[SOLVED] Surface integrals and flux

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

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

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:

$\displaystyle \int \int \int dxdydz$

If you can somehow use a circle as the inner surface you'll end up with easier line integrals.

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Originally Posted by Spec

If you use the divergence theorem

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

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

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

Laurent.

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Originally Posted by Laurent

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

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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.

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

This gives $\displaystyle \cos\theta=\frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2dt}{1+t^2}$. Start writing the integral as four times the integral on $\displaystyle [0,\frac{\pi}{2}]$ 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)

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