Question

Draw a rough sketch of the graph of the curve \[\frac{x^2}{4} + \frac{y^2}{9} = 1\]  and evaluate the area of the region under the curve and above the x-axis.

Answer 1

\[\text{ Since in the given equation }\frac{x^2}{4} + \frac{y^2}{9} = 1,\text{ all the powers of both }x\text{ and }y \text{ are even, the curve is symmetrical about both the axis }. \]
\[ \therefore\text{ Area encloed by the curve and above }x \text{ axis = area }A' BA = 2 \times \text{ area enclosed by ellipse and } x -\text{ axis in first quadrant }\]
\[(2, 0 ), ( - 2, 0) \text{ are the points of intersection of curve and }x - \text{ axis }\]
\[(0, 3), (0, - 3) \text{ are the points of intersection of curve and } y -\text{ axis }\]
\[\text{ Slicing the area in the first quadrant into vertical stripes of height }= \left| y \right| \text{ and width }= dx\]
\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx\]
\[\text{ Approximating rectangle can move between }x = 0\text{ and }x = 2 \]
\[A =\text{ Area of enclosed curve above }x - \text{ axis }= 2 \int_0^2 \left| y \right| dx\]
\[ \Rightarrow A = 2 \int_0^2 y dx\]
\[ \Rightarrow A = 2 \int_0^2 \frac{3}{2}\sqrt{4 - x^2}dx\]
\[ \Rightarrow A = 3 \int_0^2 \sqrt{4 - x^2}dx\]
\[ \Rightarrow A = 3 \left[ \frac{1}{2}x \sqrt{4 - x^2} + \frac{1}{2} 4 \sin^{- 1} x \right]_0^2 \]
\[ = 3\left[ 0 + \frac{1}{2} \times 4 \sin^{- 1} 1 \right] = 3 \times \frac{1}{2} \times 4 \times \frac{\pi}{2} = 3\pi \text{ sq . units }\]
\[ \therefore\text{ Area of enclosed region above }x -\text{ axis }= 3\pi\text{ sq . units }\]