2017 AMC 10A Problem 11

The solid is a cylinder with two hemispherical caps

Subtract the volume of the two hemispheres before solving for $AB$

Recall that all the points at most a fixed distance $r$ away from a point form a sphere.

At the end points of this line segment, we can visualize two hemispheres being formed at each end.

All the points in the middle also have spheres forming around them, but they get merged into the ones right next to them.

This means that the middle section forms a cylinder with radius $3.$ The two hemispheres form a sphere with radius $3,$ and therefore a volume of $$\dfrac{4}{3} \pi 3^3 = 27 \cdot \dfrac{4}{3} \pi = 36 \pi.$$ This means that the cylinder has a volume of $216 \pi - 36 \pi = 180 \pi.$ We know the base area is $9 \pi,$ so if $h$ is $AB,$ then the volume is $$\begin{aligned} \pi 3^2 h &= 180\pi \\9h \pi &= 180 \pi \\ h &= 20. \end{aligned}$$

Thus, D is the correct answer.