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Math Help - Surface Area of Surface

  1. #1
    Senior Member
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    Surface Area of Surface



    I got part a) b) c) however I am just unsure on part d)

    For a) I got r(\theta, \phi) = \sin(\phi)\cos(\theta)i + \sin(\phi)\sin(\theta)j+\cos(\phi)k

    b) r(x,y) = xi+yj+zk but  z^2+y^2+x^2 = 1 so z = \sqrt{1-(x^2+y^2)}

    So r(x,y) = xi+yj+\sqrt{1-(x^2+y^2)}k

    c) Not required for d)

    d) So area of surface = \int \int_D |r_x \times r_y| dA

    |r_x \times r_y| = \frac{1}{\sqrt{1-x^2-y^2}} (I've double checked this, it's correct )

    So \int \int_D \frac{1}{\sqrt{1-x^2-y^2}} dA

    But what are the bounds for the double integral?

    Thanks!
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  2. #2
    MHF Contributor

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    You are told that \pi/6\le \phi\le \pi/3.

    The bounds for the entire sphere would be 0\le \theta\le 2\pi and 0\le \phi\le \pi.

    With the restriction given, 0\le \theta\le 2\pi and \pi/6\le \phi\le \pi/3, of course.
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