Question

Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = x2 sin(y)i + x cos(y)j − xz sin(y)k, S is the "fat sphere" x8 + y8 + z8 = 8.

Answer 1

Answer:

0

Step-by-step explanation:

According to the Divergence Theorem,

$\int \int F\,dS=\int \int \int divF\,\,dV$

Here, for $F=\left ( F_1,F_2,F_3 \right )$,

$divF=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$

Take $F_1=x^2\sin y\,,\,F_2=x\cos y\,\,,F_3=-xz\sin y$

Differentiate $F_1,F_2,F_3$ with respect to $x,y,z$ respectively,

$\frac{\partial F_1}{\partial x}=\frac{\partial }{\partial x}x^2\sin y=2x\sin y\\\frac{\partial F_2}{\partial y}=\frac{\partial }{\partial y}x\cos y=-x\sin y\\\frac{\partial F_3}{\partial z}=\frac{\partial }{\partial z}-xz\sin y=-x\sin y$

Therefore,

$divF=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\\=2x\sin y-x\sin y-x\sin y\\=0$

So,

$\int \int F\,dS=\int \int \int divF\,\,dV=0$

Answer 2

$\vec F(x,y,z)=x^2\sin y\,\vec\imath+x\cos y\,\vec\jmath-xz\sin y\,\vec k$

has divergence

$\mathrm{div}\vec F(x,y,z)=2x\sin y-x\sin y-x\sin y=0$

so that the flux of $\vec F$ across $S$ is 0 by the divergence theorem.