Question
The area of the region formed by $${x^2} + {y^2} - 6x - 4y + 12 \leqslant 0,\,y \leqslant x{\text{ and }}x \leqslant \frac{5}{2}{\text{ is :}}$$
A.
$$\left( {\frac{\pi }{6} - \frac{{\sqrt 3 + 1}}{8}} \right){\text{ sq}}{\text{.}}\,{\text{unit}}$$
B.
$$\left( {\frac{\pi }{6} + \frac{{\sqrt 3 - 1}}{8}} \right){\text{ sq}}{\text{.}}\,{\text{unit}}$$
C.
$$\left( {\frac{\pi }{6} - \frac{{\sqrt 3 - 1}}{8}} \right){\text{ sq}}{\text{.}}\,{\text{unit}}$$
D.
None of these
Answer :
$$\left( {\frac{\pi }{6} - \frac{{\sqrt 3 - 1}}{8}} \right){\text{ sq}}{\text{.}}\,{\text{unit}}$$
Solution :
The required area
$$\eqalign{ & {\text{ = }}\int_2^{\frac{5}{2}} {x\,dx - } \int_2^{\frac{5}{2}} {\left[ {2 - \sqrt {1 - {{\left( {x - 3} \right)}^2}} } \right]dx} \cr & = \left[ {\frac{{{x^2}}}{2}} \right]_2^{\frac{5}{2}} - \left[ {2x} \right]_2^{\frac{5}{2}} + \left[ {\frac{{x - 3}}{2}\sqrt {1 - {{\left( {x - 3} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {x - 3} \right)} \right]_2^{\frac{5}{2}} \cr & = \frac{9}{8} - 1 + \left( { - \frac{{\sqrt 3 }}{8} + \frac{\pi }{6}} \right) \cr & = \frac{\pi }{6} - \frac{{\sqrt 3 - 1}}{8}{\text{ sq}}{\text{. unit}} \cr} $$
The required area
$$\eqalign{ & {\text{ = }}\int_2^{\frac{5}{2}} {x\,dx - } \int_2^{\frac{5}{2}} {\left[ {2 - \sqrt {1 - {{\left( {x - 3} \right)}^2}} } \right]dx} \cr & = \left[ {\frac{{{x^2}}}{2}} \right]_2^{\frac{5}{2}} - \left[ {2x} \right]_2^{\frac{5}{2}} + \left[ {\frac{{x - 3}}{2}\sqrt {1 - {{\left( {x - 3} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {x - 3} \right)} \right]_2^{\frac{5}{2}} \cr & = \frac{9}{8} - 1 + \left( { - \frac{{\sqrt 3 }}{8} + \frac{\pi }{6}} \right) \cr & = \frac{\pi }{6} - \frac{{\sqrt 3 - 1}}{8}{\text{ sq}}{\text{. unit}} \cr} $$