Find the flux of this vector field

asi123
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Homework Statement



I need to find the flux of this vector field (in the pic) that goes through this plan (in the pic) and z goes from 0 to 1.
How am I suppose to do that?


Homework Equations





The Attempt at a Solution

 

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The surface is z= \sqrt{x^2+ y^2}, the upper nappe of a cone. In cylindrical coordinates, that is z= r. So good parametic equations would be x= r cos(\theta), y= r sin(\theta), z= r which means that the vector equation would be
\vec{r}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r \vec{k}. Since z goes from 0 to 1, r goes from 0 to 1 and theta, of course, from 0 to 2\pi.

Does that help?
 
HallsofIvy said:
The surface is z= \sqrt{x^2+ y^2}, the upper nappe of a cone. In cylindrical coordinates, that is z= r. So good parametic equations would be x= r cos(\theta), y= r sin(\theta), z= r which means that the vector equation would be
\vec{r}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r \vec{k}. Since z goes from 0 to 1, r goes from 0 to 1 and theta, of course, from 0 to 2\pi.

Does that help?

Check this out, is that what you meant?
 

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Once you have done the cross product, and the integral, yes, that should be correct. I would recommend you do the integrations with respect to \theta first. Most of those trig functions, integrated from 0 to 2\pi will give 0.
 
HallsofIvy said:
Once you have done the cross product, and the integral, yes, that should be correct. I would recommend you do the integrations with respect to \theta first. Most of those trig functions, integrated from 0 to 2\pi will give 0.

10x a lot.
 
asi123 said:
10x a lot.

10-4 good buddy!
 
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