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Math Help - Ellipsoid Surface Integral

  1. #1
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    Ellipsoid Surface Integral

    I'm trying to do the following question: http://www.maths.cam.ac.uk/undergrad...erNST_IA_2.pdf Q7D and having some problems understanding what is going on in the last part of the question. The given surface area element is given by dS = (x/a^2z, y/b^2z,1) dxdy, yet surely this only suffices for the base of the elipsoid; for the rest z will be changing. I have a feeling I may need to use cylindrical polar coordinates to solve this problem? Or maybe I should use the divergence theorem to calculate F.ds over the closed surface and then evaluate it over the base and take that away?

    Many thanks!
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  2. #2
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    Quote Originally Posted by StaryNight View Post
    I'm trying to do the following question: http://www.maths.cam.ac.uk/undergrad...erNST_IA_2.pdf Q7D and having some problems understanding what is going on in the last part of the question. The given surface area element is given by dS = (x/a^2z, y/b^2z,1) dxdy, yet surely this only suffices for the base of the elipsoid; for the rest z will be changing.
    Irrelevant! This is a two dimensional surface and so can be written in terms of two parameters. Here, they have chosen to use x and y as those parameters- z is then a function of x and y. In fact, here's how you could have calculated that:
    from x^2/a^2+ y^2/b^2+ z^2= 1, we can solve for z: z= (1- x^2/a^2- y^2/b^2)^{1/2} and write the "vector equation" of the surface- r(x,y)= <x, y, z>= <x, y, (1- x^2/a^2- y^2/b^2)^{1/2}> in terms of the two "parameters" x and y.
    Now, differentiate that with respect to x and y:
    r_x= <1, 0, -x/a^2(1- x^2/a^2- y^2/b^2)^{-1/2}>= <1, 0, -x/(a^2z)>
    r_y= <0, 1, -y/b^2(1- x^2/a^2- y^2/b^2)^{-1/2}>= <0, 1, -y/(b^2z)>

    Take the cross product of those two vectors., the "fundamental vector product" for this surface
    <x/(a^2z), y/(a^2z), 1>
    That, multiplied by the differentials of the parameters, x and y, gives
    <x/(a^2z), y(a^2z), 1> dxdy as the problem says.

    I have a feeling I may need to use cylindrical polar coordinates to solve this problem? Or maybe I should use the divergence theorem to calculate F.ds over the closed surface and then evaluate it over the base and take that away?
    You are completely misunderstanding the problem and making it much harder than it is. You are NOT to integrate over the ellipsoid or even the top part of it. You are to integrate over the region S with is the part of the ellipsoid directly above the square bounded by x= 0, y= 0, x= 1, y= 1. That does NOT make a closed surface even adding the "base" (by which I take it you mean the ellipse where the ellipsoid passes through the xy plane) so the divergence theorem does not apply. And since the are of integration is a square, with no circular symmetry, cylindrical coordinates would make no sense at all.

    Just do what they say:
    integrate F dot dS as given, with limits of integration, x= 0 to 1, y= 0 to 1.

    Many thanks!
    Last edited by HallsofIvy; April 26th 2011 at 04:03 PM.
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    Quote Originally Posted by HallsofIvy View Post
    Irrelevant! This is a surface and so can be written in terms of two parameters. Here, they have chosen to use x and y as those parameters- z is then a function of x and y.


    You are completely misunderstanding the problem and making it much harder than it is. You are NOT to integrate over the ellipsoid or even the top part of it. You are to integrate over the region S with is the part of the ellipsoid directly above the square bounded by x= 0, y= 0, x= 1, y= 1. That does NOT make a closed surface even adding the "base" (by which I take it you mean the ellipse where the ellipsoid passes through the xy plane) so the divergence theorem does not apply. Just do what they say:
    integrate F dot dS as given, with limits of integration, x= 0 to 1, y= 0 to 1.

    Many thanks!
    [/QUOTE]

    Thanks for your reply. Should I also write z in terms of x and y and substitute this in? Then I just have a double integral in x and y. I definitely did misunderstand the point of the question - I haven't seen anything similar to this before.
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    Quote Originally Posted by StaryNight View Post

    Thanks for your reply. Should I also write z in terms of x and y and substitute this in? Then I just have a double integral in x and y. I definitely did misunderstand the point of the question - I haven't seen anything similar to this before.
    Yes, you are given that dS(vector) is <x/(a^2z), y/(b^2z), 1>dx dy and that F(vector)= <-y, x, 0>. The dot product of those vectors is
    (-xy/(a^2z)- xy(b^2z)) dx dy= -(1/a^2+ 1/b^2)(xy/z) dx dy and that is what you want to integrate. Of course, z=(1- x^2/a^2- y^2/b^2) so your integral is
    -(a^2+ b^2/(a^2b^2) int_0^1 int_0^1 xy/(1- x^2/a^2- y^2/b^2)^{1/2} dxdy
    I think there is an obvious substitution that will simplify the first integral.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    I think there is an obvious substitution that will simplify the first integral.
    I have tried x=asinu but I cannot get this to work.
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    To clarify I am now trying to calculate the double integral:



    Somebody suggested using a substitution, but I've never done this for a double integral before (only a change of coordinate system). Could somebody suggest what might work?

    Thanks
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  7. #7
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    One of the very first substitutions you should have learned in your first course in integration is that is you have a function involving [itex]x^2+ constant[/itex] and there is an x multiplying dx, try the substitution [itex]u= x^2+ constant[/itex].

    Here, let u= 1- y^2/b^2)- x^2/a^2. Then [itex]du= -(2/a^2)xdx[/itex] or [itex]xdx= -(a^2/2)du[/tex]. Then
    [tex]\int \frac{xy}{\sqrt{1- \frac{x^2}{a^2}- \frac{a^2}{b^2}}dx= -\frac{a^2y}{2}\int \frac{du}{\sqrt{u}}= -\frac{a^2y}{2}\int u^{-1/2}du[/itex].

    Of course, the limits of itegration will become complicated: when x= 0, [itex]u= 1- y^2/b^2[/itex] and when x= 1, [itex]u= 1- y^2/b^2- 1/a^2}[/itex]. That's to be expected when the area of integration doesn't have any simple geometric relation to the ellipsoid.
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