Different Volume Using Shell & Disk Methods

Question

I've been trying to find my mistake for a while, but I can't seem to find one. Maybe I'm missing a hole that is present?

I tried the volume of the region bounded between $x=1$, $x=3$ and a linear function $y=x-1$, also $x = 0$ rotating that region about the $y$-axis. My first attempt was to use the Disk Method, defined as

$\begin{align}V=\displaystyle{\pi \int_{y_1}^{y_2}}(f(y))^2dy=\displaystyle{\pi \int_{0}^{2}}(y+1)^2dy=\dfrac{26}{3}\pi\end{align}$

I then attempted to solve this problem using the Shell Method and got a different answer:

$V=\displaystyle{2\pi \int_{x_1}^{x_2}xf(x)dx}=2\pi\int_{1}^{3}x(x-1)dx=\dfrac{28}{3}\pi$

I have been trying to find my mistake for a while, trying different calculators and WolframAlpha, but I can't seem to get it right, maybe I'm missing some essential concept? My apologies if my formatting doesn't look good, as this is my first question here.

I would appreciate if someone could point me to the right direction.

Answer 1

By disk method we have

$$V=\pi \int_{0}^{2}\left[3^2-(f(y))^2\right]dy=\pi \int_{0}^{2}(8-y^2-2y)dy=\frac{28}{3}\pi$$

Let draw a plane sketch for the volume in order to see why the other way is wrong.

Answer 2

Your disk method is off. That works when the region lies against the $y$-axis. That's when you can slice the solid of revolution into disks and find their volume as flat cylinders.

Your region is, however, sliced into so-called washers (do an image search for "washer hardware", and you'll likely get many images of what they look like; another common visual analogy is a coin with a hole in it). And the volume of a washer is best found by filling the hole, calculate the volume of the resulting disc, then subtract the volume of the hole. This upgrades the disc method to the washer method.

The outer radius of the washers is $3$, and the inner radius is $y+1$. Which is to say, the integral should really be $$ \pi\int_0^2\big(3^2-(y+1)^2\big)dy $$