Question

Find the area of the region in the first quadrant enclosed by x-axis, the line y = \[\sqrt{3}x\] and the circle x² + y² = 16.

Answer 1

\[x^2 + y^2 = 16\]  represents a circle with centre O(0,0) and cutting the x axis at A(4,0) \[y = \sqrt{3} x\] represents straight passing through O(0,0)

Point of intersection is obtained by solving the two equations

\[x^2 + y^2 = 16\text{ and }y = \sqrt{3} x \]
\[ \Rightarrow x^2 + \left( \sqrt{3} x \right)^2 = 16\]
\[ \Rightarrow 4 x^2 = 16 \]
\[ \Rightarrow x = \pm 2\]
\[ \Rightarrow y = \pm 2\sqrt{3}\]
\[B\left( 2 , 2\sqrt{3} \right)\text{ and }B'\left( - 2 , - 2\sqrt{3} \right) \text{ are points of intersection of circle and straight line }\]
\[\text{ Shaded area }\left( OBQAO \right) =\text{ area }\left( OBPO \right) +\text{ area }\left( PBQAP \right)\]
\[ = \int_0^2 \sqrt{3} x dx + \int_2^4 \sqrt{16 - x^2} dx\]
\[ = \sqrt{3} \left[ \frac{x^2}{2} \right]_0^2 + \left[ \frac{1}{2}x\sqrt{16 - x^2} + \frac{16}{2} \sin^{- 1} \left( \frac{x}{4} \right) \right]_2^4 \]
\[ = 2\sqrt{3} + 8 \times \frac{\pi}{2} - 2\sqrt{3} - 8 \times \frac{\pi}{6}\]
\[ = 2\sqrt{3} + 4\pi - 2\sqrt{3} - \frac{4\pi}{3}\]
\[ = \frac{8\pi}{3}\text{ sq units }\]
\[\text{ Area bound by the circle and straight line above }x\text{ axis }= 2\sqrt{3} + \left( - 2\sqrt{3} + 8 \times \frac{2\pi}{6} \right) = \frac{8\pi}{3}\text{ sq units }\]