[13.6] - The flux of a vector field

Calculate $\unicode{x222F}_{S} \myv F(x,y,z)\cdot d\myv S$ for $\myv F(x,y,z)=\frac{1}{(x^2+y^2+z^2)^{3/2}}(x\uv i+y\uv j+z\uv k)$ integrated over the surface, $S$ of a sphere of radius $R$.

0. The position vector, $\myv r$ has components $\myv r=\langle x,y,z \rangle$. Its magnitude is $|\myv r|=\sqrt{x^2+y^2+z^2}=(x^2+y^2+z^2)^{1/2}$. Express $\myv F$ in terms of $\uv r$ and $r$ alone.

   Using the definitions, $F$ can be written as: $$\begineq \myv F(x,y,z)&=&\frac{1}{(x^2+y^2+z^2)^{3/2}}(x\uv i+y\uv j+z\uv k)\\ &=&\frac{1}{[(x^2+y^2+z^2)^{1/2}]^3}\myv r\\ &=&\frac{\myv r}{[r]^3}=\frac{1}{r^2}\frac{\myv r}{r}=\frac{1}{r^2}\uv r.\\ \endeq$$

1. We'll need a surface normal (unit) vector which is perpendicular to the surface of the sphere.

   Consider the sphere $x^2+y^2+z^2=R^2$ to be a level surface of the function $G(x,y,z)=x^2+y^2+z^2$. The gradient of a function is always perpendicular to a level curve or level surface, so compute the gradient $\myv \grad G(x,y,z)$ to this surface.

   $$\myv \grad G = \myc{G_x,G_y,G_z}=\myc{2x,2y,2z}$$

2. Does $\myv \grad G$ point inward or outward from the surface of the sphere?

   outward.

3. Compute an outward *unit* normal vector, $\uv n$ to the sphere.

   That gradient vector above is a normal vector, $\myv n=\myv\grad G$, perpendicular to the sphere. So, a unit normal vector would be $$\begineq \uv n&=&\frac{\myv n}{|\myv n|}= \frac{\myc{2x,2y,2z}}{\sqrt{4x^2+4y^2+4z^2}}\\ &=&\frac{2\myc{x,y,z}}{2\sqrt{x^2+y^2+z^2}}\\ &=&\frac{\myv r}{r}=\uv r.\endeq$$

4. Now consider the vector field $\myv F(x,y,z)=\frac{1}{(x^2+y^2+z^2)^{3/2}}(x\uv i+y\uv j+z\uv k)$. Where is $\myv F$ defined?

   Everywhere except at the origin (where the denominator is 0).

5. Show that the flux $\unicode{x222F}_S \myv F\cdot d\myv S = \unicode{x222F}_S\myv F\cdot \uv n\,dS=4\pi$, where $S$ is the spherical surface $x^2+y^2+z^2=R^2$. It may be useful to know that the surface area of a sphere of radius $R$ is $\unicode{x222F}_S dS=\int_0^\pi \int_0^{2\pi}(R\,d\theta)(d\phi)= 4\pi R^2$.

   We can write $\myv F=\frac 1{r^2}\uv r$ and we figured out that $\uv n=\uv r$. So, $$\myv F\cdot \uv n = \frac{1}{r^2}\uv n\cdot\uv n= \frac{1}{r^2}.$$ On the surface of the sphere of radius $R$, $r=R$ is constant. So, The flux integral is: $$\iint_S\myv F\cdot \uv n\,dS =\iint_S \frac{1}{R^2}\,dS =\frac{1}{R^2}\iint_S\,dS =\frac{1}{R^2} 4\pi R^2 = 4\pi. $$