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Math Help - [SOLVED] Surface integrals and flux

  1. #1
    Senior Member Spec's Avatar
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    [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)
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  2. #2
    Senior Member Spec's Avatar
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    No one?

    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.
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  3. #3
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    Quote Originally Posted by Spec View Post
    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.
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  4. #4
    Senior Member Spec's Avatar
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    Quote Originally Posted by Laurent View Post
    ...=-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.
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  5. #5
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    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)
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  6. #6
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    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.
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