# Surface integral solved directly and using Gauss' Theorem.

• Jul 18th 2010, 02:16 PM
rikelme23
Surface integral solved directly and using Gauss' Theorem.
To find the flow of the field $\displaystyle$F\left( {x,y,z} \right) = \left( {x^3 ,y^3 ,z^3 } \right)$$across the surface of the cone \displaystyle x^2 + y^2 = z^2$$ with 0 < z < H
a) Directly
b) Applying Gauss's theorem
(Talking)
• Jul 19th 2010, 04:10 AM
HallsofIvy
What have you tried?

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

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

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

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

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

And what is "Gauss's theorem"?