Surface integral solved directly and using Gauss' Theorem.

**rikelme23** · July 18th 2010, 02:16 PM

To find the flow of the field $F\left( {x,y,z} \right) = \left( {x^3 ,y^3 ,z^3 } \right)$
across the surface of the cone $x^2 + y^2 = z^2$ with 0 < z < H
a) Directly
b) Applying Gauss's theorem

**HallsofIvy** · July 19th 2010, 04:10 AM

What have you tried?

I will give you this much: we can write the cone in parametric coordinates using polar coordinates: $x= r cos(\theta)$ , $y= r sin(\theta)$ , $z= r$ with $r$ from 0 to H and [/tex]\theta[/tex] from 0 to $2\pi$ .

With that you can write the "position vector" of a point on the surface of the cone as $\vec{p}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r \vec{k}$ .

The deriviatives of that with respect to r and $\theta$ :
$\vec{p}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$ and
$\vec{p}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}$
are in the tangent plane to the cone and their cross product is
$-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r \vec{k}$ .

The (upward oriented) "vector differential of area" is $(-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r \vec{k})drd\theta$

Take the dot product of your field with that and integrate.

And what is "Gauss's theorem"?

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