Solids of revolution, how come we use the inverse function when we use method of cylindrical shells? Doing my second course in college calculus, and we are doing integrals and volumes by slicing/solids of revolution. The question I had trouble with was:
"Find the volume of a solid $S$, using the method of cylindrical shells, that is generated by rotating about the $x$-axis the region bounded by $y = x^2$, $y = 0$, and $x = 1$. "
I was having a real hard time figuring out why I kept getting the wrong answer when using the formula for the method of cylindrical shells, $V = 2\pi \int xf(x)\, \mathrm{d}x$
for determining a volume, until I saw that you are supposed to use the inverse function in the formula. Simply put, how come? Thanks in advance for any help! 
 A: It is the way the shells look that matters. In the usual case, i.e. rotating about the $y$-axis, the formula states:
$$
V = \int\limits_a^b \underbrace{(2 \pi x)}_{\text{circumference}} \underbrace{[f(x)]}_{\text{height}} \, \operatorname{d}x
$$
If you draw your specific problem, you will notice that your cylindrical shells end up with a radius of $y$ (circumference of $2 \pi y$) and a height of $1 - \sqrt{y}$.
A: Draw the picture.  You're rotating about the $x$-axis.  The radius of the shell is $y$.  The shell extends from the the graph rightward to the line $x=1$.  The horizontal distance from graph rightward to the line $x=1$ is the length of the segment from $(x,y)$ to $(1,y)$, where $y=f(x)$.  That means it goes from $(f^{-1}(y),y)$ to $(1,y)$.  So the surface area of the shell is
$$
\text{circumference}\times\text{(horizontal) height} = 2\pi y(1-x) = 2\pi y (1-f^{-1}(y)).
$$
The infinitely small thickness of the shell is $dy$, so you integrate
$$
\int_{y=0}^{y=1} 2\pi y (1-f^{-1}(y))\,dy.
$$
You could say the aformentioned segment goes from $(x,y)$ to $(1,y)$, i.e. from $(x,f(x))$ to $(1,f(x))$ and you have
$$
\int_{x=0}^{x=1} 2\pi f(x)\, (1-x)\  \  \underbrace{\Big(f'(x)\,dx\Big)}_{\text{This is }dy.}
$$
and that's the same thing.
At any rate, if $y=f(x)$ then $x=f^{-1}(y)$.
(All this works if $f$ is monotone, but not necessarily otherwise.)
