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Surface Integrals in Polar Coordinates

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the area cut from the surface z = 2xy by the cylinder x^2 + y^2 = 6.
    [Hint: Set up the integral using rectangular coodinates, then switch to polar coordinates.]

    2. Relevant equations
    [tex]
    A = \iint \sqrt{{z_x}^2+{z_y}^2+1}dxdy = \iint \sqrt{{r^2}+{r^2}{z_r}^2+{r^2}{z_\theta}^2} dr d\theta
    [/tex]


    3. The attempt at a solution
    [tex]

    \begin{gather*}
    z_x = 2y\\
    z_y = 2x\\
    \\
    A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{y^2}+4{x^2}+1} dx dy\\\\
    A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{r^2}+1} dx dy\\\\
    A = \int_{0}^{2\pi} \int_{0}^{\sqrt{6}} \sqrt{4{r^2}+1} dr\frac{\partial x}{\partial r} d\theta \frac{\partial y}{\partial\theta} = \int_{0}^{2\pi} \int_{0}^{\sqrt{6}} r {\cos}^2\theta \sqrt{4{r^2}+1} dr d\theta\\\\
    \end{gather*}
    [/tex]

    From there, I can work the problem down, but I'm not sure if my conversion from dx -> dr, or dy -> d theta is even a valid operation. I don't see a ready way to do this problem if I start straight in with polar (cylindrical) coordinates, since the algebra quickly becomes very complex. I guess what I need to know, is how do I appropriately convert dx*dy -> dr*dtheta?

    Any help appreciated!
     
    Last edited: Nov 29, 2009
  2. jcsd
  3. Nov 29, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The element of area dx*dy is r*dr*dtheta in polar coordinates. In a problem like this, you don't try to derive it. You just remember it and substitute.
     
  4. Nov 29, 2009 #3
    D'oh! Thanks for pointing that out :P
     
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