The area (in sq. units) of the portion lying | vedclass

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(C) The given curves are $x^2+y^2-2ax=0$ (a circle with center $(a, 0)$ and radius $a$) and $y^2=ax$ (a parabola).To find the intersection points, sub

Vedclass · Accepted Answer

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

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(C) The given curves are $x^2+y^2-2ax=0$ (a circle with center $(a, 0)$ and radius $a$) and $y^2=ax$ (a parabola).To find the intersection points, substitute $y^2=ax$ into $x^2+y^2-2ax=0$:$x^2+ax-2ax=0 \implies x^2-ax=0 \implies x(x-a)=0$.So, the curves intersect at $x=0$ and $x=a$.For the portion above the $X$-axis, the area is given by the integral of the difference between the upper curve (circle) and the lower curve (parabola) from $x=0$ to $x=a$:Area $= \int_0^a \left(\sqrt{2ax-x^2} - \sqrt{ax}\right) dx$$= \int_0^a \sqrt{a^2-(x-a)^2} dx - \sqrt{a} \int_0^a x^{1/2} dx$Using the formula $\int \sqrt{r^2-u^2} du = \frac{u}{2}\sqrt{r^2-u^2} + \frac{r^2}{2}\sin^{-1}(\frac{u}{r})$:$= \left[ \frac{x-a}{2}\sqrt{2ax-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x-a}{a}\right) - \sqrt{a} \cdot \frac{2}{3}x^{3/2} \right]_0^a$$= \left( 0 + \frac{a^2}{2}\sin^{-1}(0) - \frac{2}{3}a^2 \right) - \left( \frac{-a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(-1) - 0 \right)$$= (0 + 0 - \frac{2}{3}a^2) - (0 + \frac{a^2}{2}(-\frac{\pi}{2}) - 0)$$= -\frac{2}{3}a^2 + \frac{\pi a^2}{4} = a^2\left(\frac{\pi}{4} - \frac{2}{3}\right)$ sq. units.

The area (in sq. units) of the portion lying above the $X$-axis and enclosed between the curves $y^2=2ax-x^2$ and $y^2=ax$ is

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