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UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow:

Consider the line

x=\sqrt{3}y

and the circle

x^2+y^2=4

[NDA I 2015]

Question : 76 of 120

Marks: +1, -0

x

x=\sqrt{3}

Solution:

The area bounded by the circle in the given interval can calculated by evaluating the following definite integral.

\Delta_{\text{area}}=\int\limits_{0}^{\sqrt{3}}\sqrt{4-x^{2}}\,dx

=\left[\frac{x\sqrt{4-x^{2}}}{2}+\frac{1}{2}\times 2^{2}\times\sin^{-1}\frac{x}{2}\right]^{\sqrt{3}}_{0}

=\frac{\sqrt{3}\sqrt{4-3}}{2}+\frac{1}{2}\times 2^{2}\times\sin^{-1}\frac{\sqrt{3}}{2}

=\frac{\sqrt{3}}{2}+\frac{2\pi}{3}

Hence, option D is correct

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