The question is to find the volume of the solid formed by the revolution of the area $A$ around the $x$-axis. And the area $A$ is bounded by the curve $y=x^{\frac{1}{4}}$ and $y=x$ and $x=0$ and $x=1$.
I solved it by the washer method (I think I got an answer of $\frac{2\pi}{3}$).
But I tried to solve it using the shell method and I got lost here,
$V=\int_{0}^{1}y(x-x^{\frac{1}{4}})dy$
I know that I have to change the height of the cylinder to be in terms of $y$, but how do I do that? Which function do I use?
Both methods can be used to integrate the volume.
$$\text{washer:}\>\>\>\>\>\int_0^1 \pi(x^{1/2}-x^2)dx = \frac\pi3$$
$$\text{shell:}\>\>\>\>\> \int_0^1 2\pi y(y-y^4)dy = \frac\pi3$$