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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

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- Thread starter nb89
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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

Divergence theorem equation stated in the attempt part

- #2

CompuChip

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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 r_{max}. Instead of 0 to 3, you want to integrate each slice over r from 0 to r_{max}. You can express r_{max} 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 r_{max}(z) depends on *z* instead of being identically equal to 3 as you have now.

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 r

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- #3

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I understand what you mean about the radius changing with z, but how would i integrate dr with limits rmax(z)?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 r_{max}. Instead of 0 to 3, you want to integrate each slice over r from 0 to r_{max}. You can express r_{max}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 r_{max}(z) depends onzinstead of being identically equal to 3 as you have now.

Also what about the Fn ds double integral?

- #4

CompuChip

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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).

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