# Homework Help: Volumes of Revolution problem

1. Jan 25, 2010

### Precursor

1. The problem statement, all variables and given/known data
Find the volume of the solid generated by revolving the triangular region bounded by the lines $$y= 2x$$, $$y= 0$$ and $$x= 1$$ about the line $$x= 1$$.

2. Relevant equations
$$V= \int A(x)dx = \int \pi[R(x)]^{2}dx$$

3. The attempt at a solution
I used the disk method, in which I found the radius of the solid. I found the radius to be $$1- y/2$$.

$$V= \int \pi[1- y/2]^{2}dx$$
By the way, the upper limit is 1 and the lower limit is 0. I don't know how to put that in latex.

So I got an answer of $$7\pi/12$$. Am I right?

2. Jan 25, 2010

### Precursor

Any ideas?

3. Jan 25, 2010

### Staff: Mentor

The formula above is relevant only if your disks run horizontally. That is, if the line through their centers is horizontal.
The thickness of each disk is dy, not dx. This means that the disks range between y = 0 and y = 2, not x = 0 to x = 1. Here is your integral with limits of integration. Click on the integral expression to see how to incorporate limits of integration. Also, it's usually a good idea to move constants such as pi outside the integral.
$$V= \pi \int_{y = 0}^{2} [1- y/2]^{2}dy$$
The solid is a cone whose base has a radius of 1 and whose height is 2. There is a formula for the volume of a cone. You can use this formula to verify that your answer is correct.