# multiple integrals

• Apr 7th 2006, 03:07 PM
comp_engr
multiple integrals
I have been struggling with the following problem involving multiple integrals.
The question is "Find the volume of a right, circular cone of radius r and height h" a) as a single integral. b) as a double integral. c) as a triple integral.

The problem I have with multiple integrals is that I usually cannot figure out the limits to integrate over. With this problem no functions are given so I'm not sure where to start. Any help would be greatly appreciated.
Thanks!
• Apr 8th 2006, 02:20 PM
comp_engr
ok so I am pretty sure I solved part "a"-- pie[(h-z)r/h]^2 dz evaluated from 0 to h

Can anyone help me with the double and triple integrals?
I would asume that I am supposed to get 1/3pie r^2 h as the solution to all three integrals..
• Apr 9th 2006, 08:33 AM
TD!
A decent sketch helps to determine the limits, what have you tried so far?
• Apr 10th 2006, 11:26 AM
Rebesques
Well, let's see.

For the single integral, you can use the simple formula to give the volume of a solid of revolution. Say your function is $\displaystyle f(x), \ x\in[a,b]$. Then by revolving it round the x-axis, the resulting solid has volume
$\displaystyle \pi\int_{a}^{b}f^2(x)dx$.

Here, $\displaystyle f(x)=\frac{r}{h}x, \ 0\leq x\leq h$ - revolve this to get the cone!

For the double integral, you can use Stokes' theorem, and for the triple integral, Gauss theorem. Are you familiar with these?
• Apr 11th 2006, 07:22 AM
comp_engr
actually just learning Stoke's theorm and I can now see the application. hmm very interesting! Thank you!