July 13th, 2017, 04:59 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 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: USA Posts: 2,009 Thanks: 1042 
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: 3,240 Thanks: 884  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: 178 Thanks: 2  Quote:
 
July 14th, 2017, 02:21 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042  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: 178 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: 178 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: USA Posts: 2,009 Thanks: 1042  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.  

Tags 
circular, cone, volume 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Right Circular Cone  aaronmath  Calculus  3  October 19th, 2011 12:10 AM 
A water tank has shape of an inverted circular cone of altit  urduworld  Calculus  2  November 1st, 2009 03:06 AM 
water is poured into an inverted circular cone of base radiu  urduworld  Calculus  2  October 31st, 2009 05:31 AM 
Right Circular Cone  symmetry  Algebra  4  February 2nd, 2007 03:57 PM 
Right Circular Cone Problem  symmetry  Algebra  3  January 30th, 2007 01:04 PM 