Geometric understanding of integration / surface area of sphere

  • Thread starter Saraphim
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  • #1
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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. :-)
 

Answers and Replies

  • #2
tiny-tim
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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.
 
  • #3
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It didn't even occur to me to stop and consider the meaning of the differential. :blushing:

Thanks a lot. :)
 

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