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Miscellaneous Exercise 5(II) · Q47

Q.Find the area of the region in first quadrant bounded by the circle x2+y2=4x^2 + y^2 = 4 and the x axis and the line x=y3x = y\sqrt3.

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The line x=y3x=y\sqrt3, i.e. y=x3y=\frac{x}{\sqrt3}, has slope tan⁡30∘\tan30^\circ, so it makes a 30∘30^\circ angle with the X-axis. It meets the circle x2+y2=4x^2+y^2=4 where x2+x23=4⇒x2=3⇒x=3, y=1x^2+\frac{x^2}{3}=4\Rightarrow x^2=3\Rightarrow x=\sqrt3,\ y=1. Splitting the sector into a triangular piece under the line (from x=0x=0 to x=3x=\sqrt3) and a curved piece under the circular arc (from x=3x=\sqrt3 to x=2x=2): …

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