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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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