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Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 16.

Answers (1)

Answer:

\frac{16}{3}\; \text{ab sq. units} .

Hint:

Use basic concepts.

Given:

y^2=4ax and x^2=4by .

Solution: To find area enclosed by

y^2=4ax                ..........(1)

x^2=4by                .........(2)

Equation (1) represents a parabola with vertex (0,0) and axis as x-axis,

Equation (2) represents a parabola with vertex (0,0), and axis as y-axis,

Point of intersection of parabolas are (0, 0) and (4a^{\frac{1}{3}} b^{\frac{2}{3}} , 4a^{\frac{2}{3}}b^{\frac{1}{3}})

 

A rough sketch is given as:

The shaded region is required area  and it is sliced into rectangle of width=\Delta _x  and length(y_2-y_1)

Area of rectangle=(y_1-y_2)\Delta _x

This approximation rectangle slides from x=0 to, x= 4a^{\frac{1}{3}}b^{\frac{2}{3}} , so

Required area = Region OQAPO

\begin{aligned} &=\int_{0}^{4 a^{\frac{1}{4} b^{2}} 3}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{4 a^{\frac{1}{a} b^{2}}}\left(2 \sqrt{a} \cdot \sqrt{x}-\frac{x^{2}}{4 b}\right) d x \\ &=\left[2 \sqrt{a} \cdot \frac{2}{3} x \sqrt{x}-\frac{x^{3}}{12 b}\right]_{0}^{4 a^{\frac{1}{3} b^{2}}} \\ &=\frac{32 \sqrt{a}}{3} \cdot a^{\frac{1}{3}} b^{\frac{2}{3}} a^{\frac{1}{6}} b^{\frac{1}{3}} \\ &=\frac{64 a b^{2}}{12 b} \\ &=\frac{16}{3} a b \text { sq.units } \end{aligned}

Hence the required area is \frac{16}{3}\; \text{ab sq. units} .

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