How can we find the volume of $y= \sqrt{16-x^2}$ rotated around
A) $y$-axis
B) $x=4$
The thing I don't understand is that the graph is a semicircle that's already on the $y$-axis. Am I supposed to rotated the whole semicircle or half of it? What do I do about part B?
For A, you can revolve either the whole semicircle or the half in the first quadrant. The resulting solid is the same. In a sense, if you revolve the whole semicircle by angle $2\pi$ you double cover the hemisphere in that if you calculate the volume swept out you will count everything twice. But I would do this by the disk method.
For part B you revolve the whole thing around $x=4$ you get a shape that is half of a doughnut with the central hole shrunk to zero. Cylindrical shells seem the way to go here.