# Flux of a vector field through warped sphere

1. Feb 20, 2012

### phasic

1. The problem statement, all variables and given/known data
Consider the surface S with the graph z = 1-x$^{2}$-y$^{2}$ with z≥0, and also the unit disc in the xy plane. Give this surface an outer normal. Compute: $\int\int_{S}$$\vec{F}$$\bullet$d$\vec{S}$

where $\vec{F}$(x,y,z) = (2x,2y,z)

2. Relevant equations

$\int\int_{S}$$\vec{F}$$\bullet$d$\vec{S}$

= $\int\int_{D}$$\vec{F}$$\bullet$($\vec{T_{x}}$$\times$$\vec{T_{y}}$)dxdy

= $\int\int_{D}$ [ $F_1$(-dg/dx)+$F_2$(-dg/dy)+$F_3$ ] dxdy

Where Tx and Ty are the tangent vectors and D is the domain of the parameterization of the surface of S.

3. The attempt at a solution

I found the integrand of the 3rd relevant equation to be -5x^2 - 5y^2 + 1. But I cannot find the bounds of integration. Is the domain D the unit disc or is it just the rectangle -1<x<1 and -1<y<1? If it's the unit disc, then using polar coordinates my solution is -6*Pi/4. If it's the rectangle, my solution is -28/3.

Also, is this just flux through the open surface z = 1-(x^2)-(y^2) or is it including the bottom cap? If not, would I then have to calculate and add the flux through the disc, which should be relatively easy? Then, would these two pieces combined be equal to the integral of div(F) over the volume enclosed by these surfaces, via the divergence theorem?

Thanks for the help.

EDIT: D'oh! The integrand is actually 3x^2 +3y^2 + 1, and I have a feeling the limits of integration are over the rectangle, so the integral over that is 12.

Last edited: Feb 20, 2012
2. Feb 20, 2012

### lanedance

the limits of integration are based on the parameterisation you choose...

for this problem note that the unit disc is a much more natural domain to parameterise the inverted paraboloid with, than the unit rectangle

secondly, you will need to split the surface integral into 2 parts one for the inverted paraboloid, and one for the disc at the bottom

thirdly, though I haven't attempted it, the divergence theorem may be used as a a check, and may even simplify the problem...

3. Feb 20, 2012

### phasic

Thanks! I just realized that using the unit disc is best. My final answer was 5*pi/2.

Also, thanks for confirming that this is just for the flux through the open paraboloid, and that I also have to compute the flux through the cap at the bottom!

The flux through the cap at the bottom is 0, because the z component of F is 0.

These should sum up to the volume integral over the div(F), which is just div(F) = 5 times the volume of the paraboloid, which is Pi*(height of paraboloid)/2! For this paraboloid of height 1, the volume is Pi/2. So, the right side of divergence theorem is 5*Pi/2. These answers match!

Thanks for your help, and let me know if there is any flaw in my thinking. :D

4. Feb 20, 2012

### lanedance

thinking sounds good, and without explicitly doing the calc the fact the divergence thm gives you the same answer gives me confidence in your answer