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  • Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 29 maths textbook solution

Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 29 maths textbook solution

Answers (1)

Answer:

 (c)

Hint:

 Integration

Given:

 x^{2}+y^{2}=16,y^{2}=6x

Explanation:

Area OABC

\begin{aligned} &=2 \text { Area }O\! AB\\ &=\left [ \text { Area }O\! AD+\text { Area }AB\! D \right ] \qquad \qquad \dots(i) \end{aligned}

Intersection points are

\begin{aligned} &x^{2}+y^{2}=16 \\ &x^{2}+6x=16 \qquad \qquad \qquad \left [ \because y^{2}=6x \right ]\\ &x^{2}+6x-16=0\\ &x^{2}+8x-2x-16=0\\ &x(x+8)-2(x+8)=0\\ &x-2=0 \qquad x+8=0\\ &x=2, x=-8 \end{aligned}

Not possible in first and third quadrant

x=2

From (i)

Area OABC,

\begin{aligned} &=2\left [ \int_{0}^{2}\sqrt{6x}dx+\int_{2}^{4}\sqrt{16-x^{2}}dx \right ]\\ &=2\left [ \sqrt{6}\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{2}+2\left [ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}\frac{x}{4} \right ]_{2}^{4}\\ &=\frac{4}{3}\sqrt{6}\left [ (2)^{\frac{3}{2}-1} \right ]+2\left [ \frac{4}{2}\sqrt{16-(4)^{2}}+8sin^{-1}\frac{4}{4} \right ]-\left [ \frac{2}{2}\sqrt{16-2^{2}}+8sin^{-1}\frac{2}{4} \right ] \end{aligned}

\begin{aligned} &=\frac{4\sqrt{6}}{3}\left ( 2\sqrt{2} \right )+2\left [ 0+2\times \frac{\pi }{2}-\sqrt{12}-8sin^{-1}\frac{1}{2} \right ]\\ &=\frac{16\sqrt{3}}{3}+2\left [ 4\pi -2\sqrt{3}-8\frac{\pi }{6} \right ] \\ &=\frac{16\sqrt{3}}{3}+8\pi -4\sqrt{3}-\frac{8\pi }{3}\\ &=\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\\ &=\frac{4}{3}\left ( 4\pi +\sqrt{3} \right ) \text { sq.units } \end{aligned}

Area of circle

\begin{aligned} &=\pi r^{2}\\ &=\pi (4)^{2}\\ &=16\pi \end{aligned}

Exterior area

\begin{aligned} &=16\pi -\frac{4}{3}\left ( 4\pi +\sqrt{3} \right )\\ &=16\pi -\frac{16}{3}\pi -\frac{4}{\sqrt{3}}\\ &=\frac{4}{3}(8\pi -\sqrt{3}) \text{sq.units} \end{aligned}

Posted by

Gurleen Kaur

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