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Surface integrals to derive area of sphere

  1. May 26, 2014 #1
    Given a sphere [tex] x^2 + y^2 + z^2 = a^2[/tex] how would I derive the surface area by using surface integrals?

    The method I've tried is as follows: [tex]dA = sec\ \gamma \ dxdy[/tex] where gamma is the angle between the tangent plane at dA and the xy plane. [tex]sec \gamma = \frac{|\nabla \varphi|}{\partial \varphi /\partial z}[/tex][tex] = \frac{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}{2z}[/tex][tex] = \frac{a}{\sqrt{a^2 - x^2 - y^2}}[/tex]

    converting to polar coordinates and integrating the expression for [itex]sec \gamma[/itex][tex]\int^{a}_{0}\int^{2 \pi}_{0} \frac{a}{\sqrt{a^2 - r^2}} r \ drd \theta[/tex] and using substitution [itex] u = a^2 - r^2[/itex] yields [tex] -a \pi [2 \sqrt{a^2 - a^2} - 2 \sqrt{a^2 - 0}][/tex][tex] = 2 \pi a^2[/tex] Obviously the area of a sphere is [itex] 4 \pi a^2[/itex] so did I do the calculation wrong, or does this method only find the area of one hemisphere, in which case I should multiply the answer by 2?

    Thanks in advance
  2. jcsd
  3. May 26, 2014 #2


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    Yes, you only did the top half of the sphere. It would be much easier using spherical coordinates where$$
    dS = a^2\sin\phi~ d\phi d\theta$$
  4. May 26, 2014 #3
    Thanks, I thought as much. I know how to do it using the spherical polars, but I was just curious as to whether the surface integral method worked as well.

    Every day's a school day. Thanks again!
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