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)

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- Jul 18th 2010, 02:16 PMrikelme23Surface 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 AMHallsofIvy
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"?