The area of the region bounded by the curves | vedclass

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(C) Given curves are $x^2+y^2=8$ $(i)$ and $y^2=2x$ $(ii)$.Substituting $y^2=2x$ in $(i)$,we get $x^2+2x-8=0$.$(x+4)(x-2)=0$,which gives $x=2$ (since

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$2 \pi+\frac{4}{3}$

Vedclass · Answer

(C) Given curves are $x^2+y^2=8$ $(i)$ and $y^2=2x$ $(ii)$.Substituting $y^2=2x$ in $(i)$,we get $x^2+2x-8=0$.$(x+4)(x-2)=0$,which gives $x=2$ (since $x \ge 0$ for $y^2=2x$).For $x=2$,$y^2=4$,so $y=\pm 2$.The intersection points are $(2, 2)$ and $(2, -2)$.The required area is symmetric about the $x$-axis,so Area $= 2 	imes [	ext{Area of region bounded by } y^2=2x 	ext{ from } x=0 	ext{ to } 2 + 	ext{Area of region bounded by } x^2+y^2=8 	ext{ from } x=2 	ext{ to } 2\sqrt{2}]$.Area $= 2 \left[ \int_0^2 \sqrt{2x} \, dx + \int_2^{2\sqrt{2}} \sqrt{8-x^2} \, dx ight]$.$= 2 \left[ \sqrt{2} \left( \frac{x^{3/2}}{3/2} ight)_0^2 + \left( \frac{x}{2} \sqrt{8-x^2} + \frac{8}{2} \sin^{-1} \frac{x}{\sqrt{8}} ight)_2^{2\sqrt{2}} ight]$.$= 2 \left[ \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} + \left( (0 + 4 \sin^{-1}(1)) - (1 \cdot \sqrt{4} + 4 \sin^{-1}(1/\sqrt{2})) ight) ight]$.$= 2 \left[ \frac{8}{3} + 4(\pi/2) - 2 - 4(\pi/4) ight] = 2 \left[ \frac{8}{3} + 2\pi - 2 - \pi ight] = 2 \left[ \frac{2}{3} + \pi ight] = 2\pi + \frac{4}{3}$.

The area of the region bounded by the curves $x^{2}+y^{2}=8$ and $y^{2}=2x$ is

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