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Math Help - Question about surface elements in polar coordinates.

  1. #1
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    Question about surface elements in polar coordinates.

    I was given the equation z = x^2 - y^2 and was told to find the surface element for the surface z = x^2 - y^2 in polar coordinates. So far I have converted the equation to polar with

    z = r^2 cos(theta)^2 - r^2 sin(theta)^2.

    I simplified that to
    r^2(2cos(theta)^2 - 1) = z

    Now i think the next step would be to get 2 none parallel vectors to take the cross product of to find the surface element. I am not entirely sure though which directions I need to do them in. Would I take one holding r constant and one holding theta constant? I don't have a a lot of experience with surface integrals or polar coordinates so this has been a bit confusing and any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Enkie View Post
    I was given the equation z = x^2 - y^2 and was told to find the surface element for the surface z = x^2 - y^2 in polar coordinates. So far I have converted the equation to polar with

    z = r^2 cos(theta)^2 - r^2 sin(theta)^2.

    I simplified that to
    r^2(2cos(theta)^2 - 1) = z

    Now i think the next step would be to get 2 none parallel vectors to take the cross product of to find the surface element. I am not entirely sure though which directions I need to do them in. Would I take one holding r constant and one holding theta constant? I don't have a a lot of experience with surface integrals or polar coordinates so this has been a bit confusing and any help would be greatly appreciated.
    So you can parametrize the surface using the trivial parameterization

    \vec{s}(x,y)=x\vec{i}+y\vec{j}+z\vec{k}= x\vec{i}+y\vec{j}+f(x,y)\vec{k}=x\vec{i}+y\vec{j}+  (x^2-y^2)\vec{k}

    Now you can use the polar transfrom
    x=r\cos(\phi)y=r\sin(\phi)

    This gives

    \vec{s}(r,\phi)=(r\cos(\phi))\vec{i}+(r\sin(\phi))  \vec{j}+(r^2(\cos^2(\phi)-\sin^2(\phi)))\vec{k}

    So the surface area element is given by

    \displaystyle \bigg|\frac{\partial \vec{s}}{\partial r} \times \frac{\partial \vec{s}}{\partial \phi}\bigg|drd\phi

    The last statement is the determinant of the cross product the two vectors.
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  3. #3
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    Ok well I am going to need some time to think this over. So far we have been doing most everything using r hat phi hat and z hat but our textbook is an online one developed at OSU with very few examples which makes these problems difficult to work sometimes when the only other source readily available is stewarts book which is almost all rectangular.
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