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Homework Help: Vector calculus question - surface of ellipsoid

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]E[/itex] be the ellipsoid


    where [itex]a>\sqrt{2}[/itex] and [itex]b>\sqrt{2}[/itex]. Let S be the part of the surface of [itex]E[/itex] defined by [itex]0\le x\le1, 0\le y\le1, z>0[/itex] and let [itex]\mathbf{F}[/itex] be the vector field defined by [itex]\mathbf{F}=(-y,x,0)[/itex]. Given that the surface area element of [itex]S[/itex] is given by

    [tex]d \mathbf{S} = \left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 } \right) dxdy [/tex]

    find [itex]\int_S\mathbf{F}.d\mathbf{S}[/itex] in the case [itex]a/ne b[/itex]

    2. Relevant equations
    Scalar product

    3. The attempt at a solution
    \int_S\mathbf{F}.d\mathbf{S}=\int_S(-y,x,0).\left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 }\right) dxdy



    It's at this point I'm not sure what to do with the parameter [itex]z[/itex]. I tried continuing, treating [itex]z[/itex] as constant to get

    \int_S \mathbf{F}.d\mathbf{S}= \frac{1}{4z} \left({ \frac{1}{b^2}- \frac{1}{a^2}}\right)

    but don't like the fact there is a [itex]z[/itex] there? Would converting to spherical coordinates help? If so, how would you do it?

    With very many thanks,

    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2


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    Science Advisor

    Since you know z is not a constant, it makes no sense at all to "treat z as constant".

    You are told that
    [tex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ z^2= 1[/tex]
    [tex]z= \pm\sqrt{1- \frac{x^2}{a^2}- \frac{y^2}{b^2}[/tex]
    Do the z> 0 and z< 0 parts separately.

    I would NOT use spherical coordinates but a variation on cylindrical coordinates might help. Taking z= 0, we see that the ellipsoid projects to the ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex]. Let [itex]x= a rcos(t)[/itex], [itex]y= b rsin(t)[/itex]. The Jacobian is [tex]\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial t}\end{array}\right|= \left|\begin{array}{cc}a cos(t) & -a rsin(t) \\ b sin(t) & b rcos(t)\end{array}\right|= ab cos^2(t)+ ab sin^2(t)= ab[/tex]
    so that [itex]dxdy= ab dr dt[/itex]
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