# Another Surface Integral!

• May 13th 2011, 07:49 PM
Nguyen
Another Surface Integral!
Hey, I tried converting co-ordinates and all but I am having difficulties.

Evaluate

$\displaystyle \int \int_S F \cdot dS$

where $\displaystyle F= xi+yj +z^4 k$ and $\displaystyle S$ is the part of the cone $\displaystyle z=\sqrt{x^2+y^2}$ beneath the plane $\displaystyle z=1$ with downward orientation.

Thanks for any help.
• May 13th 2011, 08:07 PM
TheEmptySet
Quote:

Originally Posted by Nguyen
Hey, I tried converting co-ordinates and all but I am having difficulties.

Evaluate

$\displaystyle \int \int_S F \cdot dS$

where $\displaystyle F= xi+yj +z^4 k$ and $\displaystyle S$ is the part of the cone $\displaystyle z=\sqrt{x^2+y^2}$ beneath the plane $\displaystyle z=1$ with downward orientation.

Thanks for any help.

$\displaystyle d\mathbf{S}=-\left( \frac{\partial z}{\partial x} \right)\mathbf{i} - \left( \frac{\partial z}{\partial y} \right)\mathbf{j}+\mathbf{k} = -\left( \frac{x}{\sqrt{x^2+y^2}} \right)\mathbf{i} - \left( \frac{y}{\sqrt{x^2+y^2}} \right)\mathbf{j}+\mathbf{k}$

$\displaystyle \mathbf{F} \cdot d\mathbf{S}= -\left( \frac{x^2}{\sqrt{x^2+y^2}} \right) - \left( \frac{y^2}{\sqrt{x^2+y^2}} \right)+(x^2+y^2)^2$

In cylindrical coordinates this gives

$\displaystyle \mathbf{F} \cdot d\mathbf{S}= (-r+r^4)rdrd\theta$

Note we need to take the negative of the above because in the derivative I assumed and upward normal the k component is positive.

Can you finish from here?
• May 14th 2011, 05:08 AM
HallsofIvy
Or, working with cylindrical coordinates from the start, we can write the position vector of any point on the cone as $\displaystyle \vec{p}(r, \theta)= rcos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ z\vec{k}= rcos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}$.

The derivatives give vectors in the tangent plane:
$\displaystyle \vec{p}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$
$\displaystyle \vec{p}_\theta= -r sin(\theta)\vec{i}+ cos(\theta)\vec{j}$

The cross product of those two vectors gives the normal vector
$\displaystyle rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}- r\vec{k}$
where the order of multiplication has been chose to make the $\displaystyle \vec{k}$ component negative ("with downward orientation").

The "vector differential of surface area" is $\displaystyle (rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}- r\vec{k})drd\theta$

Of course, $\displaystyle \vec{F}= x\vec{i}+ y\vec{j}+ z^4\vec{k}= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ r^4\vec{k}$ so that $\displaystyle \vec{F}\cdot d\vec{S}= (r^2cos^(\theta)+ r^2sin^2(\theta)- r^5 )drd\theta= (r^2- r^5)drd\theta$ just as TheEmptySet says (with the sign changed of course).