The area enclosed by the parabola ay = 3(a2 – x2) and the x- axis is

4a2 sq. units  a2 sq. units 6a2 sq. units 5a2 sq. units

4a2 sq. units

The parabola cuts the x-axis at points (–a, 0) and (a, 0). $A = \int\limits_{-a}^ay\,dx=2\frac{3}{a}\int\limits_{0}^a(a^2-x^2)dx$ = 4a2 sq. units. Hence (A) is the correct answer.