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 July 13th, 2017, 04:59 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 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. Official Answer: $ \pi k h^2 a^2 /12$ My Solution $ \int_{0}^{2 \pi} \int_{0}^{a} \int_{0}^{rh/a} zk r dz dr d \theta = \pi k h^2 a ^2 /4$ May I ask what I did wrong? Any tips would be greatly appreciated. Thanks.
 July 13th, 2017, 05:25 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717 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.
July 14th, 2017, 04:10 AM   #3
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Quote:
 Originally Posted by zollen 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: $ \pi k h^2 a^2 /12$ My Solution $ \int_{0}^{2 \pi} \int_{0}^{a} \int_{0}^{rh/a} zk r dz dr d \theta = \pi k h^2 a ^2 /4$ 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:
$\int_0^{\pi/2}\int_0^a\int_{rh/a}^h kz r dz dr d\theta$.

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
$
\int_{0}^{2 \pi} \int_{0}^{a} \int_{0}^{h- rh/a} zk r dz dr d \theta = \pi k h^2 a ^2 /4
$

July 14th, 2017, 02:19 PM   #4
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Quote:
 Originally Posted by Country Boy 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: $\int_0^{\pi/2}\int_0^a\int_{rh/a}^h kz r dz dr d\theta$. 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 $ \int_{0}^{2 \pi} \int_{0}^{a} \int_{0}^{h- rh/a} zk r dz dr d \theta = \pi k h^2 a ^2 /4$
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.

July 14th, 2017, 02:21 PM   #5
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Quote:
 Originally Posted by zollen 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?

Last edited by skipjack; July 17th, 2017 at 09:19 PM.

July 17th, 2017, 02:48 PM   #6
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Quote:
 Originally Posted by romsek 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.

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 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: $\int_0^{\pi/2}\int_0^a\int_{rh/a}^h kz r dz dr d\theta$. 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 $ \int_{0}^{2 \pi} \int_{0}^{a} \int_{0}^{h- rh/a} zk r dz dr d \theta = \pi k h^2 a ^2 /4$

July 17th, 2017, 04:55 PM   #8
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Quote:
 Originally Posted by zollen 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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