Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 9 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 9 Maths textbook solution.

Answers (1)

Answer:

2\left [ \frac{\sqrt{2}}{3}+\frac{9 \pi} {4}-\frac{9}{2} \sin ^{-1} \left ( \frac{1}{3} \right ) \right ] square units

Hint:

Given:

\left\{(x, y): y^{2} \leq 8 x, x^{2}+y^{2} \leq 9\right\}

Solution:

To find area \left\{(x, y): y^{2} \leq 8 x, x^{2}+y^{2} \leq 9\right\}

y^2=8x.....(1)

x^2+y^2=9 .....(2)

On solving the equation (1) and (2)

\begin{aligned} &x^{2}+8 x=9 \\ &x^{2}+8 x-9=0 \\ &(x+9)(x-1)=0 \\ &x=-9 \text { or } x=1 \end{aligned}

 

And when x=1 then y= \pm 2\sqrt{2}

Equation (1) represents a parabola with vertex (0,0) and axis as x-axis equation (2) represents a circle with centre (0,0) and radius 3 units, so it meets area at (\pm 3,0),(0,\pm 3)

Point of intersection of parabola and circle is (1,2\sqrt{2}) and (1-2\sqrt{2})

The sketch of the curves is as below

https://www.sarthaks.com/?qa=blob&qa_blobid=13041479697829642655

Or, required area=(region ODCO+REGION DBCD)

\begin{aligned} &=2\left[\int_{0}^{1} \sqrt{8 x} d x+\int_{1}^{3} \sqrt{9-x^{2}} d x\right] \\ &=2\left[\left(2 \sqrt{2 \cdot \frac{2}{3} x \sqrt{x}}\right)_{0}^{1}+\left(\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right)_{1}^{3}\right] \\ &=2\left[\left(\frac{4 \sqrt{2}}{3} \cdot 1 \sqrt{1}\right)+\left\{\left(\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin ^{-1}(1)\right)-\left(\frac{1}{2} \sqrt{9-1}+\frac{9}{2} \sin ^{-1} \frac{1}{3}\right)\right\}\right] \\ &=2\left[\frac{4 \sqrt{2}}{3}+\left\{\left(\frac{9}{2} \cdot \frac{\pi}{2}\right)-\left(\frac{2 \sqrt{2}}{2}-\frac{9}{2} \sin ^{-1}\left(\frac{1}{3}\right)\right)\right\}\right] \\ &=2\left[\frac{4 \sqrt{2}}{3}+\frac{9 \pi}{4}-\sqrt{2}-\frac{9}{2} \sin ^{-1}\left(\frac{1}{3}\right)\right] \end{aligned}

Hence, the required area is 2\left [ \frac{\sqrt{2}}{3}+\frac{9 \pi} {4}-\frac{9}{2} \sin ^{-1} \left ( \frac{1}{3} \right ) \right ]  square unit

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question