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 1 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 1 Maths textbook solution.

Answers (1)

Answer:

12 sq.units.

Given:

y^2=6x

x^2=6y

Hint:

The intersecting points of the given parabolas are obtained by solving these equations for x and y , which are 0(0,0)and  (6,6).

Solution:

The given equations are

y^2=6x

y=\sqrt{6x}                                ...…(i)

And 

x=\sqrt{6y}        (ii)

Putting x  value on  x = 6

When y =0 then x=6

And when y =6 then x = 6

On solving these two equations ,we get point of intersections

The points are O(0,0) and A(6,6). These are shown in the graph below

1799432

Bounded Area ,A=[Area between the curve (i) and x -axis  from 0 to 6]-[Area between the curve (ii) and x -axis  from 0 to 6)

\begin{aligned} &A=\int_{0}^{6}\sqrt{6 x} d x-\int_{0}^{6} \frac{x^{2}}{6} d x \\ &A=\int_{0}^{6}\left(\sqrt{6 x}-\frac{x^{2}}{6}\right) d x \end{aligned}

On integration the above definite integration

\begin{aligned} &A=\int_{0}^{6}\left(\sqrt{6 x}-\frac{x^{2}}{6}\right) d x \\ &=\left[\sqrt{6} \frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{18}\right]_{0}^{6} \\ &=\left[\sqrt{6} \frac{(6)^{3 / 2}}{3 / 2}-\frac{(6)^{3}}{18}\right] \end{aligned}

=12 \; \text{sq units.}

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