The area enclosed with in the ellipse 4×2+9y2=36 is

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

1. A

$4\mathrm{\pi }$

2. B

$6\pi$

3. C

$9\pi$

4. D

$36\pi$

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### 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 $\left(3,0\right).\left(-3.0\right),\left(0,2\right),\left(0,-2\right)$'

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 $A=4{\int }_{0}^{3}\frac{2}{3}\sqrt{9-{x}^{2}}dx$

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