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..
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!
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 . Then by revolving it round the x-axis, the resulting solid has volume
.
Here, - 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?