Question

The area (in sq. units) of the part of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=36$, which is outside the parabola ${\mathrm{y}}^2=9\mathrm{x}$, is:

• A. $24\mathrm{\pi }+3\sqrt{3}$
• B. $12\mathrm{\pi }+3\sqrt{3}$
• C. $12\mathrm{\pi }-3\sqrt{3}$
• D. $24\mathrm{\pi }-3\sqrt{3}$

Answer 1

The correct option is D

$24\mathrm{\pi }-3\sqrt{3}$

Explanation for correct option:

The equation of a circle is,

${\mathrm{x}}^2+{\mathrm{y}}^2=36-1...\left(1\right)$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\left(6\right)}^2$

The radius of a given circle is $6$ units. With centre at origin.

The equation of parabola is,

${\mathrm{y}}^2=9\mathrm{x}-2...\left(2\right)$

Put Equation $\left(2\right)$ in Equation $\left(1\right)$ we get,

${\mathrm{x}}^2+4\mathrm{x}=36$

$\Rightarrow {\mathrm{x}}^2+9\mathrm{x}-36=0$

$\Rightarrow {\mathrm{x}}^2+12\mathrm{x}-3\mathrm{x}-36=0$

$\Rightarrow \mathrm{x}(\mathrm{x}+12)-3(\mathrm{x}+12)=0$

$\Rightarrow (\mathrm{x}-3)(\mathrm{x}+12)=0$

$\Rightarrow \mathrm{x}=3,-12$

The graph of the given curve is shown below,

Put, $\mathrm{x}=3$ in Equation $\left(2\right)$ we get.

$$\begin{gathered} {\mathrm{y}}^2=9\times 3 \\ \Rightarrow \mathrm{y}=\pm 3\sqrt{3} \end{gathered}$$

Therefore the point of intersection $\mathrm{P}$ is $(3,+3\sqrt{3})$

Now the area of a shaded portion is given,

$\mathrm{A}=$ Area of Circle $-2\left[{\int }_0^3\sqrt{9\mathrm{xdx}}+{\int }_3^6\sqrt{36-{\mathrm{x}}^2}\mathrm{dx}\right.$ …$[{\mathrm{y}}^2=9\mathrm{x}$ and ${\mathrm{x}}^2+{\mathrm{y}}^2=36]$

$\Rightarrow \mathrm{A}={\mathrm{\pi r}}^2-2\left[{\left(3\times \frac{{\mathrm{x}}^{\frac{3}{2}}}{2/2}\right]}_0^3+{\left[\frac{\mathrm{x}}{2}\sqrt{36-{\mathrm{x}}^2}+\frac{36}{2}{\sin }^{-1}\left(\frac{\mathrm{x}}{6}\right)\right]}_3^6\right)$

$\Rightarrow \mathrm{A}=36\mathrm{\pi }-2\left({\left[2{\mathrm{x}}^{\frac{3}{2}}\right]}_0^3+0+\frac{18\times \mathrm{\pi }}{2}-\left(\frac{3}{2}\sqrt{36-9}+18{\sin }^{-1}\left(\frac{1}{21}\right)\right)\right)$

$\Rightarrow \mathrm{A}=36\mathrm{\pi }-2\left(2\times (3{)}^{\frac{3}{2}}-2\times 0+\frac{18\mathrm{\pi }}{2}-\frac{3\sqrt{27}}{2}-18\times \frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow \mathrm{A}=36\mathrm{\pi }-2\left(2\times 3\sqrt{3}+9\mathrm{\pi }-\frac{9\sqrt{3}}{2}-3\mathrm{\pi }\right)$

$\Rightarrow \mathrm{A}=36\mathrm{\pi }-12\sqrt{3}-18\mathrm{\pi }+9\sqrt{3}+6\mathrm{\pi })$

$\Rightarrow \mathrm{A}=24\mathrm{\pi }-3\sqrt{3}$

Therefore the value of an area of a shaded portion is equal to $\mathbf{(}\mathbf{24}\mathbf{\pi }\mathbf{-}\mathbf{3}\sqrt{\mathbf{3}}\mathbf{)}$ square units.

Hence option (D) is the correct answer.