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Q.
Area common to curve $y=\sqrt{9-x^{2}}$ and $x^{2}+y^{2}=6x$ is
Application of Integrals
Solution:
$x^{2}+ y^{2} = 9$, a circle with centre $\left(0,0\right)$ and radius $3\quad\ldots\left(i\right)$
$x^{2} +y^{2}-6x=0$, a circle with centre $\left(3,0\right)$ and radius $3\quad\ldots\left(ii\right)$
On solving $\left(i\right)$ and $\left(ii\right)$, we get
$x=\frac{3}{2}, y^{2}=9-\frac{9}{4}=\frac{27}{4}\,$
$\Rightarrow \quad y=\frac{3\sqrt{3}}{2}$
$\therefore \quad A=2 \int \limits_{3/ 2}^{3}\left(\sqrt{9-x^{2}}\right)dx$
$=3\left(\pi-\frac{3\sqrt{3}}{4}\right)$ sq. units
