Question

The area given by $x+y\le 6$, $x^2+y^2\le 6y$ and $y^2\le 8x$ is

• A. $\left(\frac{27\pi +2}{12}\right)$ sq. units
• B. $\left(\frac{27\pi -2}{12}\right)$ sq. units
• C. $\left(\frac{27\pi -1}{6}\right)$ sq. units
• D. $\left(\frac{27\pi +1}{6}\right)$ sq. units

Answer 1

Answer: (B)

We have, $x+y\le 6$ or $x+y=6$, a st. line $...\left(i\right)$
$x^2+y^2\le 6y$ 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 $\left(0,0\right)$ $\dots \left(iii\right)$
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
$A={\int }_0^3\sqrt{6y-y^2}dy+{\int }_3^4\left(6-y\right)dy-{\int }_0^4\frac{y^2}{8}dy$
$=\underset{0}{\overset{3}{\int }}\sqrt{-\left(y^2-6y+9-9\right)}dy+\underset{3}{\overset{4}{\int }}\left(6-y\right)dy-\underset{0}{\overset{4}{\int }}\frac{y^2}{8}dy$
$=\underset{0}{\overset{3}{\int }}\sqrt{-{\left(y-3\right)}^2+3^2}dy+{\left|6y-\frac{y^2}{2}\right|}_3^4-{\left|\frac{y^3}{24}\right|}_0^4$
$={\left|\frac{y-3}{2}\sqrt{3^2-{\left(y-3\right)}^2}+\frac{9}{2}si{n}^{-1}\left(\frac{y-3}{3}\right)\right|}_0^3$
$+\left(\left(24-18\right)-\left(8-\frac{9}{2}\right)\right)-\left(\frac{64}{24}\right)$
$=\frac{9\pi }{4}+\frac{5}{2}-\frac{8}{3}$
$=\frac{9\pi }{4}-\frac{1}{6}$
$=\left(\frac{27\pi -2}{12}\right)$ sq. units