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Q.
The area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ is
Application of Integrals
Solution:
The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
The given ellipse is symmetrical about both axis as it contains only even powers of $y$ and $x$.
Now, $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1 \Rightarrow \frac{y^{2}}{9}=1-\frac{x^{2}}{16} \Rightarrow y^{2}=\frac{9}{16}\left(16-x^{2}\right)$
$\Rightarrow y =\pm \frac{3}{4}\left(\sqrt{16- x ^{2}}\right)$
Now, area bounded by the ellipse $=4$ (area of ellipse in first quardant)
$=4($ area $OAC )$
$=4 \int\limits_{0}^{4} ydx =\int\limits_{0}^{4} \frac{3}{4} \sqrt{16- x ^{2}} dx [\because y \geq 0$ in first quadrant $]$
Put $x=4 \sin \theta$ so that $d x=4 \cos \theta d \theta$
Now when $x=0, \theta=0$ and when $x=4, \theta=\frac{\pi}{2}$
$\therefore $ Req. area $=\frac{4 \times 3}{4} \int_{0}^{\frac{\pi}{2}} \sqrt{16-16 \sin ^{2}} \theta \cdot 4 \cos \theta d \theta$
$=3 \int\limits_{0}^{\frac{\pi}{2}} 4 \sqrt{1-\sin ^{2} \theta} \cdot 4 \cos \theta d \theta$
$=48 \int\limits_{0}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta=48 \int_{0}^{\frac{\pi}{2}}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta$
$=24 \int\limits_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta=24\left[\theta+\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}}$
$=24\left[\left(\frac{\pi}{2}-0\right)+\frac{1}{2}(0-0)\right]=12 \pi$ sq. units.
