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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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Last edited by skipjack; October 2nd, 2018 at 07:00 AM.
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October 2nd, 2018, 07:04 AM   #2
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Can you post the work you've already done on the problem, so that we can see how you've already used integration?
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October 2nd, 2018, 03:03 PM   #3
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This is how I worked out the cylinder section.
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Last edited by skipjack; October 2nd, 2018 at 03:30 PM.
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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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Last edited by skipjack; October 2nd, 2018 at 04:05 PM.
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October 2nd, 2018, 03:59 PM   #5
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By the rules of integration, $\displaystyle \pi\!\int_0^{45}\! ((2/9)x + 20)^2dx = \pi{\large[}(3/2)((2/9)x + 20)^3{\large]}_0^{45} = (40500 - 12000)\pi = 28500\pi$.
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