Find Volume of Right Circular Cone: Setup Integrals

[sunglasboy]

I am currently having trouble solving this problem "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.

My difficulty lies in setting up the integrals. I usually have trouble with problems that do not give any functions like this one. Once an integral is setup, I am pretty decent at doing the actual integration.
For this problem I am pretty sure all three integrals should yield 1/3pier^2h
Any help on this would be greatly appreciated.

 

[Galileo]

For the first, find the area of a cross-section perpendicular to the cone-axis as a function of height z.

Then V=\int_0^h A(z)dz

 

[sunglasboy]

thanks for the suggestion. So basically my single integral would be-
pie[(h-z)r/h]^2 dz evaluated from 0 to h ?
Is this correct?
Still unsure how to proceed with the double and triple integrals.
Thanks

 

[Galileo]

You got the first integral right.

The triple integral is not hard. I would choose cylindrical coordinates to parametrize the cone. The only hard stuff is figuring out the bounds on the integral.

For the double integral you have to figure out something similar. You could for example do something like rotate the graph of a line around an axis to get your cylinder. Or notice that once you've integrated one iterated integral in the triple integral, you're left with a double integral and you could use the expression to find a geometrical interpretation.

 

[sunglasboy]

hmm its still not very clear to me. I am having lots of trouble thinking in 3d..

 

[Galileo]

Your integral will just be \int_{cone} \int \int dV right? The only problem will be finding the bounds.
If you take an arbitrary point (r,theta,z) in your cone, by fixing theta and z, how can you vary r? (It will depend on z). If z=0 you can vary r from 0 to R. If z=h/2, r can vary between 0 and R/2 etc. This will tell you the bounds which you'll need if you integrate wrt to r first.

And draw a picture.