Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves $y=\sqrt{x-1}$, $y=0$, $x=2$, and $x=3$ about the $x$-axis.
To me it seems like this question is asking for the upper half of the solid because it specifies $y=0$.
Setting it up, then: $\dfrac{\pi}{2}\displaystyle\int_{2}^{3}\left(\sqrt{x-1}\right)^2dx$ and the answer is $3\pi/4$.
That is not the answer webwork is after, though.
What don't I understand about how $y=0$ figures in?
TIA
The question is not asking for just the upper half as it's of the solid rotated completely around the $x$-axis. Note that to rotate a region with certain bounds around the $x$-axis, you need to know what those bounds are. In this case, the lower vertical bound is specified as being $y = 0$.
However, as imranfat's question comment indicates, if this lower limit was replaced by something else, say like $y = 1$ instead, you would still rotate it completely around the $x$-axis. However, because the rotation region is smaller, the resulting volume would be smaller as well. Also, because the lower bound would then no longer coincide with the line you're rotating around, i.e., the $x$-axis which is the same as $y = 0$, it would change from being a disc problem to being a washer problem instead.
Note the disc method to determine volume rotations is a special of the washer method, where the internal circle is a degenerate one with a radius of $0$. Thus, your current volume is double what your integral specifies, but with the lower bound squared being implicitly subtracted, i.e., it would be $\pi\int_{2}^{3}\left((\sqrt{x-1})^2 - (0)^2\right)dx$. Also, in my alternate example above of a lower bound of $y = 1$, the integral would instead be $\pi\int_{2}^{3}\left((\sqrt{x-1})^2 - (1)^2\right)dx$.
A more general different case would be if the lower bound of the region was $y = -1$ and rotating around the line $y = -2$. In this case, the region being rotated would include parts of both the upper & lower halves of the Cartesian co-ordinate plane being rotated. Also, since the height of the function would then be $\sqrt{x - 1} - (-2) = \sqrt{x - 1} + 2$, but with the height still being $-1 - (-2) = 1$ for the lower bound, the volume in this case would then be given by $\pi\int_{2}^{3}\left((\sqrt{x-1} + 2)^2 - (1)^2\right)dx$.