# Solids of Revolution Question (Method of Cylinders vs Disc/Washers)

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?

• What can I do to improve the quality of this question? I wasn't aware of anything wrong about it besides the improper formatting. Thank you. – bandicoot12 Jan 19 '15 at 21:03
• The problem is that $\int_1^2\pi\sqrt{y-1}^2dy$ represents the volume formed by spinning the region bounded by $y=x^2+1, y=0,$ and $y=2$ about the y-axis. – John Joy Jan 19 '15 at 21:08

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: $$V = \int_{y=0}^1 \pi(1^2) \, dy + \int_{y=1}^2 \pi (1^2 - (\sqrt{y-1})^2) \, dy = \pi + \pi \int_{y=1}^2 2-y \, dy = \frac{3\pi}{2}.$$