[SOLVED] Surface integrals and flux

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September 24th 2008, 11:15 AM
Spec

[SOLVED] Surface integrals and flux

1)
$S: x^2 + 4y^2 = 4$ , $0 \leq z \leq 1$ . Calculate the flux of the vector field $\overline{A}(x,y,z) = \left(\frac{-6x}{x^2+y^2},\frac{-6y}{x^2+y^2}, z+1\right)$ out through the z-axis.

2)
$S: z=1 - \sqrt{x^2 +y^2}$ , $0 \leq z \leq \frac{1}{2}$ and the vector field $\overline{A}(x,y,z)=\frac{1}{\sqrt{x^2+y^2}}(x, 1, 0)$ are given. Calculate the flux of $\overline{A}$ through S in the direction $N \cdot ze_z > 0$ (the z part of the normal is > 0)
September 29th 2008, 02:41 AM
Spec

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:

$\int \int \int dxdydz$

If you can somehow use a circle as the inner surface you'll end up with easier line integrals.
September 29th 2008, 01:00 PM
Laurent

Quote:

Originally Posted by Spec

If you use the divergence theorem

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 $(u,v)\mapsto \phi(u,v)$ , a normal vector at the point $\phi(u,v)$ is $\partial_u\phi\wedge\partial_v\phi$ (partial derivatives with respect to $u$ and $v$ ), and the "infinitesimal surface element" at this point is $\|\partial_u\phi\wedge\partial_v\phi\|$ , so that the flux of $\vec{A}$ through the surface (oriented with the above normal vector) is: $\int\int \vec{A}\cdot (\partial_u\phi\wedge\partial_v\phi)du\,dv$ .

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 $(x,y)$ is possible but leads to an integral that requires a polar change of coordinates, so let's use polar coordinate since the beginning: $(r,\theta)\mapsto\phi(r,\theta)=(r\cos\theta,r\sin \theta,1-r)$ where $0\leq\theta< 2\pi$ and $\frac{1}{2}\leq r\leq 1$ . Using this, I get $\partial_r \phi\wedge\partial_\theta \phi=(r\cos\theta,r\sin\theta,r)$ (this is oriented towards the greater values of $z$ , as wanted), so that the flux is $\int_{1/2}^1\int_0^{2\pi} \frac{1}{r}(r^2\cos^2\theta+r\sin\theta)d\theta\,d r=\int_{1/2}^1\int_0^{2\pi}(r\cos^2\theta+\sin\theta)d\theta \,dr$ $=\int_{1/2}^1\int_0^{2\pi}r\cos^2\theta d\theta\,dr=\int_{1/2}^1 rdr\int_0^{2\pi}\cos^2\theta d\theta$ , 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: $x=2\cos\theta$ , $y=\sin\theta$ . Then $\phi(\theta,z)=(2\cos\theta,\sin\theta,z)$ gives $\partial_\theta\phi\wedge\partial_z\phi=(\cos\thet a,2\sin\theta,0)$ (this is directed to the "outside") and the flux is (up to errors) $\int_0^{2\pi}\int_0^1 \frac{-6\cos^2\theta-6\sin^2\theta}{4\cos^2\theta+\sin^2\theta}dz\,d\th eta=-6\int_0^{2\pi}\frac{d\theta}{1+3\cos^2\theta}$ , and there are standard methods to compute that (my calculator pretends it is $-6\pi$ ).

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

Laurent.
September 29th 2008, 02:06 PM
Spec

Quote:

Originally Posted by Laurent

$...=-6\int_0^{2\pi}\frac{d\theta}{1+3\cos^2\theta}$ , and there are standard methods to compute that (my calculator pretends it is $-6\pi$ )

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.
September 29th 2008, 02:30 PM
Laurent

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 $t=\tan\frac{\theta}{2}$ , which transforms the problem into integrating a rational function.

This gives $\cos\theta=\frac{1-t^2}{1+t^2}$ and $d\theta=\frac{2dt}{1+t^2}$ . Start writing the integral as four times the integral on $[0,\frac{\pi}{2}]$ 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)
September 30th 2008, 02:03 AM
Laurent

Just an additional hint for the integral (to compute it simply): it leads to $2\int_0^1 \frac{t^2+1}{1-t^2+t^4}dt$ . You have $1-t^2+t^4=(t^2-1)^2-3t^2=(t^2-1+t\sqrt{3})(t^2-1-t\sqrt{3})$ and the partial fraction decomposition is $\frac{t^2+1}{1-t^2+t^4}=\frac{1}{2}\left(\frac{1}{t^2-t\sqrt{3}+1}+\frac{1}{t^2+t\sqrt{3}+1}\right)$ . Now write $t^2-t\sqrt{3}+1=(t-\frac{\sqrt{3}}{2})^2+\frac{1}{4}$ and procede to the change of variable $t\mapsto u=t-\frac{\sqrt{3}}{2}$ so you find arctangents.