Proving area of cone.

craigmain · #1 $Old$ 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.

malaygoel · #2 $Old$ July 4th, 2009, 09:38 AM

I am finding it hard to visualise how your integral gives the volume of cone...are you sure that you set up the right integral?

craigmain · #3 $Old$ July 4th, 2009, 09:52 AM

Hi,

I read the problem incorrectly, it was a right circular cone, not a vertical cone, sorry to waste your time.

The radius is ${(cb)}^2$ and the altitude is $b$ giving the correct formula.

Thanks
Regards
Craig.