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October 2nd, 2018, 04:26 AM  #1 
Newbie Joined: Oct 2018 From: Australia Posts: 4 Thanks: 0  volume of frustum using integration solid of revolution
Hi, I am trying to find the volume of the attached solid using integration and solid of revolution. I have worked out the cylinder section but I'm having trouble with the frustum cone section. Q, The metal cover for a piece of machinery is 0.90 m in length, the radius of one end is 20 cm and the radius of the other end is 30cms as shown in the following diagram: Any help would be great. Last edited by skipjack; October 2nd, 2018 at 06:00 AM. 
October 2nd, 2018, 06:04 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,759 Thanks: 2138 
Can you post the work you've already done on the problem, so that we can see how you've already used integration?

October 2nd, 2018, 02:03 PM  #3 
Newbie Joined: Oct 2018 From: Australia Posts: 4 Thanks: 0 
This is how I worked out the cylinder section.
Last edited by skipjack; October 2nd, 2018 at 02:30 PM. 
October 2nd, 2018, 02:05 PM  #4 
Newbie Joined: Oct 2018 From: Australia Posts: 4 Thanks: 0 
And this where I can get to with the frustum section, but I'm really stuck at this point and not sure whether I'm on the right track.
Last edited by skipjack; October 2nd, 2018 at 03:05 PM. 
October 2nd, 2018, 02:59 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,759 Thanks: 2138 
By the rules of integration, $\displaystyle \pi\!\int_0^{45}\! ((2/9)x + 20)^2dx = \pi{\large[}(3/2)((2/9)x + 20)^3{\large]}_0^{45} = (40500  12000)\pi = 28500\pi$.


Tags 
frustum, integration, revolution, solid, volume 
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