Question

Find the area common to the circle x² + y² = 16 a² and the parabola y² = 6 ax.
                                   OR
Find the area of the region {(x, y) : y² ≤ 6ax} and {(x, y) : x² + y² ≤ 16a²}.

Answer 1

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations
\[x^2 + y^2 = 16 a^2\text{ and }y^2 = 6ax\]
\[ \Rightarrow x^2 + 6ax = 16 a^2 \]
\[ \Rightarrow x^2 + 6ax - 16 a^2 = 0\]
\[ \Rightarrow \left( x + 8a \right)\left( x - 2a \right) = 0\]
\[ \Rightarrow x = 2a\text{ or }x = - 8a , x = - 8a\text{ is not the possible solution . }\]
\[ \therefore\text{ When }x = 2a, y = \pm \sqrt{6a \times 2a} = \pm \sqrt{12}a = \pm 2\sqrt{3}a\]
\[ \therefore B\left( 2a , 2\sqrt{3a} \right)\text{ and }B'\left( 2a , - 2\sqrt{3}a \right)\text{ are points of intersection of the parabola and circle . }\]
\[\text{ Now, Required area = area }\left( OBAB'O \right) \]
\[ = 2 \times\text{ area }\left( OBAO \right)\]
\[ = 2\left\{ \text{ area }\left( OBDO \right) +\text{ area }\left( DBAD \right) \right\}\]
\[ = 2 \times \left[ \int_0^{2a} \sqrt{6ax}dx + \int_{2a}^{4a} \sqrt{16 a^2 - x^2} dx \right]\]
\[ = 2 \times \left\{ \left[ \sqrt{6a}\frac{x^\frac{3}{2}}{\frac{3}{2}} \right]_0^{2a} + \left[ \frac{1}{2}x\sqrt{16 a^2 - x^2} + \frac{1}{2} \times 16 a^2 \sin^{- 1} \left( \frac{x}{4a} \right) \right]_{2a}^{4a} \right\}\]
\[ = 2 \times \left\{ \left( \sqrt{6a} \times \frac{2}{3} \times \left( 2a \right)^\frac{3}{2} - 0 \right) + \left( \frac{1}{2} \times 4a\sqrt{16 a^2 - \left( 4a \right)^2} + \frac{1}{2} \times 16 a^2 \sin^{- 1} \frac{4a}{4a} - \frac{1}{2} \times 2a\sqrt{16 a^2 - \left( 2a \right)^2} - \frac{1}{2} \times 16 a^2 \sin^{- 1} \frac{2a}{4a} \right) \right\}\]
\[ = 2 \times \left\{ \left( \sqrt{6a} \times \frac{2}{3} \times 2a\sqrt{2a} \right) + 0 + 8 a^2 \sin^{- 1} \left( 1 \right) - 2\sqrt{3} a^2 - 8a \sin^{- 1} \left( \frac{1}{2} \right) \right\}\]
\[ = 2 \times \left[ \frac{8 a^2 \sqrt{3}}{3} + 8 a^2 \times \frac{\pi}{2} - 2\sqrt{3} a^2 - 8 a^2 \frac{\pi}{6} \right]\]
\[ = 2 \left\{ \left( \frac{8\sqrt{3} - 6\sqrt{3}}{3} \right) a^2 + 8\left( \frac{\pi}{2} - \frac{\pi}{6} \right) a^2 \right\}\]
\[ = 2\left\{ \frac{2\sqrt{3}}{3} a^2 + 8 a^2 \left( \frac{2\pi}{6} \right) \right\}\]
\[ = \frac{4\sqrt{3}}{3} a^2 + \frac{16\pi}{3} a^2 \]
\[ = \frac{4 a^2}{3}\left( 4\pi + \sqrt{3} \right)\]