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  • Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 40 maths

Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 40 maths

Answers (1)

Answer:

 (a)

Hint:

 \int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c

Given:

 y= \sqrt{16-x^{2}}, x-axis(y=0)

Explanation:

\begin{aligned} &y=0\\ &0=\sqrt{16-x^{2}}\\ &16-x^{2}=0\\ &x^{2}=16\\ &x=\pm 4\\ \end{aligned}

Area bounded by the curve

\begin{aligned} &=\int_{-4}^{4}\sqrt{16-x^{2}}dx\\ &=\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{4^{2}}{2}sin^{-1}\frac{x}{4} \right ]_{-4}^{4}\\ &=\left [ \frac{4}{2}\sqrt{16-4^{2}}+\frac{4^{2}}{2}sin^{-1}\frac{4}{4} \right ]-\left [ \frac{-4}{2}\sqrt{16-(-4)^{2}}+\frac{4^{2}}{2}sin^{-1}\left ( \frac{-4}{4} \right ) \right ] \qquad \qquad \left [ \because sin^{-1}1=\frac{\pi }{2} \right ]\\ &=0+8sin^{-1}1-0+8sin^{-1}1\\ &=8\times 2\times \frac{\pi }{2}\\ &=8\pi \text { sq.units } \end{aligned}

Posted by

Gurleen Kaur

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