That seems to me to be the hard way to integrate.
take a small rectangular box dV somewhere along the height.
dV= dh*darea
darea= dw dl=(H-h)(b)*(H-h)2b /H^2
dV=1/H^2 (H-h)^2 2b^2 dh h from 0 to H
V= 2b^2/H^2 INT (H-h)^2 dh= 2b^2 /3 H^2 *(H-h)^3 from zero to H
=2b^2/3 H
which you knew, the volume of any rectangular prism is 1/3 area base*height
Find the volume of a pyramid with height h and rectangular base with dimensions b and 2b.
Essentially, I'm trying to derive the volume of a pyramid using integrals and a graph and revolving the plane about an axis. I hope that's kind of clear and you guys understand what I'm asking.
Thank you very much
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