JEE Main On 8 April 2017 Question 19
Question: The area (in sq. units) of the smaller portion enclosed between the curves, $ x^{2}+y^{2}=4 $ and $ y^{2}=3x, $ is: [JEE Online 08-04-2017]
Options:
A) $ \frac{1}{\sqrt{3}}+\frac{4\pi }{3} $
B) $ \frac{1}{\sqrt{3}}+\frac{2\pi }{3} $
C) $ \frac{1}{2\sqrt{3}}-\frac{\pi }{3} $
D) $ \frac{1}{2\sqrt{3}}+\frac{2\pi }{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ x^{2}+3x-4=0 $ $ (x+4)(x-1)=0 $ $ x=-4,x=1 $ Area $ =( \int\limits_0^{1}{\sqrt{3}.\sqrt{x}dx} )+\int\limits_0^{1}{( \int\limits_0^{1}{\sqrt{3}.\sqrt{x}dx}+\int\limits_1^{2}{\sqrt{4-x^{2}}dx} )\times 2} $ $ =( \sqrt{3}( \frac{{x^{3/2}}}{3/2} )_0^{1}+( \frac{x}{2}\sqrt{4-x^{2}}+2{{\sin }^{-1}}\frac{x}{2} )_1^{2} )\times 2 $ $ =( \sqrt{3}( \frac{2}{3} )+{ 2.\frac{\pi }{2}-\frac{\sqrt{3}}{2}+\frac{\pi }{3} } )\times 2 $ $ ( \frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi }{3} )\times 2 $ $ =( \frac{2}{2\sqrt{3}}+\frac{2\pi }{3} )\times 2=\frac{1}{\sqrt{3}}+\frac{4\pi }{3} $ d
