle\int\int_{{S}}{F}.{N}{d}{S}={0}\).
Step 3
So the final answer is zero by the Gauss Divergence theorem.
0

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