The area (in sq. units) of the smaller region | vedclass

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(B) Given the curves are $x^2+y^2=2ax$ (which is $(x-a)^2+y^2=a^2$) and $y^2=ax$.To find the intersection points,substitute $y^2=ax$ into the circle e

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the curves are $x^2+y^2=2ax$ (which is $(x-a)^2+y^2=a^2$) and $y^2=ax$.To find the intersection points,substitute $y^2=ax$ into the circle equation: $x^2+ax=2ax \Rightarrow x^2-ax=0 \Rightarrow x(x-a)=0$. Thus,$x=0$ or $x=a$.For $x=a$,$y^2=a^2 \Rightarrow y=a$ (since we consider the region above the $X$-axis).The area of the shaded region is the area under the circle minus the area under the parabola from $x=0$ to $x=a$.Area $= \int_0^a (y_{circle} - y_{parabola}) dx = \int_0^a (\sqrt{a^2-(x-a)^2} - \sqrt{ax}) dx$.Area $= \int_0^a \sqrt{a^2-(x-a)^2} dx - \int_0^a \sqrt{ax} dx$.The first integral represents the area of a quarter circle of radius $a$,which is $\frac{\pi a^2}{4}$.The second integral is $\sqrt{a} \int_0^a x^{1/2} dx = \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_0^a = \sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{2a^2}{3}$.Therefore,the required area is $\frac{\pi a^2}{4} - \frac{2a^2}{3} = a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$ sq. units.

The area (in sq. units) of the smaller region lying above the $X$-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is

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