Volume of the solid - one more

I want to make sure I understand and vizualise it right - Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x, y = 0, x = 2, and x=4 about the line x = 1 ;

the outer radius would be (4-y) - 1 = 3 - y, and the inner radius (2-y) - 1 = 1 - y

The area A(y) = pi(R^2 - r^2) = pi(3-y)^2 - pi(1-y)^2 = pi[(9 - 6y + y^2) - (1 - 2y + y^2)] = pi(8 - 4y) which I need to integrate from 0 to 4, right? It's

confusing and I am not sure I get the right result - maybe it's something in the area that I'm missing...

thanks

Re: Volume of the solid - one more

Have you drawn a graph of this? The line y= x crosses the line x= 2 at, of course, (2, 2). But the region you are rotating around the line x= 1 goes down to y= 0. If you are using "disks or washers" you will have to do this in two parts: for y= 0 to 2, The outer radius is 4- 1= 3 and the inner radius is 2- 1= 1. For y from 2 to 4, the inner radius is y- 1 and the outer radius is 3.

Re: Volume of the solid - one more

oh, in this case I will have A1 = pi(3-1)^2 = 4pi; the A2 = pi(9 - (y-1)^2) = pi(9 - y^2 - 2y +1) = pi(10 - y^2 - 2y)

and I supose the integration would be also separate - A1 from 0 to 2 and A2 from 2 to 4 , right?

Re: Volume of the solid - one more

Quote:

Originally Posted by

**dokrbb** oh, in this case I will have A1 = pi(3-1)^2 = 4pi; the A2 = pi(9 - (y-1)^2) = pi(9 - y^2 - 2y +1) = pi(10 - y^2 - 2y)

and I supose the integration would be also separate - A1 from 0 to 2 and A2 from 2 to 4 , right?

For $\displaystyle A_1$, the inner radius is $\displaystyle r=(2-1) = 1$ and the outer radius is $\displaystyle r=(4-1)=3$. So, $\displaystyle A_1$ should be the area of a washer, not a disk.

For $\displaystyle A_2$, the inner radius is $\displaystyle r = (y-1)$ and the outer radius is $\displaystyle r = (4-1)=3$. So, just as $\displaystyle A_1$ should be the area of a washer, so too should $\displaystyle A_2$.