Area enclosed between the curve y^{2} (2a – x) = x^{3} and the line x = 2a above x-axis is

y^{2}(2a – x) = x^{3}

⇒

⇒

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 =

=

=

=

=

=

=

=

=

= 4a^{2}(π/2 – 0 – 0 + 0) – a^{2}(π/2 – 0 – 0 + 0)

= πa^{2} (Ans)

1