Instead of using shells, can I do this problem via horizontal integration and subtracting the smaller circle from the larger circle and use washers instead?
Is the volume just:
$\int$ Area of larger circle - Area of smaller circle \, dy ?
outer circle equation: $x = \sqrt{y} $ inner circle equation: $x = y$ $$\int_0^1 \pi (outer radius)^2 - \pi (inner radius)^2 \, dy$$
$$\int_0^1 \pi y - \pi y^2\, dy$$
$$ \left [ \pi \frac{y^2}{2} - \pi \frac{y^3}{3} \right ]_0^1$$
$$\left [ \pi \frac{1}{2} - \pi \frac{1}{3} \right ]$$
$$\frac{3\pi}{6} - \frac{2 \pi}{15} = \frac{\pi}{6}$$
This right?
Yes this is a correct application of the disk method.
Maybe it is just a typo, but note that we are integrating with repect to $y$ then the integral is
$$\int_0^1 \pi y - \pi y^2\,\color{red}{dy}$$