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Q.
Area bounded by the curves $x^{2} + y^{2} \le\, 8x$ and $y^{2} \ge\, 4x$ lies in the first quadrant is equal to
Application of Integrals
Solution:
Given $x^{2} + y^{2} \le\, 8x$
Let $x^{2} + y^{2} = 8x$, a circle with centre $(4,0)$ and radius $4\quad\ldots\left(i\right)$
$y^{2}\ge\,4x$, $y^{2}=4x$ a parabola with vertex $(0,0)$ $\quad\ldots\left(ii\right)$
$\therefore \quad$ From $(i)$ and $(ii)$, we have $x^{2}=4x $
$ \Rightarrow \quad x=0, x=4$
Required area = area of shaded region
$=\int\limits_{0}^{4} \sqrt{4^{2}-\left(x-4\right)^{2}}-\int\limits_{0}^{4}2\sqrt{x}\,dx$
$=\left[\frac{\left(x-4\right)}{2}\sqrt{4^{2}-\left(x-4\right)^{2}}+\frac{16}{2}sin^{-1}\left(\frac{x-4}{4}\right)\right]_{0}^{4}-2\times\frac{2}{3}\left[x^{3/ 2}\right]_{0}^{4}$
$=8\times\frac{\pi}{2}-\frac{4}{3}\times\left(4\right)^{3/ 2}$
$=\left(4\pi-\frac{32}{3}\right)=\frac{32}{3}\left(\frac{3\pi}{8}-1\right)$ sq. units
