# Volumes of revolution

• Apr 10th 2013, 03:31 PM
Purdue1
Volumes of revolution
A region http://www.ilrn.com/ilrn/formulaImag...%7D%5C%3A&ns=0 is bounded by the given curves. Find the volume of the solid of revolution obtained by rotating the region http://www.ilrn.com/ilrn/formulaImag...%7D%5C%3A&ns=0 about the http://www.ilrn.com/ilrn/formulaImag...%7D%5C%2C&ns=0-axis. (give your answer in terms of http://www.ilrn.com/ilrn/formulaImage?f=%5Cpi&ns=0 or approximated to the nearest tenth).
y=3x x=1 x=2 x axis.
Im having alot of trouble with this assignment because I just dont understand this stuff, any help would be great!

Thanks
Greg
• Apr 10th 2013, 05:18 PM
HallsofIvy
Re: Volumes of revolution
You don't need to use Calculus at all. The line y= 3x, from x= 0 to 1, rotated around the x-axis, gives a cone with "height" 1 and base radius 3(1)= 3. The lline y= 2x, from x= 0 to 1, rotated around the x-axis, cone with "height" 1 and base radius 2(1)= 2. Look up the formula for volume of a cone, find each of those areas, and subtract the inner cone from the outer cone.

If you really want to use Calculus, you can use the "washer method". The a vertical line between those two lines, rotated around the x-axis will form a "washer", a disk with a hole in it. Its area is the are between the two cirles, $\pi(3x)^2- \pi(2x)^2$. Its thickness, dx, gives volume $\pi(9- 4)x^2 dx= 5\pi x^2 dx$. Integrate that from x= 0 to 1.
• Apr 10th 2013, 05:26 PM
Prove It
Re: Volumes of revolution
Quote:

Originally Posted by HallsofIvy
You don't need to use Calculus at all. The line y= 3x, from x= 0 to 1, rotated around the x-axis, gives a cone with "height" 1 and base radius 3(1)= 3. The lline y= 2x, from x= 0 to 1, rotated around the x-axis, cone with "height" 1 and base radius 2(1)= 2. Look up the formula for volume of a cone, find each of those areas, and subtract the inner cone from the outer cone.

If you really want to use Calculus, you can use the "washer method". The a vertical line between those two lines, rotated around the x-axis will form a "washer", a disk with a hole in it. Its area is the are between the two cirles, $\pi(3x)^2- \pi(2x)^2$. Its thickness, dx, gives volume $\pi(9- 4)x^2 dx= 5\pi x^2 dx$. Integrate that from x= 0 to 1.

You have read the question wrong HallsOfIvy. The region is bounded by y = 3x, x = 1, x = 2 and the x axis. Not y = 3x and y = 2x. The OP needs to realise that the region will be DISCS or CYLINDERS (not washers) of radius 3x, so the volume of revolution will be $\displaystyle V = \int_1^2{ \pi \left( 3x \right) ^2 \, dx }$.