Register to reply

Flux Integral Help through abnormal cone

by mgibson
Tags: abnormal, cone, flux, integral
Share this thread:
mgibson
#1
Apr29-08, 03:50 AM
P: 29
1. The problem statement

The problem requires me to calculate the flux of F=x^2 i + z j + y k out of the closed cone, x=sqrt(y^2 + z^2) with x between 0 and 1.

I am having trouble approaching this problem because most of the problems I have done give the curve as z=f(x,y) instead of x=f(y,z) and I am therefore confused as to how to apply the below equation.

2. Relevant equations

For the flux through a surface given by z=f(x,y)

Flux = int(F . dA) = int( [ F(x,y, f(x,y)) dot (-df/dx i - df/dy j + k) ]dxdy

where df/dx is the partial derivative of f with respect to x and df/dy is the partial derivative of f with respect to y.

How can I modify/apply this formula (if I even can) when given a surface as a function of x=f(y,z) as opposed to z=f(x,y) to find the flux through the horizontally opening cone?

Any help would be greatly appreciated! Thanks so much.


3. The attempt at a solution

I tried putting f in terms of z and going that route but ran into some nasty integrals.

I tried replacing z with x and x with z (for F and f) as to simulate the same vector field and cone in a way that better applied to the given formula but once again ran into some nasty integrals.

Suggestions?
Phys.Org News Partner Science news on Phys.org
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history
HallsofIvy
#2
Apr29-08, 05:23 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363
If you have x= f(y,z) rather than z= f(x,y), then just swap x and z!
With x= f(y,z), x- f(y,z)= 0 can be treated as a level curve for a function whose gradient is then normal to the surface. That is, [itex](\vec{i}- \partial f/\partial y \vec{j}- \partial f/\partial z\vec{k})dydz[/itex] is the vector differential of surface area.

You can parametrize [itex]x= \sqrt{y^2+ z^2}[/itex] by taking [itex]y= r cos(\theta), z= r sin(\theta)[/itex]- in other words, use polar coordinates but in the yz-plane rather than the xy-plane. The equation of the cone becomes x= r so the position vector for any point on the plane is [itex]\vec{r}= r\vec{i}+ r cos(\theta)\vec{j}+ r sin(\theta)\vec{k}[/itex].

Then [itex]\vec{r}_r= \vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}[/itex], [itex]\vec{r}_\theta= -r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}[/itex] and the "fundamental vector product, their cross product, is [itex]r\vec{i}- r cos(\theta)\vec{j}- r sin(\theta)\vec{k}[/itex]. Since you said the "flux out of the closed cone", you want this oriented by the outward pointing normals and so the vector differential of surface area is [itex]d\vec{\sigma}= (-\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k})r dr d\theta[/itex].

In this parameterization, [itex]F(x,y,z)= x^2\vec{i}+ z\vec{j}+ y\vec{k}= + r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}[/itex]. I think that gives a very simple integral.

Oh, and since this is a "closed cone", don't forget to integrate over the x= 1 base.


Register to reply

Related Discussions
Flux through a Cone Introductory Physics Homework 12
Electric Flux question [Surface integral] Introductory Physics Homework 3
Flux integral help Calculus & Beyond Homework 0
Double integral of mass of circular cone Calculus & Beyond Homework 6
Flux through surface of half a cone Introductory Physics Homework 2