Is this example in the Leithold's calculus book wrong? I'm trying to understand this example from Leithold's Calculus book. 
The volume of the solid generated by the rotation about the line $x=1$ of the region bounded by the curve $(x-1)^2=20-4y$ and by the lines $x=1,y=1$ and $y=3$ and in the right of $x=1$ is given by $$\pi\int_1^3(20-4y)dy$$

In order to check if this method is valid I tried to use it in a simpler case which I know the volume:

It's easy to see that the cylinder with height $h$ and radius $5$ without the cylinder with height $h$ and radius $2$ has volume of $21\pi$. However using his technique I'm getting $V=\pi\int_0^h(5-2)^2dy=9\pi h$.
Where is my mistake in my example? following the arguments in the comments, the volume in this example should be $\pi\int ((\sqrt{20-4y}+1)^2-1^2)dy$, instead of $\pi\int (20-4y)dy$
Thanks
 A: For the exercise in your book, you should use the circular-disk method about a vertical line where your function needs to return the horizontal radius with respect to $y$
$$
(F(y) - 1)^2 = 20 - 4y \\
F(y) - 1 = \pm \sqrt{20 - 4y} \\
F(y) = 1 \pm \sqrt{20 - 4y}
$$
Since you are revolving about the line $x = 1$, the volume of the circular disk is given by
$$
\Delta_i V = \pi(F(y_i) - 1)^2 \Delta_i y
$$
Thus,
$$
\begin{eqnarray}
V &=& \lim_{||\Delta|| \to 0} \sum_{i=1}^n \pi(F(y_i) - 1)^2 \Delta_i y \\
&=& \pi \int_1^3 (1 \pm \sqrt{20 - 4y} - 1)^2 \, \textrm{d}y \\
&=& \pi \int_1^3 (20 - 4y) \, \textrm{d}y \\
\end{eqnarray}
$$
The integrand happens to be the same expression with respect to $(x - 1)^2$ which is probably what the author wanted to demonstrate in this exercise.
For your simple case, you should use the circular-ring method about a vertical line where you need one function to return the outer radius and another to return the inner radius with respect to $y$
$$
F(y) = 5 \\
G(y) = 2
$$
Since you are now revolving about the $y$-axis, the volume of the circular ring is given by
$$
\Delta_i V = \pi([F(y_i)]^2 - [G(y_i)]^2) \Delta_i y
$$
Thus,
$$
\begin{eqnarray}
V &=& \lim_{||\Delta|| \to 0} \sum_{i=1}^n \pi([F(y_i)]^2 - [G(y_i)]^2) \Delta_i y \\
&=& \pi \int_0^h (5^2 - 2^2) \, \textrm{d}y \\
&=& 21 \pi \int_0^h \textrm{d}y \\
&=& 21 \pi h
\end{eqnarray}
$$
Note that the circular-disk method is essentially the same as the circular-ring method where $G(y) = axis$. In both cases, the integral for the volume can be defined as
$$
V = \pi \int_a^b ([F(y) - axis]^2 - [G(y) - axis]^2) \, \textrm{d}x
$$
For the exercise in your book, $axis = 1$ and $G(y) = axis$. For your simple case, $axis = 0$ and $G(y) = 2$.
