1)

, . Calculate the flux of the vector field out through the z-axis.

2)

, and the vector field are given. Calculate the flux of through S in the direction (the z part of the normal is > 0)

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- Sep 24th 2008, 11:15 AMSpec[SOLVED] Surface integrals and flux
1)

, . Calculate the flux of the vector field out through the z-axis.

2)

, and the vector field are given. Calculate the flux of through S in the direction (the z part of the normal is > 0) - Sep 29th 2008, 02:41 AMSpec
No one? (Worried)

Both of them are tricky since the flux is not defined in the origin. I know how to solve problems like these, but the integrals I end up with here are very difficult. If you use the divergence theorem, you end up with:

If you can somehow use a circle as the inner surface you'll end up with easier line integrals. - Sep 29th 2008, 01:00 PMLaurent
This theorem expresses the flux through a

*closed*surface; one idea could be to close the surfaces first, but it looks tricky.

Why not use the definition of the flux as a surface integral?

As far as I can remember, if a surface is given by a parametrization , a normal vector at the point is (partial derivatives with respect to and ), and the "infinitesimal surface element" at this point is , so that the flux of through the surface (oriented with the above normal vector) is: .

It turns out that the integrand has a simple expression in both of your questions, with appropriate parametrizations.

For 2) [the surface is a piece of cone, draw it], a parametrization by is possible but leads to an integral that requires a polar change of coordinates, so let's use polar coordinate since the beginning: where and . Using this, I get (this is oriented towards the greater values of , as wanted), so that the flux is , etc.

For 1), the problem is not very clear: what does the flux "through the z-axis" mean? Let's suppose it is the flux outing the piece of "cylinder". I use the following parametrization of the ellipse: , . Then gives (this is directed to the "outside") and the flux is (up to errors) , and there are standard methods to compute that (my calculator pretends it is ).

I hope there aren't too many errors in my computations. At least, the idea is there.

Laurent. - Sep 29th 2008, 02:06 PMSpec
That's as far as I got as well, but I can't solve that integral. Is that really a standard integral? And I'm well aware that the divergence theorem requires a closed surface – That's why I mentioned the line integrals you need to subtract because the surface wasn't originally closed.

- Sep 29th 2008, 02:30 PMLaurent
Then you should perhaps have asked for the integral instead... Anyway, the usual method for integrating rational expressions in trigonometric functions is to procede to the change of variable , which transforms the problem into integrating a rational function.

This gives and . Start writing the integral as four times the integral on to avoid problems with the image of the interval by the change of variable. I let you try that. If you can integrate rational functions, then you can do it (but I don't pretend it is quick and simple, I did not give it a try).

(And I'm not sure about line integrals being possibly involved in any way, but it doesn't matter) - Sep 30th 2008, 02:03 AMLaurent
Just an additional hint for the integral (to compute it simply): it leads to . You have and the partial fraction decomposition is . Now write and procede to the change of variable so you find arctangents.