My Math Forum volume of frustum using integration solid of revolution

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October 2nd, 2018, 05:26 AM   #1
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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.
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 Capture.PNG (13.1 KB, 4 views)

Last edited by skipjack; October 2nd, 2018 at 07:00 AM.

 October 2nd, 2018, 07:04 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,302 Thanks: 1974 Can you post the work you've already done on the problem, so that we can see how you've already used integration? Thanks from ProofOfALifetime and maths noob
October 2nd, 2018, 03:03 PM   #3
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This is how I worked out the cylinder section.
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 2018-10-03 07.54.49.jpg (83.4 KB, 5 views)

Last edited by skipjack; October 2nd, 2018 at 03:30 PM.

October 2nd, 2018, 03:05 PM   #4
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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.
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 2018-10-03 07.57.00.jpg (85.4 KB, 5 views)

Last edited by skipjack; October 2nd, 2018 at 04:05 PM.

 October 2nd, 2018, 03:59 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,302 Thanks: 1974 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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