Find Volume of Right Circular Cone: Setup Integrals

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Homework Help Overview

The problem involves finding the volume of a right circular cone with radius r and height h using different types of integrals: a single integral, a double integral, and a triple integral. The original poster expresses difficulty in setting up these integrals, particularly in the absence of explicit functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss finding the area of a cross-section of the cone as a function of height and suggest using this to set up a single integral. There is also mention of using cylindrical coordinates for the triple integral and the need to determine appropriate bounds for the integrals.

Discussion Status

Some participants have provided suggestions for setting up the single integral and have confirmed its correctness. There is ongoing exploration of the double and triple integrals, with participants sharing thoughts on how to approach the bounds and the geometric interpretation of the problem.

Contextual Notes

The original poster notes a general difficulty with problems lacking explicit functions and expresses uncertainty about visualizing the problem in three dimensions.

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

Then [tex]V=\int_0^h A(z)dz[/tex]
 
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
 
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 :rolleyes: and you could use the expression to find a geometrical interpretation.
 
hmm its still not very clear to me. I am having lots of trouble thinking in 3d..
 
Your integral will just be [itex]\int_{cone} \int \int dV[/itex] 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.
 

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