Find the volume of the submerged part of the ring.?

Question

The outer radius of the ring is 1.98 m and its inner radius is 1.82 m and it is 0.153 m thick. The ring is set to stand so that the submerged part is shaped like an arc. The ring is submerged to a depth of 0.308 m of water. Find the volume of the submerged part of the ring.

Answer 1

Volume of submerged part = Area x thickness

Area = Area(segment1) - Area(segment2)

Let a = angle of 2 pts. along the circle w/ the center of the circle

Area(segment of a circle) = 0.5(a - sin a)r^2

a(outer) = 2 arccos[(1.98-.308)/1.98)] = 64.775º

a(inner) = 2 arccos[(1.98-.308)/1.82)] = 46.5319º

Area = Area(segment1) - Area(segment2)

         = 0.5(64.775π/180 - sin 64.775)(1.98^2) - 0.5(46.5319π/180 - sin 46.5319)(1.82^2)

        = 0.2997m²

Volume of submerged part = Area x thickness

                                                 = (0.2997m²)(0.153)

                                                 = 0.04586 m³