Q21 of 123 Page 22

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)

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