Find the volume of the shape created when rotating the region(s) bounded by $y=\sqrt{x+1}, y=0, x=0, x=1$, about the x-axis.
I know this is a rudimentary question. My issue is that I tried to test myself using the shell method instead of using the disk-method. I have been taught that both methods work, yet what I assumed to be correct usage of the shell method in this instance garnered me an incorrect answer. I am not asking to see the disk method or why it would have been easier, I am asking what went wrong with my usage of the shell method.
\begin{align*} V &= \int_0^\sqrt{2}2\pi(y)({y^2}-1)dy \\ &=2\pi\int_0^\sqrt{2}({y^3}-y)dy \\ &=2\pi\left[\frac{y^4}4-\frac{y^2}2\right]{\sqrt{2} \choose 0} \\ &=2\pi\left(\frac44-\frac22\right) \\ &=2\pi(0) \\ &=0 \end{align*} ... where of course $x = {y^2}-1$ and $y=\sqrt{1+1}$ = $\sqrt{2}$
Thanks!
As Andre Nicolas pointed out, you need two integrals using the shell method;
and you also need a different expression for the height:
$\displaystyle V=\int_0^1 2\pi r(y)h(y)dy+\int_1^{\sqrt{2}}2\pi r(y)h(y)dy=\int_0^1 2\pi y(1)dy+\int_1^{\sqrt{2}} 2\pi y(1-(y^2-1))dy$