The area enclosed with in the ellipse 4x2+9y2=36 is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$4 π$
B
$6 π$
C
$9 π$
D
$36 π$

Solution:

The given curve is $4 x^{2} + 9 y^{2} = 36$

This can be written as $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

This is an ellipse which intersect the coordinate axes at $(3, 0) . (- 3.0), (0, 2), (0, - 2)$ '

The rough sketch of requried area is as below

The area of the sahded region is four times the area of the region bounded by the curve $y = \frac{2}{3} \sqrt{9 - x^{2}}$ and the lines $x = 0, x = 3$ and $x -$ axix

Hence, the area of the shaded region is $A = 4 \int_{0}^{3} \frac{2}{3} \sqrt{9 - x^{2}} d x$

$\begin{array}{rcl} A & = & \frac{8}{3} \int_{0}^{3} \sqrt{9 - x^{2}} d x \\ = & \frac{8}{3} {(\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{x}{3}))}_{0}^{3} \\ = & \frac{8}{3} (\frac{9}{2} \cdot \frac{π}{2}) \\ = & 6 π \end{array}$

The area enclosed with in the ellipse $4 x^{2} + 9 y^{2} = 36$ is

Solution:

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