integration (area)

y = a sin 2t = 2a sin t cos t
= 2x √(1 - sin2 t)
= 2x √[1 - (x/a)2]
= (2x/a) √(a2 - x2)

So the curve is an odd function with domain of x from - a to a inclusive.

Thus, the area of the loop is twice the area of the loop covered in positive x.

For the loop covered in positive x, taking parametric integration, has an area:

A = ∫(t = 0 → π/2) y(t) d[x(t)]
= a2∫(t = 0 → π/2) sin 2t cos t dt
= 2a2∫(t = 0 → π/2) sin t cos2 t dt
= - 2a2∫(t = 0 → π/2) cos2 t d(cos t)
= - (2a2/3) [cos3 t] (t = 0 → π/2)
= 2a2/3

Hence total area = 4a2/3

Please see the 2 loops...
http://img14.imageshack.us/img14/208/picur.png