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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 44 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 44 Maths textbook solution.

Answers (1)

Answer: \left(\frac{3 \pi}{2}-3\right) \text { sq. units }

Hints:

Use concept.

Given:

\left\{(x, y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\right\}

Solution:

To find the area of a region

\left\{(x, y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\right\}

Here,

\frac{x^2}{9}+\frac{y^2}{4}=1    …..(1)

\frac{x}{3}+\frac{y}{2}=1        …..(2)

Equation (1) represents an ellipse with centre at origin and meets axis at (\pm 3,0) ,(0,\pm 2) .

Equation (2) is a line that meets axis at (3,0) ,(0,2)

A rough sketch is as under:

Shaded region represents required area. This is sliced into rectangles with area (y_2-y_2)x  which slides from x=0  to  x=3, so

Required area = Region APBQA

\begin{aligned} &A=\int_{0}^{3}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{3}\left[\frac{2}{3} \sqrt{9-x^{2}} d x-\frac{2}{3}(3-x) d x\right] \\ &=\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)-3 x+\frac{x^{2}}{2}\right]_{0}^{3} \\ &=\frac{2}{3}\left[\left\{0+\frac{9}{2} \cdot \frac{\pi}{2}-9+\frac{9}{2}\right\}-\{0\}\right] \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right] \\ &A=\left(\frac{3 \pi}{2}-3\right) \text { sq. units } \end{aligned}

Therefore, the area of the region is \left(\frac{3 \pi}{2}-3\right) \text { sq. units }  

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