Definite Integration Question 300
Question: The area of region $ {(x,y):x^{2}+y^{2}\le 1\le x+y} $ is
[Kerala (Engg.) 2002]
Options:
A) $ \frac{{{\pi }^{2}}}{5} $
B) $ \frac{{{\pi }^{2}}}{2} $
C) $ \frac{{{\pi }^{2}}}{3} $
D) $ \frac{\pi }{4}-\frac{1}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ x^{2}+y^{2}=1,x+y=1 $ meet when $ x^{2}+{{(1-x)}^{2}}=1\Rightarrow x^{2}+1+x^{2}-2x=1 $
$ \Rightarrow 2x^{2}-2x=0\Rightarrow 2x(x-1)=0 $
$ \Rightarrow x=0,x=1 $
therefore $ y=1,y=0 $ , i.e., $ A(1,0);B(0,1) $
Required area $ =\int_0^{1}{[\sqrt{1-x^{2}}-(1-x)]} $ dx $ =[ \frac{x\sqrt{1-x^{2}}}{2}+\frac{1}{2}{{\sin }^{-1}}x-x+\frac{x^{2}}{2} ]_0^{1} $
$ =\frac{1}{2}.\frac{\pi }{2}-1+\frac{1}{2}=\frac{\pi }{4}-\frac{1}{2} $ .
BETA
