# Flux Integral Help through abnormal cone

by mgibson
Tags: abnormal, cone, flux, integral
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,533 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.