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Please refer the below image for solution.  Solutionu(y)=3sqrt(y+3),quad v(y)=y+3To find the integral for the required area, first we have to find the point of intersection bet ween the given curves u(y)=3sqrt(y+3), and v(y)=y+3:.3sqrt(y+3)=y+3Squaring we get, (3sqrt(y+3))^(2)=(y+3)^(2){:[9(y+3)=y^(2)+6y+9],[9y+27=y^(2)+6y+9],[:.y^(2)+6y-9y+9-27=0],[y^(2)-3y-18=0 ... See the full answer