$\begin{aligned}
& x\left(1+y^2\right)=1 \
& y^2=2 x
\end{aligned}Fromequation(1)&(2)\begin{aligned}
x(1+2 x)=1 & \Rightarrow 2 x^2+x-1=0 \
& \Rightarrow x=\frac{1}{2}, x=-1 \text { (Reject) } \
& \Rightarrow y^2=2\left(\frac{1}{2}\right) \
& \Rightarrow y= \pm 1
\end{aligned}$ 
Area bounded =∫−11(1+y21−2y2)dy=(tan−1y−6y3)−11=2π−31