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: My Solution 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  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,656 Thanks: 681  Quote:
. Or, represent the cone as having its base in the xyplane and its vertex at (0, 0, h). The sides then are given by z= h rh/a so that your integral is  
July 14th, 2017, 02:19 PM  #4  
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2  Quote:
 
July 14th, 2017, 02:21 PM  #5  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  Quote:
It did produce the answer your book did. Last edited by skipjack; July 17th, 2017 at 09:19 PM.  
July 17th, 2017, 02:48 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2  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  
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 
You were correct. I made a mistake. z = rh/a is incorrect. The correct equation for z = (a  r) * (h / a) Quote:
 
July 17th, 2017, 04:55 PM  #8  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717  Quote:
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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