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  • Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 32 Maths.

Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 32 Maths.

Answers (1)

Answer:

 4 sq. Units

Hint:

Use concept.

Given:

Equation of the given curve x=y^2  and x=3-y^2

Solution: To find area bounded by

x=y^2 …(i)

And x-3=y^2  …(ii)

On solving the equation (i) and (ii),

y^2=3-2y^2

Or 3y^2=3

Or  y=\pm 1

when y =1 then x =1  and when y = -1 then x =1

Equation (i) represents an upward parabola with vertex (0, 0) and axis -y

Equation (ii) represents a parabola with vertex (3, 0) and axis as  x-axis

They intersect at A (1, – 1) and C (1, 1)

These are shown in the graph below:

Required area = Region OABCO = 2

= 2 Region OBCO

= 2[Region ODCO + Region BDCB]

\begin{aligned} &=2\left[\int_{0}^{1} y_{1} d x+\int_{1}^{3} y_{2} d x\right] \\ &=2\left[\int_{0}^{1} \sqrt{x} d x+\int_{1}^{3} \sqrt{\frac{3-x}{2} d x}\right] \\ &\left.=2\left[\left(\frac{2}{3} x \sqrt{x}\right)_{0}^{1}+\left(\frac{2}{3} \cdot\left(\frac{3-x}{2}\right) \sqrt{\frac{3-x}{2}} \cdot(-2)\right)^{2}\right]_{1}\right] \\ &=2\left[\left(\frac{2}{3}-0\right)+\left\{(0)-\left(\frac{2}{3}\right) .1 .1 .(-2)\right\}\right] \\ &=2\left[\frac{2}{3}+\frac{4}{3}\right] \end{aligned}

= 4 sq.units

Posted by

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