Let $R$ be the region in the first quadrant enclosed by $f(x)=\sqrt x$, $g(x) = −x + 2$ and $y = 0$, as shown in the figure. What is the volume of the solid generated when $R$ is rotated about the $x$-axis?
Wouldn't the integral for this be $\pi\int_0^2[x-(-x+2)^2]dx$? I don't get what I'm doing wrong.
Hint:
$$V_{x}=\pi \int_0^1 (\sqrt{x})^2 dx + \pi \int_1^2 (-x+2)^2 dx.$$
Figure: