Question

The area of the region formed by $x^2+y^2-6x-4y+12\le 0$, $y\le x$ and $x\le \frac{5}{2}$ is?

• A. $(\frac{1}{16}+{\sin }^{-1}\frac{\pi }{6}\frac{5\sqrt{3}}{7})$ sq. units
• B. $(\frac{1}{8}+\frac{5\sqrt{3}}{7}+\frac{\pi }{6})$ sq. units
• C. $(\frac{1}{8}+\frac{5\sqrt{3}}{8}+\frac{\pi }{6})$ sq. units
• D. None of these

Answer 1

Answer: (D)

None of these

$x^2+y^2-6x-4y+12\le 0$
$\Rightarrow {(x-3)}^2+{(y-2)}^2\le 1$
$\mathrm{Re}quiredArea=\underset{2}{\overset{5/2}{\int }}xdx-\underset{2}{\overset{5/2}{\int }}2+\sqrt{1+{(x-3)}^2}dx$
$={\left[\frac{x^2}{2}\right]}_2^{5/2}-{\left[2x\right]}_2^{5/2}-\underset{2}{\overset{5/2}{\int }}\sqrt{1+{(x-3)}^2}dx$
$=(\frac{25}{8}-2)-(5-4)-{[\left(\frac{x-3}{2}\right)\sqrt{1+{(x-3)}^2}+\frac{1}{2}{\sin }^{-1}(x-3)]}_2^{5/2}$
$=\frac{25}{8}-2-1-[(\frac{-1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}{\sin }^{-1}\left(\frac{-1}{2}\right))-(0+\frac{1}{2}{\sin }^{-1}(-1))]$
$=\frac{1}{8}-[(\frac{-\sqrt{3}}{8}-\frac{\pi }{12})-(0-\frac{\pi }{4})]$
$=\frac{1}{8}+\frac{\sqrt{3}}{8}+\frac{\pi }{12}-\frac{\pi }{4}$
$=\frac{1+\sqrt{3}}{8}-\frac{\pi }{6}$