Please see attached for the volume of the purple region. I want to find in question using the disk method.
The region is bounded by $ y= \sqrt{x+1} ,\; y=0 ,\; y=-x+1 \;$ about the line $x=3$.
My attempt: I split up the integral into the sum of two integrals since I noticed that the $r(y)$ (radius) changes after $y=1$. Solving for $r(y)$'s from the $y$'s, I get that $ x=y^2-1 ,\ x=1-y ,\ $ $ \ x=3 $
So, the integral comes out as:
$$ \pi \int_{0}^1 \left[\left(2+y^2\right)^2-\left(2+y\right)^2 \right]\mathrm dy + \pi \int_{1}^2 \left(2-y^2\right)\mathrm dy = \frac{\pi}{15}$$
Is this correct? It'd be great getting some peace of mind if it is. If it's not, please point out where I went wrong. Thank you.
I don't understand why you are splitting it into two integrals. If you are integrating with respect to $y$, you shouldn't need to split it up because $x = y-1$ is always to the right of $x = y^2 - 1$ for $y \in [0,1]$.
Thus, by using the washer method, you should just have this:
$$\pi\int_0^1 \left(\left(4-y^2\right)^2 - \left(2+y\right)^2\right)\,\mathrm dy$$
This is because the greater radius is $3-\left(y^2 - 1\right)$ and the smaller is $3-(1-y)$.