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Double integral and polar coord

  1. Mar 15, 2005 #1
    Please help. Thank you.
     

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    Last edited: Mar 16, 2005
  2. jcsd
  3. Mar 15, 2005 #2
    Alright. The unit disk is

    [tex]S = \{(x, \ y) | x^2 + y^2 \leq 1 \}[/tex]

    Changing to polar coordinates,

    [tex] S = \{ (\rho \cos{\theta}, \ \rho \sin{\theta}) | \rho \leq 1, \ 0 \leq \theta \leq 2\pi \}[/tex]

    ie. [tex] x = \rho \cos{\theta}, \ y = \rho \sin{\theta} [/tex].

    Thus

    [tex] \int \int_S xy\sqrt{x^2 + y^2} dA = \int_0^{2\pi} \int_0^1 \rho \sin{\theta}\rho \cos{\theta} \sqrt{\rho^2 \sin^2{\theta} + \rho^2 \cos^2{\theta}} \ \left|\frac{\partial (x, \ y)}{\partial(\rho, \ \theta)}\right| \ d\rho \ d\theta = \int_0^{2\pi} \int_0^1 \rho^4\cos{\theta}\sin{\theta} \ d\rho \ d\theta[/tex]

    I trust you can work out the rest.

    I will note that it is quite easy to evaluate this integral without actually doing any calculations, by appealing to symmetry.
     
    Last edited: Mar 15, 2005
  4. Mar 15, 2005 #3
    What if D is a closed disk with radius 1 and center (1,0).
    That would make the polar coord. x=(r cos(theta)-1) and y=(r sin (theta)).
     
    Last edited: Mar 15, 2005
  5. Mar 16, 2005 #4
    It's actually still the same answer. You can make the same symmetry argument.

    Here's a hint:

    Obviously, if [tex]f[/tex] is [tex]2\pi[/tex]-periodic, then [tex] \int_0^{2\pi} f(x)\sin{x} \ dx = \int_{-\pi}^{\pi} f(x)\sin{x} \ dx[/tex]. Thus if [tex]f[/tex] is an even function, since [tex]\sin[/tex] is odd, then [tex]f(x)\sin{x}[/tex] is odd and

    [tex]\int_0^{2\pi} f(x)\sin{x} \ dx = 0[/tex]
     
    Last edited: Mar 16, 2005
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