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Q.
The area given by $x+y \, \le\,6$, $x^{2}+y^{2}\,\le\,6y$ and $y^{2}\,\le\,8x$ is
Application of Integrals
Solution:
We have, $x + y \,\le\, 6$ or $x + y = 6$, a st. line $\quad...\left(i\right)$
$x^{2} + y^{2}\,\le\, 6y$ or $x^{2} + y^{2} - 6y = 0$, a circle $...\left(ii\right)$ with centre $\left(0,3\right)$ and $ r=3$
and $y^{2} \,\le\, 8x$ or $y^{2} = 8x$, a parabola with vertex $\left(0, 0\right)$ $\quad\ldots\left(iii\right)$
The line and circle meet at $\left(3,3\right)$ and the line and parabola meet at $\left(2,4\right)$
Required area = area of shaded region
$A=\int_{0}^{3}\sqrt{6y-y^{2}}\, dy+\int_{3}^{4}\left(6-y\right)dy-\int_{0}^{4}\frac{y^{2}}{8} dy$
$=\int\limits_{0}^{3}\sqrt{-\left(y^{2}-6y+9-9\right)}\, dy+\int\limits_{3}^{4}\left(6-y\right) dy-\int\limits_{0}^{4}\frac{y^{2}}{8} dy$
$=\int\limits_{0}^{3}\sqrt{-\left(y-3\right)^{2} +3^{2}} dy+\left|6y-\frac{y^{2}}{2}\right|_{3}^{4} -\left|\frac{y^{3}}{24}\right|_{0}^{4}$
$=\left|\frac{y-3}{2}\sqrt{3^{2}-\left(y-3\right)^{2}}+\frac{9}{2}sin^{-1}\left(\frac{y-3}{3}\right)\right|_{0}^{3}$
$+\left(\left(24-18\right)-\left(8-\frac{9}{2}\right)\right)-\left(\frac{64}{24}\right)$
$=\frac{9\pi}{4}+\frac{5}{2}-\frac{8}{3}$
$=\frac{9\pi}{4}-\frac{1}{6}$
$=\left(\frac{27\,\pi-2}{12}\right)$ sq. units
