Finding the flux (Divergence Theorem)

In summary, the conversation is about using the divergence theorem to find the flux of a vector field out of the surface of a paraboloid. The conversation includes a discussion about integrating each slice over a changing radius and finding the maximum radius for a given z, as well as a question about the surface integral.
  • #1
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Homework Statement


By using divergence theorem find the flux of vector F out of the surface of the paraboloid z = x^2 + y^2, z<=9, when F = (y^3)i + (x^3)j + (3z^2)k


Homework Equations


Divergence theorem equation stated in the attempt part


The Attempt at a Solution


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  • #2
Suppose you are doing the z-integral first, so we are looking at a slice of fixed z. Then the slice looks like a disk of radius rmax. Instead of 0 to 3, you want to integrate each slice over r from 0 to rmax. You can express rmax in terms of z.
So you will get
[tex]\int_0^9 \int_0^{r_\mathrm{max}(z)} \int_0^{2\pi} 6 z r \, \mathrm{d}\varphi \, \mathrm{d}r \, \mathrm d{z}[/tex]
where rmax(z) depends on z instead of being identically equal to 3 as you have now.
 
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  • #3
CompuChip said:
Suppose you are doing the z-integral first, so we are looking at a slice of fixed z. Then the slice looks like a disk of radius rmax. Instead of 0 to 3, you want to integrate each slice over r from 0 to rmax. You can express rmax in terms of z.
So you will get
[tex]\int_0^9 \int_0^{r_\mathrm{max}(z)} \int_0^{2\pi} 6 z r \, \mathrm{d}\varphi \, \mathrm{d}r \, \mathrm d{z}[/tex]
where rmax(z) depends on z instead of being identically equal to 3 as you have now.
I understand what you mean about the radius changing with z, but how would i integrate dr with limits rmax(z)?
Also what about the Fn ds double integral?
 
  • #4
For some given z, what is the maximum radius (i.e. the upper boundary of your r-integration)?

I haven't looked into the other approach (where you first apply the divergence theorem) but it does look a bit more complicated to me (finding the correct normal unit vector and all).
 
  • #5
Just thought id bump this question. I understand my error when calculating the volume integral, the problem I'm left with is the surface integral. I still can't spot the mistake there. Any suggestions? Thanks.
 

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