volume of frustum using integration solid of revolution

maths noob

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.

Attachments

• 13.1 KB Views: 5
Last edited by a moderator:

skipjack

Forum Staff
Can you post the work you've already done on the problem, so that we can see how you've already used integration?

2 people

maths noob

This is how I worked out the cylinder section.

Attachments

• 83.4 KB Views: 6
Last edited by a moderator:

maths noob

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.

Attachments

• 85.4 KB Views: 6
Last edited by a moderator:

skipjack

Forum Staff
By the rules of integration, $$\displaystyle \pi\!\int_0^{45}\! ((2/9)x + 20)^2dx = \pi{\large[}(3/2)((2/9)x + 20)^3{\large]}_0^{45} = (40500 - 12000)\pi = 28500\pi$$.