Analysis: for example, the solution graph is used as an auxiliary line, and the symmetry of the graph is used to solve it. The key to solve the problem is to find the area of arch OmC.

Answer:

Solution: As shown in the figure, let point O be the intersection of arcs.

When OA and OB are connected, △OAB is an equilateral triangle, ∴∠obc = 30°.

If point O passes through EF⊥CD, AB and CD pass through point E and point F respectively, OE is the height of equilateral △OAB.

∴OE=(√3/2)AB=√3,∴OF=2-√3.

If point O is PQ⊥BC, and AD and BC are handed in at points P and Q respectively, then OQ = 1.

S-shaped OmC=S-shaped OBC-S △ OBC = [(30× π× 2 2)/360]-1/2× 2×1= (π/3)-1.

∴S shadow =4(S△OCD-2S bow OmC)=4[( 1/2)×2×(2-√3)-? 2×((π/3)- 1)]= 16-4√3-(8π/3).

So the answer is: 16-4 √ 3-(8 π/3).