First find intersection by equating values of y^2,
x+1=7-x
Solve for x
x=3
consequently
y^2=7-x=4 =>
[tex]y=\pm2[/tex]
The intersection points of the two curves are therefore (3,2) and (3,-2).
The area enclosed by the two curves is therefore obtained by integration within appropriate limits.
[tex]Area=\int_{-2}^{+2}\int_{y^2-1}^{7-y^2}1dxdy[/tex]
[tex]=\int_{-2}^{+2} [x]_{y^2-1}^{7-y^2}dy[/tex]
[tex]=\int_{-2}^{+2} (8-2y^2)dy[/tex]
[tex]=[8y-\frac{2y^3}{3}]_{-2}^{2}[/tex]
[tex]=8(2-(-2))-\frac{32}{3}[/tex]
[tex]=32-\frac{32}{3}[/tex]
[tex]=\frac{64}{3}[/tex]
Answer: Area of between the two curves is [tex]\frac{64}{3}[/tex]
