I don't understand a part of the solution using the method of shells on this problem

Question

\begin{array} { l } { 4 \mathrm { C } - 3 \text { Find the volume of the region } \sqrt { x } \leq y \leq 1 , x \geq 0 \text { revolved around the } y \text { -axis } } \\ { \text { by both the method of shells and the method of disks and washers. } } \end{array}

That is the problem. It is from MIT's OCW. It has a solution, which is

\begin{array} { l } { 4 \mathrm { C } - 3 \text { Shells: } \quad \int _ { 0 } ^ { 1 } 2 \pi x ( 1 - y ) d x = \int _ { 0 } ^ { 1 } 2 \pi x ( 1 - \sqrt { x } ) d x = \pi / 5} \end{array}

I don't understand why it's (1−y) and not just y.

Answer 1

The region integrated, in the y-direction, is between $\sqrt{x}$ and $1$. If the region being revolved was $0\leq y\leq\sqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.

Answer 2

If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=\sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $\sqrt r$, which would be below the first surface, but $1-\sqrt r$ which is between the two surfaces.