# Flux Integral Help through abnormal cone

1. ### mgibson

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?

2. ### HallsofIvy

40,939
Staff Emeritus
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, $(\vec{i}- \partial f/\partial y \vec{j}- \partial f/\partial z\vec{k})dydz$ is the vector differential of surface area.

You can parametrize $x= \sqrt{y^2+ z^2}$ by taking $y= r cos(\theta), z= r sin(\theta)$- 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 $\vec{r}= r\vec{i}+ r cos(\theta)\vec{j}+ r sin(\theta)\vec{k}$.

Then $\vec{r}_r= \vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$, $\vec{r}_\theta= -r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}$ and the "fundamental vector product, their cross product, is $r\vec{i}- r cos(\theta)\vec{j}- r sin(\theta)\vec{k}$. 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 $d\vec{\sigma}= (-\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k})r dr d\theta$.

In this parameterization, $F(x,y,z)= x^2\vec{i}+ z\vec{j}+ y\vec{k}= + r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}$. 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.