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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 43 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 43 Maths textbook solution.

Answers (1)

Answer:

\pi -2  sq. units

Hints:

Use concept.

Given:

\left\{(x, y): x^{2}+y^{2} \leq 4, x+y \geq 2\right\}

Solution: The equation of the given curves are

x^2+y^2=4 ……(1)

x+y=2  …….(2)

Clearly x^2+y^2=4 represents a circle and x+y=2 is the equation of a straight line cutting x and y axis at (0,2) and (2,0) respectively.

The smaller region bounded by thesetwo curves is shaded in the following figure.

Length = y_2-y_1

Width = x and

Area = (y_2-y_1)x

Since the approximating rectangle can move from x=0 to x=2, the required area is given by

\begin{aligned} &\mathrm{A}=\int_{0}^{2}\left(y_{2}-y_{1}\right) d x\\ &\text { We have } y_{1}=2-x \text { and } y_{2}=\sqrt{4-x^{2}}\\ &\text { Thus, } A=\int_{0}^{2}\left(\sqrt{4-x^{2}}-2+x\right) d x\\ &\Rightarrow \mathrm{A}=\int_{0}^{2}\left(\sqrt{4-x^{2}}\right) d x-2 \int_{0}^{2} d x+\int_{0}^{2} x d x \end{aligned}

\begin{aligned} &\Rightarrow A=\left[\frac{x \sqrt{4-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{2}-2(x)_{0}^{2}+\left(\frac{x^{2}}{2}\right)_{0}^{2} \\ &\Rightarrow A=\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)-4+2 \\ &\Rightarrow A=2 \sin ^{-1}(1)-2 \\ &\Rightarrow A=2 \times \frac{\pi}{2}-2 \\ &\Rightarrow A=\pi-2 \end{aligned}

Therefore, the area of the region is \pi -2  sq. units

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