The base of a solid is the region in the xy-plane bounded by the curves $ y = \sqrt x, x=4,$ and $y = 0$. Cross-sections of the solid perpendicular to the x-axis are semicircles. Find its volume.
I solved the following integral, adding a coefficient of $\frac1 2$ because the cross section is a semicircle, thus a half rotation. I get a final answer of $4\pi$ as follows:
$$ V = \frac \pi2 \int_0^4 \sqrt x ^2 \,dx = \frac \pi2 \int_0^4 x dx = \frac \pi4 x^2 \bigg|_0^4 = 4\pi $$
The computer says this answer is incorrect. Am I misinterpreting the axis of rotation?
Since the solid has semicircular cross sections perpendicular to the $x$-axis, the area of one of these cross sections is $$\frac{1}{2}\pi r^{2} = \frac{1}{2}\pi \left(\frac{\sqrt{x}}{2}\right)^{2} = \frac{\pi x}{8}.$$ The total volume then is given by $$V = \int_{0}^{4}\frac{\pi x}{8}\,dx = \frac{\pi}{16}x^{2}\bigg|_{0}^{4} = \pi.$$ Your error was in that you said that the radius of the semicircles was $\sqrt{x}$, where in reality this is the diameter.