Question

Area common to the curve $y=\sqrt{9-x^2}$ and $x^2+y^2=6x$ is $\frac{9\sqrt{3}}{2}+n\pi$. Find $n$

Answer 1

$y=\sqrt{9-x^2}$ & $x^2+y^2=6x\Rightarrow y=y=\sqrt{6x-x^2}$

$\Downarrow$

$x^2+y^2=9$

Area in the common region in the both side point of intersection of circles :

$x^2+y^2=9$.... $\left(i\right)$

$x^2+y^2=6x$.... $\left(ii\right)$

from $\left(i\right)$ & $\left(ii\right)$

$6x=9$

$x=\frac{3}{2}$

$y=\sqrt{9-x^2}=\sqrt{9-\frac{9}{4}}$

$=\sqrt{\frac{36-9}{4}}=\pm \frac{3\sqrt{3}}{2}$

Firstly we are going to calculate the area above $x$-axis then we will multiply by $2$ to get the total area.

Form $(0,0)$ to $(\frac{3}{2},\frac{3\sqrt{3}}{2})$ the shaded region is in circle $II$.

$={\int }_0^{3/2}\sqrt{6x-x^2}dx={\int }_0^{3/2}\sqrt{9-9-x^2+6x}dx$

$={\int }_0^{3/2}\sqrt{(9-(x^2-6x+9\left)\right)}dx={\int }_0^{3/2}\sqrt{3^2-(x-3{)}^2}dx$

$={\int }_{-3/2}^{-3}\sqrt{3^2-t^2}dt$

From the special integral theorem substituting the value of $t$ :

Substituting

$x=3=t$

$dx=dt$

Limits

$-3t-3/2$

$=(\frac{(-3)}{4}\sqrt{9-\frac{9}{4}}+\frac{9}{2}{\sin }^{-1}\left(\frac{-1}{2}\right))-(\frac{(-3)}{2}\sqrt{9-9}+\frac{9}{2}{\sin }^{-1}\frac{(-3)}{3})$

$=\frac{-3}{4}\sqrt{\frac{27}{4}}+\frac{9}{2}(\frac{-\pi }{6}+\frac{3}{2}\times 0-\frac{9}{2}\times \left(\frac{-\pi }{2}\right))$

$=\frac{-3}{4}\times \frac{3\sqrt{3}}{2}+\frac{(-3)}{4}\pi +\frac{9\pi }{4}=\frac{6\sqrt{3}+3}{4}\frac{3\pi }{2}+\frac{(-9\sqrt{3})}{8}$

From $(\frac{3}{2},\frac{3\sqrt{3}}{2})$ to $(3,0)$

$A_2={\int }_{3/2}^3\sqrt{9-x^2}dx$ as the region has in the circle $I$.

$={\int }_{3/2}^3\sqrt{9-x^2}dx={\int }_{3/2}^3\sqrt{3^2}-x^{2}dx$

$=[\frac{x}{2}\sqrt{3^2-x^2}+\frac{3^2}{2}{\sin }^{-1}\frac{x}{3}]$ : from the special integrals substituting the values of $x$ :

$(\frac{3}{2}\sqrt{3^2-3^2}+\frac{3^2}{2}{\sin }^{-1}\frac{3}{3})-(\frac{3}{4}\sqrt{3^2-{\left(\frac{3}{2}\right)}^2}+\frac{3^2}{2}{\sin }^{-1}\left(\frac{1}{2}\right))$

$=\frac{3}{2}\times 0+\frac{9}{2}\times \frac{\pi }{2}-(\frac{3}{4}\sqrt{\frac{27}{4}}+\frac{9}{2}\left(\frac{\pi }{6}\right))$

$=\frac{9\pi }{4}-\frac{9\sqrt{3}}{8}-\frac{3\pi }{4}=\frac{3\pi }{2}-\frac{\left)\right(-9\sqrt{3})}{8}$

$=\frac{12\sqrt{3}-6-9\sqrt{3}}{8}+3\pi$

$=3\pi -\frac{9\sqrt{3}}{4}$

So, the total area is :

$2(A_1+A_2)=2\times (3\pi -\frac{9\sqrt{3}}{4})=6\pi -\frac{9\sqrt{3}}{2}$

Hence $n$ is $6$.