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  • Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 21 maths textbook solution

Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 21 maths textbook solution

Answers (1)

Answer:

 (b)

Hint:

 Integration

Given:

 y^{2}(2a-x)=x^{3} \text { and } x=2a \text { above }x-axis

Explanation:

\begin{aligned} &x=2a\\ &y^{2}(2a-x)=x^{3}\\ &y^{2}=\frac{x^{3}}{2a-x}\Rightarrow y=\sqrt{\frac{x^{3}}{2a-x}}\\ \end{aligned}

Let

\begin{aligned} &x=2a\: sin^{2}\theta \\ &dx=4a\: sin\: \theta\: cos\: \theta \end{aligned}

Area

\begin{aligned} &=\int_{0}^{2a}\sqrt{\frac{x^{3}}{2a-x}}dx\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{(2a)^{3}(sin^{2}\theta )^{3}}{2a-2asin^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [W\! hen\: x=0, \theta=0\: and\: x=2a,\theta =\frac{\pi }{2} ]\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{8a^{3}sin^{6}\theta }{2a\, cos^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [\because 1-sin^{2}\theta =cos^{2}\theta ] \end{aligned}

\begin{aligned} &=8a^{2}\int_{0}^{\frac{\pi }{2}}\sqrt{sin^{6}\theta }(sin\, \theta d\theta )\\ &=8a^{2}\int_{0}^{\frac{\pi }{2}}sin^{4}\theta d\theta \qquad \qquad [\because \int sin^{4}xdx=\frac{3x}{8}-\frac{1}{4}sin2x+\frac{1}{32}sin4x+c]\\ &=8a^{2}\left [ \frac{3\theta }{8}-\frac{1}{4}sin2\theta + \frac{1}{32}sin4\theta \right ]_{0}^{\frac{\pi }{2}}\\ &=8a^{2}\left [ \frac{3\left ( \frac{\pi }{2} \right )}{8}-\frac{1}{4}sin\pi +\frac{1}{32}sin2\pi \right ]-0\\ &=8a^{2}\times \frac{3\left ( \frac{\pi }{2} \right )}{8}\\ &=\frac{3}{2}\pi a^{2} \end{aligned}

Posted by

Gurleen Kaur

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