Question

Sketch the region {(x, y) : 9x² + 4y² = 36} and find the area of the region enclosed by it, using integration.

Answer 1

We have, 

\[9 x^2 + 4 y^2 = 36 . . . . . \left( 1 \right)\]

\[ \Rightarrow 4 y^2 = 36 - 9 x^2 \]

\[ \Rightarrow y^2 = \frac{9}{4}\left( 4 - x^2 \right)\]

\[ \Rightarrow y = \frac{3}{2}\sqrt{\left( 4 - x^2 \right)} . . . . . \left( 2 \right)\]

\[\text{ From }\left( 1 \right)\text{ we get }\]

\[ \Rightarrow \frac{x^2}{4} + \frac{y^2}{9} = 1\]

\[\text{ Since in the given equation }\frac{x^2}{4} + \frac{y^2}{9} = 1,\text{ all the powers of both }  x\text{ and }y\text{ are even, the curve is symmetrical about both the axes }. \]

\[ \therefore\text{ Area enclosed by the curve and above }x\text{ axis }= 4 \times\text{ area enclosed by ellipse and }x - \text{ axis in first quadrant }\]

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

\[(0, 3), (0, - 3) \text{ are the points of intersection of curve and } y -\text{ axis }\]

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

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

\[\text{ Approximating rectangle can move between }x = 0 \text{ and }x = 2\]

\[A =\text{  Area of enclosed curve }= 4 \int_0^2 \left| y \right| dx\]

\[ \Rightarrow A = 4 \int_0^2 \frac{3}{2}\sqrt{4 - x^2} d x ................\left[ \text{ From }\left( 2 \right) \right]\]

\[ = 4 \times \frac{3}{2} \int_0^2 \sqrt{4 - x^2} d x\]

\[ = 6 \int_0^2 \sqrt{2^2 - x^2} d x\]

\[ = 6 \left[ \frac{x}{2}\sqrt{2^2 - x^2} + \frac{1}{2} 2^2 \sin^{- 1} \frac{x}{2} \right]_0^2 \]

\[ = 6\left\{ 0 + \frac{1}{2}4 \sin^{- 1} 1 \right\}\]

\[ = 6\left\{ \frac{1}{2} \times 4\left( \frac{\pi}{2} \right) \right\}\]

\[ \Rightarrow A = 6\pi\text{ sq units }\]