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July 13th, 2017, 04:59 PM   #1
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volume of a right circular cone...

Find the mass of a right circular cone of height h and base radius a if the density is proportional to the distance from the base.

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May I ask what I did wrong? Any tips would be greatly appreciated. Thanks.
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July 13th, 2017, 05:25 PM   #2
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Well, the immediate question that comes to mind is what is $k$?

Oh I see, it's just a proportionality constant.

I think this should be set up as

$\displaystyle \int_0^{2\pi}\int_0^h \int_0^{a\left(1-\frac z h\right)}~k z ~r dr ~dz~d\theta = \dfrac{1}{12} \pi k a^2 h^2$

Last edited by skipjack; July 17th, 2017 at 09:12 PM.
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July 14th, 2017, 04:10 AM   #3
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Quote:
Originally Posted by zollen View Post
Find the mass of a right circular cone of height h and base radius a if the density is proportional to the distance from the base.

Official Answer:


My Solution


May I ask what I did wrong? Any tips would be greatly appreciated. Thanks.
Your error: taking the sides of the cone to be given by z= rh/a, you are representing the cone as having its vertex at the origin and its base at z= h But then you are integrating taking z from 0 to rh/z but that is the region below the cone. Representing the cone as you have it, take the z integral from the sides, z= rh/a up to h:
.

Or, represent the cone as having its base in the xy-plane and its vertex at (0, 0, h). The sides then are given by z= h- rh/a so that your integral is
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July 14th, 2017, 02:19 PM   #4
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Quote:
Originally Posted by Country Boy View Post
Your error: taking the sides of the cone to be given by z= rh/a, you are representing the cone as having its vertex at the origin and its base at z= h But then you are integrating taking z from 0 to rh/z but that is the region below the cone. Representing the cone as you have it, take the z integral from the sides, z= rh/a up to h:
.

Or, represent the cone as having its base in the xy-plane and its vertex at (0, 0, h). The sides then are given by z= h- rh/a so that your integral is
Why did you say 0 <= z <= rh/a is below the cone?? I thought 0 <= z <= rh/a is the 'exact height' of the clone.
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July 14th, 2017, 02:21 PM   #5
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Quote:
Originally Posted by zollen View Post
Why did you say 0 <= z <= rh/a is below the cone?? I thought 0 <= z <= rh/a is the 'exact height' of the clone.
Do you understand my formulation of this integral?

It did produce the answer your book did.
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Last edited by skipjack; July 17th, 2017 at 09:19 PM.
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July 17th, 2017, 02:48 PM   #6
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Do you understand my formulation of this integral?

It did produce the answer your book did.
It takes me some time to review all of the comments and to experiment on my own. Your approach is actually easier to understand than the standard $\displaystyle dz dr d \theta $.

Last edited by skipjack; July 17th, 2017 at 09:17 PM.
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July 17th, 2017, 02:49 PM   #7
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You were correct. I made a mistake. z = rh/a is incorrect.

The correct equation for z = (a - r) * (h / a)

Quote:
Originally Posted by Country Boy View Post
Your error: taking the sides of the cone to be given by z= rh/a, you are representing the cone as having its vertex at the origin and its base at z= h But then you are integrating taking z from 0 to rh/z but that is the region below the cone. Representing the cone as you have it, take the z integral from the sides, z= rh/a up to h:
.

Or, represent the cone as having its base in the xy-plane and its vertex at (0, 0, h). The sides then are given by z= h- rh/a so that your integral is
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July 17th, 2017, 04:55 PM   #8
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Quote:
Originally Posted by zollen View Post
It takes me some time to review all of the comments and to experiment on my own. Your approach is actually easier to understand than the standard $\displaystyle dz dr d \theta $.
You can integrate in any order you like.

If you wish to integrate over $z$ first, you would have

$\displaystyle \int_0^{2\pi} \int_0^a \int_0^{h \left(1- \frac r a \right) }~z r ~dz~dr~d\theta = \dfrac{1}{12} \pi a^2 h^2$

Last edited by skipjack; July 17th, 2017 at 09:18 PM.
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