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, $\displaystyle \pi(3x)^2- \pi(2x)^2$. Its thickness, dx, gives volume $\displaystyle \pi(9- 4)x^2 dx= 5\pi x^2 dx$. Integrate that from x= 0 to 1.

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, $\displaystyle \pi(3x)^2- \pi(2x)^2$. Its thickness, dx, gives volume $\displaystyle \pi(9- 4)x^2 dx= 5\pi x^2 dx$. 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 $\displaystyle \displaystyle V = \int_1^2{ \pi \left( 3x \right) ^2 \, dx }$.