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  • Provide solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 2

Provide solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 2

Answers (1)

Answer:

 (c)

Hint:

 Find the bounded region between the parabolas

Given:

y2 = 4x and x2 = 4y

Explanation:

Intersection point is

\begin{aligned} &y^{2}=4x \\ &=4(2\sqrt{y}) \qquad \qquad \qquad [\because x^{2}=4y] \\ &y^{2}=8\sqrt{y} \\ &y^{4}=64y \\ &y^{4}-64y=0 \\ &y(y^{3}-64)=0 \\ &y=0, \: \: y^{3}=64,\: \: y=4 \\ &\therefore x=4 \end{aligned}

Area between the parabolas

\begin{aligned} &=\int_{0}^{4}\left ( 2\sqrt{x}-\frac{x^{2}}{4} \right )dx \\ &=\frac{1}{4}\int_{0}^{4}(8\sqrt{x}-x^{2})dx \\ &=\frac{1}{4}\left [ \frac{8x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{3} \right ]_{0}^{4} \\ &=\frac{1}{4}\left [ \frac{16}{3}(4)^{\frac{3}{2}}-\frac{4^{3}}{3}-0+0 \right ] \\ &=\frac{1}{4}\left [ \frac{16}{3}(2^{2})^{\frac{3}{2}}-\frac{64}{3} \right ] \\ &=\frac{1}{4}\left [ \frac{128}{3}-\frac{64}{3} \right ]\\ &=\frac{16}{3}sq.units \end{aligned}

Posted by

Gurleen Kaur

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