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Q.
The area bounded by the curve $y^{2}\left(a^{2}+x^{2}\right)=x^{2}\left(a^{2}-x^{2}\right)$ is
Application of Integrals
Solution:
Since, the curve is symmetrical about both the axes. Therefore, the required area
$=4 \int\limits_{0}^{a} y \,d x=4 \int\limits_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x=4 \int\limits_{a}^{\sqrt{2} a} \sqrt{(\sqrt{2} a)^{2}-z^{2}} d z$
Let $x^{2}+a^{2}=z^{2} $
$\Rightarrow x \,d x=z\, d z .$
Also, $a^{2}-x^{2}=2 a^{2}-z^{2}$
$=4\left[\frac{z}{2} \sqrt{(\sqrt{2} a)^{2}-z^{2}}+\frac{(\sqrt{2} a)^{2}}{2} \sin ^{-1} \frac{z}{\sqrt{2} a}\right]_{a}^{\sqrt{2} a}$
$=a^{2}(\pi-2)$ sq. unit
