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craigmain · #1 July 4th, 2009, 09:26 AM

Given $f(x)=cx$
I have solved the volume of a cone under the line from zero to b as:

$\int_0^b \pi{(cx)}^2 dx = \pi \frac{c^2b^3}{3}$

I need to show this is a third the area of the base times height.

Surely the radius of the base must be $\frac{b}{2}$ giving the area of the base times the height as.

$\pi {(\frac{b}{2})}^2$ as the base, and
$\frac{cb}{2}$ as the altitude (height), giving

$\frac{\pi cb^3}{8}$ as base times height. Multiplying this by a third doesn't get me back to my integral.