The area of the region in the first quadrant | vedclass

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Question

(B) First,find the intersection points of the circle $x^2+y^2=8$ and the parabola $y^2=2x$.Substitute $y^2=2x$ into the circle equation: $x^2+2x-8=0$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) First,find the intersection points of the circle $x^2+y^2=8$ and the parabola $y^2=2x$.Substitute $y^2=2x$ into the circle equation: $x^2+2x-8=0$.$(x+4)(x-2)=0$,so $x=2$ (since $x \ge 0$ in the first quadrant).At $x=2$,$y^2=4$,so $y=2$.The required area is the area under the circle from $x=0$ to $x=2$ minus the area under the parabola from $x=0$ to $x=2$.Area $= \int_0^2 \sqrt{8-x^2} dx - \int_0^2 \sqrt{2x} dx$$= \left[ \frac{x}{2}\sqrt{8-x^2} + \frac{8}{2}\sin^{-1}\left(\frac{x}{\sqrt{8}}\right) \right]_0^2 - \sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_0^2$$= \left( \frac{2}{2}\sqrt{8-4} + 4\sin^{-1}\left(\frac{2}{2\sqrt{2}}\right) \right) - \frac{2\sqrt{2}}{3}(2^{3/2})$$= (1 \cdot 2 + 4 \cdot \frac{\pi}{4}) - \frac{2\sqrt{2}}{3}(2\sqrt{2})$$= 2 + \pi - \frac{8}{3} = \pi - \frac{2}{3}$.

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

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