A cylinder rests on top of a table, with a cone inscribed within, vertex up. Both heights and radii are 8 cm. A hemispherical bowl of radius 8 cm rests nearby on the same table, its circular rim parallel to the table. Consider that part of the cylinder that is outside (above) the cone. Slice this region by a plane that is parallel to the table and 3 cm from it. The intersection is a ring between two concentric circles. Calculate its area. The same plane slices the hemisphere, creating a disk. Show that the disk has the same area as the ring. The diagram shows both a top view and a side view of the ring, the disk, and the hemisphere.

Answer 1

To find the area of the ring, we need to find the areas of the outer circle and the inner circle and then subtract them.

The radius of the outer circle is the distance from the center of the cylinder to the parallel plane, which is 8 + 3 = 11 cm.
The area of the outer circle is π times the square of its radius: π(11^2) = 121π cm^2.

The radius of the inner circle is the distance from the center of the cone to the parallel plane, which is 8 cm.
The area of the inner circle is π times the square of its radius: π(8^2) = 64π cm^2.

Therefore, the area of the ring is 121π - 64π = 57π cm^2.

Now let's look at the disk that is created by slicing the hemisphere.

The radius of the disk is the same as the radius of the hemisphere, which is 8 cm.
The area of the disk is π times the square of its radius: π(8^2) = 64π cm^2.

Therefore, the disk has the same area as the ring, which is 57π cm^2.

So, the area of the ring is equal to the area of the disk.