The question comes from the 1991 AP Calculus test.
Let $R$ be the region between the graphs of $y = 1 + sin(\pi x) $ and $y = x^2$ from $x = 0$ to $x = 1$.
c) Set up, but do not integrate an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis
The answers are
$$V = 2\pi\int_{0}^{1}x(1 + sin(\pi x) - x^2)dx $$
or
$$ V = \pi\int_{0}^{1}y dy + \pi\int_{1}^{2}(1 - \frac{1} \pi arcsin(y-1))^2 - (\frac{1} \pi arcsin(y-1))^2dy $$
I am confused on how they reached these two answers. I was able to do the question before this which involved rotating the same area but around the x-axis. In school, we learned to write the equations as $x = ...$ when dealing with rotations around the y-axis but I'm not sure how that applies here.