The area bounded by the parabola y^{2}=x,the | vedclass

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(B) The line $y=4$ meets the parabola $y^{2}=x$ at point $A$. Substituting $y=4$ into the equation of the parabola,we get $4^{2}=x$,so $x=16$. Thus,po

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$64 / 3$ sq. unit

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(B) The line $y=4$ meets the parabola $y^{2}=x$ at point $A$. Substituting $y=4$ into the equation of the parabola,we get $4^{2}=x$,so $x=16$. Thus,point $A$ is $(16, 4)$.The required area is bounded by the parabola $x=y^{2}$,the $y$-axis $(x=0)$,and the line $y=4$ from $y=0$ to $y=4$.Required Area $= \int_{0}^{4} x \, dy = \int_{0}^{4} y^{2} \, dy$$= \left[ \frac{y^{3}}{3} ight]_{0}^{4}$$= \frac{4^{3}}{3} - \frac{0^{3}}{3} = \frac{64}{3} 	ext{ sq. units}$.

The area bounded by the parabola $y^{2}=x$,the straight line $y=4$,and the $y$-axis in square units is:

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