I'm just confused whether I'm going about solving the above problem correctly. We're asked to set up the integrals only and simplify the integrand. Here's my process:
$$= \pi \int_{0}^{1}(1+\sqrt{y})^2\,dy - \pi \int_{0}^{1}(1+y)^2\,dy$$ $$= \pi \int_{0}^{1}1+2\sqrt{y}+y\,dy - \pi \int_{0}^{1}1+2y+y^2\,dy$$ $$= \pi \int_{0}^{1}1+2\sqrt{y}+y-1-2y-y^2\,dy$$ $$= \pi \int_{0}^{1}2\sqrt{y}-y-y^2\,dy$$
Yes your work is correct using washer method. If you were to use shell method,
height of the shell at a given $x$ is $ ~(x - x^2)$
At the intersection of curves $y = x$ and $y = x^2$,
$y = x = x^2 \implies x = 0 ~$ or $~x = 1$
Also distance of shell from the axis of rotation is $~x - (-1) = 1 + x$
So the integral to find volume using shell method will be,
$ \displaystyle \int_0^1 2 \pi (1+x) ~ (x-x^2) ~ dx = 2 \pi\int_{0}^{1} (x - x^3) ~dx$