How does Volume work with integration? Using a cross section suppose, as described here: Area formula Paul Notes

Suppose this is: $y = f(x)$.  
He says the volume is:
$$\int_{a}^{b} A(x) dx$$ 
But how does area over that interval give the volume? I mean, 
You are not getting the area of a cross section (the YELLOW part) are you? You are just taking the area of The 2D region, which isnt the cross section?

Then how does summing that area give the volume when its not the cross section? The actual cross section is the yellow part. 
 A: $A(x)$ represents the area of the cross section. It is a circle of radius $f(x)$. Thus $A(x)=\pi\,f(x)^2$.
To see that the integral gives the volume imagine the solid of revolution is a salami and you cut it in thin slices of width $dx$. The volume of each slice is $A(x)\,dx$, and when you sum them all you get the volume of the salami. 
A: It is as simple as 
\begin{align}
V 
&= 
\int\limits_V dV 
\quad 
\mbox{(adding up tiny Volume elements over the volume)} \\ 
&= 
\int\limits_a^b A(x) \, dx 
\quad 
\mbox{(adding up slices $dV = A(x)\,dx$ along the $x$-axis)}\\ 
&= 
\int\limits_a^b \int\limits_0^{f(x)} \int\limits_0^{2\pi} y\, dx \, dy \, d\varphi \\
\end{align}
Where
$$
A(x) 
= 
\int\limits_0^{f(x)} dA 
= 
\int\limits_0^{f(x)} 2\pi y dy = \pi f(x)^2
$$ 
results from a summation of circular strips $dA$
and the circular strips result from summing up the elements
$y\,dy\,d\varphi$ over the circle
$$
dA = \int\limits_0^{2\pi} y \, dy \, d\phi = 2\pi y\, dy
$$
A: Note that $f(x)$ going from left to right of the graph simply defines the shape. With the $\pi$ constant it makes it into a circle, that is to say you're rotating the shape around the $x$-axis by $360$ degrees. Therefore, $f(x) \;  \textbf{becomes}$ the radius at each point $x$. You're simply adding up the circles with the varying radius $f(x)$.
Hope that helps.
A: During rotation or sweeping the green area you are summing up many slices of the thin yellow circle pancakes along axis , but there is no connection to the green meridional cross section you sketched. 
However if you want to find the volume of straight prismatic piece of cheese of cross section shaped that way, just multiply the green area by the distance moved normal to the green area along $z$ to find the linearly swept volume.
