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Area section of a sphere used for integration?

  1. Oct 21, 2008 #1
    I cannot figure out why the area of a section of a sphere used for integration is:

    (r * dtheta) (r * sin(theta) * dphi) ???

    where dtheta and dphi are the differential angles that subtend the arcs, which make up the sides of the rectangle used for the differential area.

    I did an integral of theta from 0 to pi and phi from 0 to 2*pi and the result was the area of a sphere. That is the correct result if the above formula was the correct differential area.

    I see that this formula is made up of the angle that subtends the arc (the dtheta or dphi) multiplied by the length of one of the triangle's legs. The "triangle" that I mention here is the triangle that has two legs from the center of the sphere to the arc that is one side of the area rectangle. I just don't know why this would be the correct differential area and not some other formula?

  2. jcsd
  3. Oct 21, 2008 #2


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    See the wiki page on the change of variable forumla:


    Basically, it says that as you change from cartesian to polar coordinate, in order for the result of the integral to be the same, you have to multiply by the absolute value of the determinant of the jacobian matrix of the transformation map. This is just the multivariable analogue of the know change of variable theorem for functions of 1 variable:

    [tex]\int_g^{-1}(a)^g^{-1}(b)f(g(t))g'(t)dt = \int_a^bf(t)dt[/tex]

    So that r^2sin(theta) from dr * (r * dtheta) (r * sin(theta) * dphi) is the absolute value of the determinant of the jacobian matrix of the transformation map from polar to cartesian coordinate.

    Of course it is also possible to give an informal but intuitively satisfying argument involving "infinitesimal quantities" justifying the formula dr * (r * dtheta) (r * sin(theta) * dphi) for the volume element in polar coordinates. You can find this argument in some classical mechanics textbooks and multivaribale calculus textbooks.
    Last edited: Oct 21, 2008
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