Height formula of cylindrical shell method With the equations y = sqrt(x) and y=2, rotated about the x axis, why would the height of the shell be just (y^2) and not (2 - Y^2)? Isn't the formula for the height the curve on the top - the curve on the bottom?
 A: Though it is not stated explicitly, presumably we are rotating the region below $y=2$, above the curve $y=\sqrt{x}$, and to the right of the $y$-axis, about the $x$-axis. We could use slicing parallel to the $y$-axis ("washers") and integrate with respect to $x$. Or else we can use cylindrical shells, and integrate with respect to $y$. If we use washers, then we end up with a washer with outer radius $2$, and inner radius $y$. Thus the area of cross-section is $\pi(2^2-y^2)$. We replace $y$ by $\sqrt{x}$, and integrate from $x=0$ to $x=\sqrt{2}$.
But that's not the question, we are using shells.  Draw a thin horizontal strip of width "$dy$" at height $y$, and imagine rotating it about the $x$-axis. The strip is at height about $y$, so it sweeps out a thin cylindrical shell, of radius $y$. The "height" of the shell is the length of the strip. It is just $x$. So the volume of the shell is approximately $(2\pi y)x\,dy$.  Now add up (integrate) from $y=0$ to $y=2$. 
To do the integration, we need to express $x$ in terms of $y$. Since $y=\sqrt{x}$, we have $x=y^2$.
Remark: Despite having done this sort of calculation many times (too many), I still always draw a picture, and go through the reasoning, each time. All formulas tend to look the same. I don't trust my memory, but do trust my reasoning powers. Memorizing formulas can be an excellent short term strategy. If you have a decent memory, need to know certain things for a test next week, and will never need to know those things again, one cannot fault it as a strategy. In the long term, the strategy doesn't work. 
