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Geometric understanding of integration / surface area of sphere

  1. May 6, 2010 #1
    Hi everyone,

    I've browsed around the forum a bit and found that others have had the same problem as me, however, none of the answers help me a lot, so I thought to post a more specific question, I hope you don't mind.

    I'm having a problem with the surface area of a sphere, probably because I'm extending "backwards", ie. if the derivative of sphere volume is sphere surface, why is the integral over the circumference of a circle not the surface area of a sphere? More formally, why is this:

    [tex]2\int_0^R 2\pi r dr \neq 4\pi R^2[/tex]

    I know plenty of other ways of obtaining the surface area of a sphere, also through integration, but I simply lack the geometric understanding to show me why this particular integral does NOT describe the surface area of a (hemi)sphere.

    Can anyone enlighten me? Bear in mind, I am by no means very good at math. I know a bunch of tools, and how to use them, but my understanding of them is vague at best, and that is what I am trying to rectify.

    Thank you in advance. :-)
  2. jcsd
  3. May 6, 2010 #2


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    Hi Saraphim! :smile:

    2πrdr is the area of a ring of inner radius r and outer radius r + dr.

    So ∫0R 2πrdr is the area of a whole disc of radius R.

    To find the surface area of a sphere, you would divide the sphere into tiny slices, but they would all have the same r, wouldn't they? :wink:

    (the expression inside the ∫ would be the area of a typical tiny slice)

    They would have different θ instead, and you would integrate over θ.
  4. May 6, 2010 #3
    It didn't even occur to me to stop and consider the meaning of the differential. :blushing:

    Thanks a lot. :)
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