The area of the region bounded by the curves | vedclass

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(C) Given curves are $x(1+y^2)=1$ and $y^2=2x$.From the second equation,$x = \frac{y^2}{2}$.Substituting this into the first equation: $\frac{y^2}{2}(

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$\frac{\pi}{2}-\frac{1}{3}$

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(C) Given curves are $x(1+y^2)=1$ and $y^2=2x$.From the second equation,$x = \frac{y^2}{2}$.Substituting this into the first equation: $\frac{y^2}{2}(1+y^2) = 1 \Rightarrow y^2 + y^4 = 2 \Rightarrow y^4 + y^2 - 2 = 0$.Let $y^2 = t$,then $t^2 + t - 2 = 0 \Rightarrow (t+2)(t-1) = 0$.Since $y^2 = t \ge 0$,we have $t = 1$,so $y^2 = 1 \Rightarrow y = \pm 1$.For $y = \pm 1$,$x = \frac{1}{2}$.The area bounded by the curves is given by $\int_{-1}^{1} \left( \frac{1}{1+y^2} - \frac{y^2}{2} ight) dy$.$= \left[ 	an^{-1}(y) - \frac{y^3}{6} ight]_{-1}^{1}$.$= \left( 	an^{-1}(1) - \frac{1}{6} ight) - \left( 	an^{-1}(-1) - \frac{(-1)^3}{6} ight)$.$= \left( \frac{\pi}{4} - \frac{1}{6} ight) - \left( -\frac{\pi}{4} + \frac{1}{6} ight)$.$= \frac{\pi}{4} + \frac{\pi}{4} - \frac{1}{6} - \frac{1}{6} = \frac{\pi}{2} - \frac{1}{3}$.

The area of the region bounded by the curves $x(1+y^2)=1$ and $y^2=2x$ is :

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