Question

Find the area of the region {(x, y) : y² ≤ 8x, x² + y² ≤ 9}.

Answer 1

\[\text{ Let }R = \left\{ \left( x, y \right): y^2 \leq 8x, x^2 + y^2 \leq 9 \right\}\]
\[ R_1 = \left\{ \left( x, y \right): y^2 \leq 8x \right\}\]
\[ R_2 = \left\{ \left( x, y \right): x^2 + y^2 \leq 9 \right\}\]
\[\text{ Thus, }R = R_1 \cap R_2 \]
\[\text{ Now, }y^2 = 8x\text{ represents a parabola with vertex O(0, 0) and symmetrical about }x -\text{ axis }\]
\[\text{ Thus, }R_1\text{ such that }y^2 \leq 8x\text{ is the area inside the parabola }\]
\[\text{ Also, }x^2 + y^2 = 9\text{ represents with circle with centre O(0, 0) and radius 3 units .} \]
\[\text{ The circle cuts the x axis at C(3, 0) and C'( - 3, 0 ) and }Y -\text{ axis at B(0, 3) and B'(0, - 3 ) }\]
\[\text{ Thus, }R_2 \text{ such that }x^2 + y^2 \leq 9\text{ is the area inside the circle }\]
\[ \Rightarrow R = R_1 \cap R_2 =\text{ Area OACA'O }= 2 \left( \text{ shaded area OACO }\right) . . . \left( 1 \right)\]
The point of intersection between the two curves is obtained by solving the two equations

\[ y^2 = 8x\text{ and }x^2 + y^2 = 9 \]
\[ \Rightarrow x^2 + 8x = 9 \]
\[ \Rightarrow x^2 + 8x - 9 = 0\]
\[ \Rightarrow \left( x + 9 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x = - 9\text{ or }x = 1\]
\[\text{ Since, parabola is symmetric about + ve }x - \text{ axis, }x = 1\text{ is the correct solution }\]
\[ \Rightarrow y^2 = 8 \]
\[ \Rightarrow y = \pm 2\sqrt{2}\]
\[\text{ Thus, A}\left( 1, 2\sqrt{2} \right)\text{ and A' }\left( 1, - 2\sqrt{2} \right)\text{ are the two points of intersection }\]
\[\text{ Area OACO = area OADO + area DACD }. . . \left( 2 \right)\]
\[\text{ Area OADO }= \int_0^1 \sqrt{8x} dx .............\left[ \text{ Area bound by curve }y^2 = 8x\text{ between }x = 0\text{ and }x = 1 \right]\]
\[ = 2\sqrt{2} \left[ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right]_0^1 \]
\[ \Rightarrow\text{ Area OADO }= \frac{4\sqrt{2}}{3} . . . \left( 3 \right)\]
\[ \therefore\text{ Area DACD = area bound by }x^2 + y^2 = 9 \text{ between
}x = 1\text{ to }x = 3\]
\[ \Rightarrow A = \int_1^3 \sqrt{9 - x^2} dx \]
\[ = \left[ \frac{1}{2}x\sqrt{9 - x^2} + \frac{1}{2}9 \sin^{- 1} \left( \frac{x}{3} \right) \right]_1^3 \]
\[ = 0 + \frac{9}{2} si n^{- 1} \left( \frac{3}{3} \right) - \frac{1}{2}\sqrt{9 - 1^2} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right)\]
\[ = \frac{9}{2} si n^{- 1} 1 - \frac{1}{2}\sqrt{8} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right)\]
\[ = \frac{9}{2} \frac{\pi}{2} - \frac{1}{2}2\sqrt{2} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right)\]
\[ \Rightarrow\text{ Area DACD }= 9 \frac{\pi}{4} - \sqrt{2} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right) . . . \left( 4 \right)\]
\[\text{ From }\left( 1 \right), \left( 2 \right), \left( 3 \right)\text{ and }\left( 4 \right)\]
\[R =\text{ Area OACA'O }\]
\[ = 2\left( \frac{4\sqrt{2}}{3} + 9 \frac{\pi}{4} - \sqrt{2} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right) \right)\]
\[ = 2\left( \frac{4\sqrt{2}}{3} - \sqrt{2} + 9 \frac{\pi}{4} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right) \right)\]
\[ \therefore\text{ Area OACA'O }= 2\left( \frac{\sqrt{2}}{3} + \frac{9\pi}{4} - \frac{9}{2} si n^{- 1} \left( \frac{1}{3} \right) \right) \text{ sq . units }\]