Area enclosed between the curve y2 (2a – x) = x3 and the line x = 2a above x-axis is
y2(2a – x) = x3
⇒
⇒
Since we are concerned with the area above the x – axis, we’ll be considering the positive root.
We can see that at x = 0, y = 0.
So,
A =
Putting √x = u
We get du = (1/2√x)dx
or, dx = 2u du
So,
A =
Putting u = √2√a sin t
We get du = √2√a cos t dt
A =
=
=
=
=
=
=
=
=
= 4a2(π/2 – 0 – 0 + 0) – a2(π/2 – 0 – 0 + 0)
= πa2 (Ans)