A cylindrical tank with radius $6$ feet and length $5$ feet is lying on its side. The tank is filled with water to a depth of $9$ feet. What is the volume of water, in cubic feet?

1. The cross section of the water is a circle of radius $6$ with the part above a chord removed. The depth is $9=6+3$, so the chord lies $3$ above the center, and the radii to its endpoints make an angle of $120^\circ$ on the dry side.
2. The wet region is therefore the sector of central angle $240^\circ$ plus the isosceles triangle formed by the two radii and the chord:
   \[
   \frac{240}{360}\pi(6)^2+\frac12(6)^2\sin120^\circ=24\pi+9\sqrt3.
   \]
3. Multiplying by the length $5$ gives the volume $5(24\pi+9\sqrt3)=\boxed{120\pi+45\sqrt3}$ cubic feet.