The area of the region lying above the X-axis | vedclass

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(B) Given,the equation of the circle is $x^{2}+y^{2}=2ax$,which can be written as $(x-a)^{2}+y^{2}=a^{2}$.The equation of the parabola is $y^{2}=ax$,w

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$a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

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(B) Given,the equation of the circle is $x^{2}+y^{2}=2ax$,which can be written as $(x-a)^{2}+y^{2}=a^{2}$.The equation of the parabola is $y^{2}=ax$,where $a>0$.To find the intersection points,substitute $y^{2}=ax$ into the circle equation:$x^{2}+ax=2ax$$x^{2}-ax=0$$x(x-a)=0$Thus,$x=0$ or $x=a$.The intersection points are $(0,0)$ and $(a, a)$ (considering the region above the $X$-axis).The required area is the area of the quarter circle in the first quadrant minus the area under the parabola from $x=0$ to $x=a$.Area $= \int_{0}^{a} \sqrt{2ax-x^{2}} dx - \int_{0}^{a} \sqrt{ax} dx$$= \frac{\pi a^{2}}{4} - \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$$= \frac{\pi a^{2}}{4} - \frac{2}{3} \sqrt{a} (a^{3/2})$$= \frac{\pi a^{2}}{4} - \frac{2a^{2}}{3}$$= a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

The area of the region lying above the $X$-axis and included between the circle $x^{2}+y^{2}=2ax$ and the parabola $y^{2}=ax$,$a>0$ is

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