Question

Use cylindrical shells:

Answer 1

You don't really want to use shells for this.

Answer 2

Sadly, I have to.. it's method they told us to use.

Answer 3

\[\begin{array}{rcl}
1<&y&<2 \\
-\sqrt{2-y}<&x&<\sqrt{2-y}
\end{array}
\]

Answer 4

\You have to use \[
2\pi \int rh~dy
\]

Answer 5

In this case \(r = |y-1|\), however this will always be positive so it simplifies to \(r=y-1\)

Answer 6

And \(h= \sqrt{2-y} - (-\sqrt{2-y}) = 2\sqrt{2-y}\)

Answer 7

To get \(h\), I used \[
y= 2-x^2 \implies x^2=2-y \implies x=\pm \sqrt{2-y}
\]

Answer 8

All together: \[
V = 2\pi \int_1^2(y-1)(2\sqrt{y-2})~dy
\]

Answer 9

Alright, just needed to check the set up, thanks wio :)