The area of the region bounded by the curves | vedclass

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(A) Given curves are $y=x^{2}$ and $x=y^{2}$,which are parabolas.To find the points of intersection,substitute $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{2}

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$1/3$

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(A) Given curves are $y=x^{2}$ and $x=y^{2}$,which are parabolas.To find the points of intersection,substitute $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{2}$$x=x^{4}$$x^{4}-x=0$$x(x^{3}-1)=0$$x(x-1)(x^{2}+x+1)=0$Since $x^{2}+x+1=0$ has no real roots,we have $x=0$ and $x=1$.When $x=0$,$y=0$. When $x=1$,$y=1$.Thus,the points of intersection are $(0,0)$ and $(1,1)$.The area of the bounded region is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:$	ext{Area} = \int_{0}^{1} (\sqrt{x} - x^{2}) \, dx$$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^{3}}{3} ight]_{0}^{1}$$= \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^{3} ight]_{0}^{1}$$= (\frac{2}{3}(1) - \frac{1}{3}(1)) - (0 - 0)$$= \frac{2}{3} - \frac{1}{3} = \frac{1}{3} 	ext{ sq units}$.

The area of the region bounded by the curves $y=x^{2}$ and $x=y^{2}$ is

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