Find the volume of the solid formed by revolving the region bounded by y=x^2+1, y=0, x=0, and x=1 about the y-axis.
I was practicing this concept and I came across this problem. I did it using the shell method and got 2pi*Integral of x(x^2+1) from 0 to 1, which yielded the answer 3pi/2.
I tried checking this with the disc/washer method, but this gave me a different answer. Pi*Integral of [Sqrt(y-1)]^2 from 1 to 2 = pi/2.
Likewise, I tried a similar problem with the functions: y = Sqrt(x), x=axis, and x=4 roated about the x=8
Cylinders: 2piIntegral of Sqrt(x)(8-x) from 0 to 4 = 896pi/15 Disk/Washers: pi*Integral of (8-y^2)^2 from 0 to 2 = 1376pi/15
Which ones are correct? Where did I make mistakes?
The general rule is that if the axis of rotation is perpendicular to the interval of integration, then you will use cylindrical shells. If the axis of rotation is parallel to the interval of integration, then you will use either discs or washers.
Sometimes, you can use either method, but one method will usually be easier than the other.
For the particular example you have, it clearly seems that cylindrical shells will be easier. The interval of integration will be $x \in [0,1]$. For a particular representative radius $x$ from the $y$-axis, the height will be $f(x) = y = x^2 + 1$. The circumference is $2\pi x$, and the differential thickness of the shell is $dx$. So the differential volume is $$dV = 2 \pi x f(x) \, dx = 2 \pi x (x^2 + 1) \, dx,$$ and the total volume is $$V = \int_{x=0}^1 2 \pi x (x^2 + 1) \, dx = \frac{3\pi}{2}.$$
Now using the method of washers, the problem you have is that on the interval $y \in [0,1]$, the volume is just a cylinder of radius $1$, whereas on the interval $y \in (1,2]$, we have washers with outer radius $1$ and inner radius $g(y) = \sqrt{y-1}$, since the inverse function of $y = x^2 + 1$ is $x = \sqrt{y-1}$.
So combined, we would have: $$V = \int_{y=0}^1 \pi(1^2) \, dy + \int_{y=1}^2 \pi (1^2 - (\sqrt{y-1})^2) \, dy = \pi + \pi \int_{y=1}^2 2-y \, dy = \frac{3\pi}{2}.$$