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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.