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  • The area enclosed by the curves <img alt="y=\sqrt{4-x^{2}},y\geq \sqrt{2}sin\left ( \frac{x\pi}{2\sqrt{2}} \right )" src="https://learn.careers360.com/latex-image/?y%3D%5Csqrt%7B4-x%5E%7B2%7D%

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

Option: 1

\frac{\pi^{2}-8}{\pi^{2}+8}


Option: 2

\frac{\pi^{2}-4}{\pi^{2}+4}


Option: 3

\frac{\pi-4}{\pi-4}


Option: 4

\frac{2\pi^{2}}{2\pi+\pi^{2}-8}


Answers (1)

best_answer

 

Area between two curves -

If we have two functions intersection each other.First find the point of intersection.  Then integrate to find area

\int_{o}^{a}\left [ f\left ( x \right )-9\left ( x \right ) \right ]dx

- wherein

 

 

y=\sqrt{4-x^{2}},y= \sqrt{2}sin\left ( \frac{x\pi}{2\sqrt{2}} \right )

intersect at x=\sqrt{2}

Area of the left of y-axis is \pi

Area of the right of y-axis = \int_{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 )^{\sqrt{2}}_0+\frac{4}{\pi }\cos \frac{x\pi }{2\sqrt{2}}\mid _{0}^{\sqrt{2}}

=\left ( 1+2.\frac{\pi }{4} \right )+\frac{4}{\pi}(0-1)\Rightarrow 1+\frac{\pi}{2}-\frac{4}{\pi}=\frac{2\pi+\pi^{2}-8}{2\pi}

\therefore ratio=\frac{2\pi^{2}}{2\pi+\pi^{2}-8}

Posted by

Rishabh

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