Definite Integration Question 172
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
Options:
A) $ ( \frac{\pi }{6}-\frac{\sqrt{3}+1}{8} )squnit $
B) $ ( \frac{\pi }{6}+\frac{\sqrt{3}-1}{8} )squnit $
C) $ ( \frac{\pi }{6}-\frac{\sqrt{3}-1}{8} )squnit $
D) None of theses
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The required area $ =\int_2^{5/2}{xdx-\int_2^{5/2}{[ 2-\sqrt{1-{{(x-3)}^{2}}} ]dx}} $
$ =[ \frac{x^{2}}{2} ]_2^{5/2}-[2x]_2^{5/2}+ $
$ [ \frac{x-3}{2}\sqrt{1-{{(x-3)}^{2}}}+\frac{1}{2}{{\sin }^{-1}}(x-3) ]_2^{5/2} $
$ =\frac{9}{8}-1+( -\frac{\sqrt{3}}{8}+\frac{\pi }{6} )=\frac{\pi }{6}-\frac{\sqrt{3}-1}{8} $ sq. unit.
BETA
