Help setting up a problem

[IsaaKC]

It is y=(x-2) ^2, y=1
It's finding the solid of a revolution rotated around the x axis, so I believe with respect to y using the shell method.

What I've come up with is 2? times the integral of y[(?y)+1] with the limits of 1 and 0. This isn't getting to the right answer.
or
2? times the integral of y[(?y)+1 - (?y)-1 ] with the limits of 1 and 0, but this one can't be right, so I didn't use it.

 

[galactus]

If you solve \(\displaystyle y=(x-2)^{2}\) for x, you should get \(\displaystyle x=2+\sqrt{y}, \;\ x=2-\sqrt{y}, \;\ y\geq 0\).

 

[soroban]

Hello, IsaaKC!

Did you make a sketch?

\(\displaystyle y \:=\x-2) ^2,\;\;y\,=\,1,\;\text{ revolved about the }x\text{-axis.}\)

  |
  |   *               *
  |
  |    *             *
 1 + - - * - - - - - * -
  |       *:::::::*
--+-----*-----*-----*-----
  |     1     2     3
  |

I would use Washers.

\(\displaystyle V \;=\;\pi\int^3_1 \bigg[1^2 - (x-2)^4\bigg]\,dx\)

 

[galactus]

You can also do it with shells. See if you get the same result as with washers.

\(\displaystyle 4\pi\int_{0}^{1}y^{\frac{3}{2}}dy\)

I simplified this down. Perhaps beyond recognition.

But, it comes from \(\displaystyle x=2+\sqrt{y}, \;\ x=2-\sqrt{y}\)

\(\displaystyle 2+\sqrt{y}-(2-\sqrt{y})=2\sqrt{y}\)