Question 26
Using integration, find the area in the first quadrant bounded by the curve y = ๐ฅ|๐ฅ|, the circle x2 + y2 = 2 and the y-axis
Given
y = ๐ฅ|๐ฅ|
y = {โ(๐ฅร๐ฅ, ๐ฅโฅ0@๐ฅ ร(โ๐ฅ), ๐ฅ<0)โค
y = {โ(๐ฅ^2, ๐ฅโฅ0@โ๐ฅ^2, ๐ฅ<0)โค
And
x2 + y2 = 2
(x โ 0)2 + (y โ 0)2 = (โ2)^2
So, it is a circle with center (0, 0)
and Radius = โ2
Question 26
Using integration, find the area in the first quadrant bounded by the curve y = ๐ฅ|๐ฅ|, the circle x2 + y2 = 2 and the y-axis
Given
y = ๐ฅ|๐ฅ|
y = {โ(๐ฅร๐ฅ, ๐ฅโฅ0@๐ฅ ร(โ๐ฅ), ๐ฅ<0)โค
y = {โ(๐ฅ^2, ๐ฅโฅ0@โ๐ฅ^2, ๐ฅ<0)โค
And
x2 + y2 = 2
(x โ 0)2 + (y โ 0)2 = (โ2)^2
So, it is a circle with center (0, 0)
and Radius = โ2
๐=ใโ๐ใ^๐ for x < 0
๐=ใโ๐ใ^๐ for x < 0
Combining, we form graph
We need to find Area OAB
First we find point B
Point B is point of intersection of circle and parabola
Now,
equation of circle is
๐ฅ^2+๐ฆ^2=2
Putting x2 = y
๐ฆ+๐ฆ^2=2
๐ฆ^2+๐ฆโ2=0
๐ฆ^2+2๐ฆโ๐ฆโ2=0
๐ฆ(๐ฆ+2)โ1(๐ฆ+2)=0
(yโ1)(๐ฆ+2)=0
So, y = 1, y = โ2
So, y = 1, y = โ2
Since y is in 1st quadrant, it is positive
โด y = 1
Now,
y = x2
1 = x2
x2 = 1
x = ยฑ โ1
x = ยฑ 1
โด x = 1, โ1
Since x is in 1st quadrant
x = 1
โด x = 1, y = 1
So, point B = (1, 1)
Area Required = Area AODB โ Area ODB
Area AODB
Area AODB = โซ_0^1โใ๐ฆ ๐๐ฅใ
For circle
x2 + y2 = 2
y2 = 2 โ x2
y = ยฑ โ(2โ๐ฅ^2 )
Since AODB is in 1st quadrant
y =โ(2โ๐ฅ^2 )
โด Area AODB = โซ_0^1โโ(2โ๐ฅ^2 ) ๐๐ฅ
= โซ_0^1โโ((โ2)^2โ๐ฅ^2 ) ๐๐ฅ It is of form
โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ
Here, a = โ2
= [๐ฅ/2 โ((โ2)^2โ๐ฅ^2 )+(โ2)^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1
= [๐ฅ/2 โ(2โ๐ฅ^2 )+2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1
= [๐ฅ/2 โ(2โ๐ฅ^2 )+ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1
= [1/2 โ(2โ1^2 )+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0/2 โ(2โ0^2 )+ใ๐ ๐๐ใ^(โ1)โกใ 0/โ2ใ " " ] = [1/2 โ(2โ1)+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0+ใ๐ ๐๐ใ^(โ1)โก0 " " ] = [1/2+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0+0" " ]
= 1/2+๐/4
Area ODB
Area ODB = โซ_0^1โใ๐ฆ ๐๐ฅใ
For parabola
y = x2
โด Area ODB = โซ_0^1โ๐ฅ^2 ๐๐ฅ
= [๐ฅ^3/3]_0^1
=1^3/3โ0^3/3 = 1/3
Thus,
Area Required = Area AODB โ Area ODB
= 1/2+๐/4 โ 1/3
= 1/2โ1/3+๐/4
= ๐/๐+๐ /๐ square units

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!