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  • Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 24 maths

Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 24 maths

Answers (1)

Answer:

 (a)

Hint:

 Integration

Given:

 y^{2}=8x,x^{2}=8y

Explanation:

\begin{aligned} &y^{2}=8x\\ &x=\frac{y^{2}}{8}\\ &x^{2}=8y\\ &\left ( \frac{y^{2}}{8} \right )^{2}=8y\\ &\left ( \frac{y^{4}}{64} \right )=8y\\ &y^{4}-512y=0\\ &y(y^{3}-512)=0\\ &y=0,\: \: y^{3}=512\\ &y^{3}=8^{3}\\ &y=8 \end{aligned}

\begin{aligned} &\text { Area }=\int_{0}^{8}\left ( \sqrt{8x}-\frac{x^{2}}{8} \right )dx\\ &=\sqrt{8x}\int_{0}^{8}\sqrt{x}dx-\frac{1}{8}\int_{0}^{8}x^{2}dx\\ &=\sqrt{8}\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{8}-\frac{1}{8}\left [ \frac{x^{3}}{3} \right ]_{0}^{8}\\ &=\frac{2}{3}\sqrt{8}\left [ 8^{\frac{3}{2}}-0 \right ]-\frac{1}{24}\left [ 8^{3} -0\right ]\\ &=\frac{2\times 8^{2}}{3}-\frac{8^{2}}{3}\\ &=\frac{64}{3} \text { sq.units } \end{aligned}

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Gurleen Kaur

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