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Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 5

Answers (1)

\frac{3}{5}a^{2}\left ( 2^{\frac{5}{3}} -1\right )

Hint:

Area of the bounded region = Area under (y=2a) and (ay^{2}=x^{3})- Area under (y=2a) and (ay^{2}=x^{2})

Given:           ay^{2}=x^{3}, y=a, y=2a

Solution:

 

Area under (y=2a) and (ay^{2}=x^{2})  

 

               \begin{aligned} &=\int_{0}^{1.58 a}\left(2 a-\sqrt{\frac{x^{3}}{a}}\right) d x \\ \end{aligned}

\begin{aligned} &=\left[2 a x-\frac{2 x^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]_{0}^{1.58 a} \\ \end{aligned}

\begin{aligned} &=2 a(1.58 a)-\frac{2(1.58 a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}} \end{aligned}

Area under (y=a) and (ay^{2}=x^{2})

               \begin{aligned} &=\int_{0}^{a}\left(a-\sqrt{\frac{x^{2}}{a}}\right) d x \\ \end{aligned}

\begin{aligned} &=\left[a x-\frac{2 x^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]_{0}^{a} \\ \end{aligned}

\begin{aligned} &=a(a)-\frac{2(a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}} \end{aligned}

Required area

               2 a(1.58 a)-\frac{2(1.58 a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}-\left[a^{2}-\frac{2(a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]

Solving further,

        

\begin{aligned} &\frac{3}{5} a^{2}\left(2(2)^{\frac{2}{3}}-1\right) \text { sq. Units } \\ \end{aligned}

\begin{aligned} &=\frac{3}{5} a^{2}\left(2^{1+\frac{2}{3}}-1\right) \text { sq. Units } \\ \end{aligned}

\begin{aligned} &=\frac{3}{5} a^{2}\left(2^{\frac{5}{3}}-1\right) \text { sq. Units } \end{aligned}

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