Question

Find the area of the region bounded by the curve y = \[\sqrt{1 - x^2}\], line y = x and the positive x-axis.

Answer 1

\[y = \sqrt{1 - x^2} \]
\[ \Rightarrow y^2 = 1 - x^2 \]
\[ \Rightarrow x^2 + y^2 = 1\]
Hence,
\[y = \sqrt{1 - x^2}\]  represents the upper half of the circle x² + y² = 1 a circle with centre O(0, 0) and radius 1 unit.
y = x   represents equation of a straight line passing through O(0, 0)
Point of intersection is obtained by solving two equations
\[y = x\]
\[y = \sqrt{1 - x^2}\]
\[ \Rightarrow x = \sqrt{1 - x^2}\]
\[ \Rightarrow x^2 = 1 - x^2 \]
\[ \Rightarrow 2 x^2 = 1 \]
\[ \Rightarrow x = \pm \frac{1}{\sqrt{2}} \]
\[ \Rightarrow y = \pm \frac{1}{\sqrt{2}}\]
\[D\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\text{ and }D'\left( \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}} \right)\text{ are two points of intersection between the circle and the straight line }\]
\[\text{ And }D\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\text{ is the intersection point of }y = \sqrt{1 - x^2}\text{ and }y = x . \]
\[\text{ Required area = Shaded area }\left( ODAEO \right) \]
\[ = \text{ Area }\left( ODEO \right) \hspace{0.167em} + \text{ area }\left( EDAE \right) . . . . . \left( 1 \right)\]
\[\text{ Now, area }\left( ODEO \right) = \int_0^\frac{1}{\sqrt{2}} x dx\]
\[ = \left[ \frac{x^2}{2} \right]_0^\frac{1}{\sqrt{2}} \]
\[ = \frac{1}{2} \left( \frac{1}{\sqrt{2}} \right)^2 \]
\[ = \frac{1}{4}\text{ sq units }. . . . . \left( 2 \right)\]
\[\text{ Area }\left( \text{ EDAE }\right) = \int_\frac{1}{\sqrt{2}}^1 \sqrt{1 - x^2} dx\]
\[ = \left[ \frac{1}{2}x\sqrt{1 - x^2} + \times \frac{1}{2} \times \sin^{- 1} \left( \frac{x}{1} \right) \right]_\frac{1}{\sqrt{2}}^1 \]
\[ = 0 + \frac{1}{2} \sin^{- 1} \left( 1 \right) - \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \sqrt{1 - \left( \frac{1}{\sqrt{2}} \right)^2} - \frac{1}{2} \sin^{- 1} \left( \frac{1}{\sqrt{2}} \right)\]
\[ = \frac{1}{2} \times \frac{\pi}{2} - \frac{1}{4} - \frac{1}{2} \times \frac{\pi}{4} ................\left\{\text{ using, }\sin^{- 1} \left( 1 \right) = \frac{\pi}{2}\text{ and }\sin^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \right\}\]
\[ = \frac{\pi}{4} - \frac{\pi}{8} - \frac{1}{4}\]
\[ = \frac{\pi}{8} - \frac{1}{4}\text{ sq units }. . . . . \left( 3 \right)\]
\[\text{ From }\left( 1 \right), \left( 2 \right)\text{ and }\left( 3 \right),\text{ we get }\]
\[\text{ Area }\left(\text{ODAEO}\right) = \frac{1}{4} + \frac{\pi}{8} - \frac{1}{4} = \frac{\pi}{8}\text{ sq . units }\]