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 Areas of Bounded Region exercise 20.1 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 13 maths textbook solution

Answers (1)

Answer:

\frac{\pi a^{2}}{4} s q \cdot \text { unit }

Hint:

y=\sqrt{a^{2}-x^{2}} \text { at line } x=0, x=a

Given: Determine the area under curve y=\sqrt{a^{2}-x^{2}} include between line x=0 and x=a

Solution:

\begin{aligned} &y=\sqrt{a^{2}-x^{2}} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &x^{2}+y^{2}=a^{2} \end{aligned}

Required area =  area of shaded region

\begin{aligned} &=\int_{0}^{a} y d x \\\\ &=\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \end{aligned}

\begin{aligned} &=\left[\frac{a}{2} \sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a}{a}\right]-\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{a^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \\\\ &=\frac{a^{2}}{2} \sin ^{-1} 1 \\\\ &=\frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ \end{aligned}

=\frac{\pi a^{2}}{4} s q \cdot u n i t

      

Posted by

infoexpert26

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

CBSE Board Exams 2027 LIVE: Class 12, 10 date sheet shortly; LOC submission rules, total subjects
anam-khan

CBSE organises webinar on LOC registration, correction queries for Class 10, 12 board exam 2027
anam-khan

CBSE Board Exam Date 2027 Highlights: Class 10, 12 exam schedule soon at cbse.gov.in

Latest Question