What is the Volume of the Solid Generated by Revolving a Triangular Region?

[Precursor]

Homework Statement

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 $$x= 1$$.

Homework Equations

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

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?

 

[Precursor]

Any ideas?

 

[Mark44]

Homework Statement

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 $$x= 1$$.

Homework Equations

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

The formula above is relevant only if your disks run horizontally. That is, if the line through their centers is horizontal.

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.

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.
$$V= \pi \int_{y = 0}^{2} [1- y/2]^{2}dy$$

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

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.