Question

Draw a rough sketch of the graph of the function y = 2 \[\sqrt{1 - x^2}\] , x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.

Answer 1

We have, 

\[y = 2\sqrt{1 - x^2}\]

\[ \Rightarrow \frac{y}{2} = \sqrt{1 - x^2}\]

\[ \Rightarrow \frac{y^2}{4} = 1 - x^2 \]

\[ \Rightarrow x^2 + \frac{y^2}{4} = 1\]

\[ \Rightarrow \frac{x^2}{1} + \frac{y^2}{4} = 1\]

\[\text{ Since in the given equation } \frac{x^2}{1} + \frac{y^2}{4} = 1,\text{ all the powers of both }x\text{ and }y\text{ are even, the curve is symmetrical about both the axes }. \]

\[ \therefore\text{ Required area = area enclosed by ellipse and }x - \text{ axis in first quadrant }\]

\[(1, 0 ), ( - 1, 0) \text{ are the points of intersection of curve and }x - \text{ axis }\]

\[(0, 2), (0, - 2) \text{ are the points of intersection of curve and }y - \text{ axis }\]

\[\text{ Slicing the area in the first quadrant into vertical stripes of height }= \left| y \right|\text{ and width }= dx\]

\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx\]

\[\text{ Approximating rectangle can move between }x = 0\text{ and }x = 1 \]

\[\text{ A = Area of enclosed curve above }x -\text{ axis }= \int_0^1 \left| y \right| dx\]

\[ \Rightarrow A = \int_0^1 2\sqrt{1 - x^2} d x\]

\[ = 2 \int_0^1 \sqrt{1 - x^2} d x\]

\[ = 2 \left[ \frac{1}{2}x\sqrt{1 - x^2} + \frac{1}{2} \sin^{- 1} x \right]_0^1 \]

\[ = 2\left\{ \frac{1}{2} \sin^{- 1} 1 \right\}\]

\[ = 2\left\{ \frac{1}{2}\left( \frac{\pi}{2} - 0 \right) \right\}\]

\[ \Rightarrow A = \frac{\pi}{2}\text{ sq . units }\]