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Geometry of a Frustum

  1. Nov 1, 2011 #1

    ElijahRockers

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    when calculating the surface area of curves rotated about axes, you supposedly integrate 2∏f(x)√(1+f '(x)2) dx. This has been explained to me as being derived from the geometry of a frustum, and I understand how the integration works, but I am confused as the process of deriving this formula - actually churning out the derivation. I feel like most explanations I read on the geometry are skipping a step or two, and it's confusing me.

    http://www.analyzemath.com/Geometry/conical_frustum.html

    The above link seems to have the most concise method, and I understand how to find all the required measurements in terms of r, R, and H, but they don't actually show how to get the surface area in terms of r, R, and H. I tried a bit of circle geometry to figure it out, but the math started getting pretty hairy.

    I figured the surface area of the frustum should be the area of the 'complete' sector minus the area of the 'little' sector. perhaps I still don't completely understand the area of a circle yet?

    any insight or links appreciated.
     
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