Find the flux of this vector field

[asi123]

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

 

[HallsofIvy]

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?

 

[asi123]

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?

 

[HallsofIvy]

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.

 

[asi123]

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.

 

[HallsofIvy]

10x a lot.

10-4 good buddy!