Definite Integration Question 421
Question: The area of the region bounded by $ x^{2}+y^{2}-2x-3=0 $ and $ y=| x |+1 $ is
Options:
A) $ \frac{\pi }{2}-1 $ sq. units
B) $ 2\pi $ sq. units
C) $ 4\pi $ sq. units
D) $ \pi /2 $ sq. units
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Answer:
Correct Answer: A
Solution:
[a] $ x^{2}+y^{2}-2x-3=0 $ or $ {{(x-1)}^{2}}+y^{2}=4 $
$ A=\int\limits _{1-\sqrt{2}}^{0}{(\sqrt{4-{{(x-1)}^{2}}}-(-x+1))}dx $
$ +\int\limits_0^{1}{(\sqrt{4-{{(x-1)}^{2}}}-(x+1))}dx $ $ =\frac{x-1}{2}\sqrt{4-{{(x-1)}^{2}}}. +\frac{4}{2}{{\sin }^{-1}}\frac{x-1}{2}+\frac{x^{2}}{2}-x | _{1-\sqrt{2}}^{0} $
$ +\frac{x-1}{2}\sqrt{4-{{(x-1)}^{2}}}+. \frac{4}{2}{{\sin }^{-1}}\frac{x-1}{2}-\frac{x^{2}}{2}-x | $ $ =( -\frac{\sqrt{3}}{2}-\frac{\pi }{3} )-( \frac{-\sqrt{2}}{2}\sqrt{2}-\frac{\pi }{2}+\frac{3-2\sqrt{2}}{2}-1+\sqrt{2} ) $
$ +( -\frac{1}{2}-1 )-( -\frac{\sqrt{3}}{2}-\frac{\pi }{3} ) $ $ =-( -1-\frac{\pi }{2}+\frac{3}{2}-\sqrt{2}-1+\sqrt{2} )-\frac{3}{2} $ $ =\frac{\pi }{2}-1sq.units. $
BETA
