Let the equation of curve y2(2a−x)=x3
and equation of line x=2a The given curve is symmetrical about x -axis and passes through origin.
From
(i) we have, y2=2a−xx3
But 2a−xx3<0 for x>2a and x<0 So, curve does not lie in the portion x>2a and x<0, therefore curve lies in 0≤x≤2a . ∴ Area bounded by the curve and line =0∫2aydx=0∫2a2a−xx3/2dx
Put x=2asin2θ and dx=4asinθcosθdθ ∴I=0∫π/28a2sin4θdθ=8a2[43⋅21⋅2π]
using 0∫π/2sinmxcosnxdx =2Γ2m+n+2Γ2m+1Γ2n+1=23πa2 sq. unit