MvCalc36

Given domains are $3x=4-y^2,x=y^2\,$ . we have to find the limits for evaluating the integration $A=\iint_D dx dy\,$

Solving the given equations for y, we get $3y^2=4-y^2,4y^2=4,y^2=1,y=1,-1\,$

Hence the limits of y are from -1 to 1.

The limits of x are from $y^2\,$ to $\frac{4-y^2}{3}\,$

Therefore the area of the plane A= $\int_{-1}^{1}[\int_{x=y^2}^{\frac{4-y^2}{3}} dx]dy\,$

= $\int_{-1}^{1}[x]_{x=y^2}^{\frac{4-y^2}{3}} dy\,$

= $\int_{-1}^{1}[\frac{4-y^2}{3}-y^2]dy\,$

= $\int_{-1}^{1}[\frac{4}{3}-\frac{4y^2}{3}]dy\,$

= $[\frac{4}{3}y-\frac{4y^3}{9}]_{-1}^{1}=[\frac{4}{3}-\frac{4}{9}]-[-\frac{4}{3}+\frac{4}{9}]\,$

= $\frac{8}{3}-\frac{8}{9}=\frac{16}{9}\,$