question_answer

The area enclosed by the curve \[y=\sqrt{4-{{x}^{2}}},\] \[y\ge \sqrt{2}\sin \left( \frac{x\pi }{2\sqrt{2}} \right)\], and the x-axis is divided by the y-axis in the ratio

A) \[\frac{{{\pi }^{2}}-8}{{{\pi }^{2}}+8}\]       

B) \[\frac{{{\pi }^{2}}-4}{{{\pi }^{2}}+4}\]

C) \[\frac{\pi -4}{\pi -4}\]   

D) \[\frac{2{{\pi }^{2}}}{2\pi +{{\pi }^{2}}-8}\]

Correct Answer: D

Solution :

[d] \[y=\sqrt{4-{{x}^{2}}},y=\sqrt{2}\sin \left( \frac{x\pi }{2\sqrt{2}} \right)\]intersect at \[x=\sqrt{2}\] Area to the left of y-axis is \[\pi \] Area to the right of y-axis \[=\int\limits_{0}^{\sqrt{2}}{\left( \sqrt{4-{{x}^{2}}}-\sqrt{2}\sin \frac{x\pi }{2\sqrt{2}} \right)dx}\] \[=\left( \frac{x\sqrt{4-{{x}^{2}}}}{2}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right)_{0}^{\sqrt{2}}+\left( \frac{4}{\pi }\cos \frac{x\pi }{2\sqrt{2}} \right)_{0}^{\sqrt{2}}\]\[=\left( 1+2\times \frac{\pi }{4} \right)+\frac{4}{\pi }(0-1)=1+\frac{\pi }{2}-\frac{4}{\pi }\] \[=\frac{2\pi +{{\pi }^{2}}=8}{2\pi }\] Sq. units. \[\therefore \] Ratio \[=\frac{2{{\pi }^{2}}}{2\pi +{{\pi }^{2}}-8}\]