Question 26
Using integration, find the area in the first quadrant bounded by the curve y = π₯|π₯|, the circle x2 + y2 = 2 and the y-axis
Given
y = π₯|π₯|
y = {β(π₯Γπ₯, π₯β₯[email protected]π₯ Γ(βπ₯), π₯<0)β€
y = {β(π₯^2, π₯β₯[email protected]βπ₯^2, π₯<0)β€
And
x2 + y2 = 2
(x β 0)2 + (y β 0)2 = (β2)^2
So, it is a circle with center (0, 0)
and Radius = β2
Question 26
Using integration, find the area in the first quadrant bounded by the curve y = π₯|π₯|, the circle x2 + y2 = 2 and the y-axis
Given
y = π₯|π₯|
y = {β(π₯Γπ₯, π₯β₯[email protected]π₯ Γ(βπ₯), π₯<0)β€
y = {β(π₯^2, π₯β₯[email protected]βπ₯^2, π₯<0)β€
And
x2 + y2 = 2
(x β 0)2 + (y β 0)2 = (β2)^2
So, it is a circle with center (0, 0)
and Radius = β2
π=γβπγ^π for x < 0
π=γβπγ^π 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
π₯^2+π¦^2=2
Putting x2 = y
π¦+π¦^2=2
π¦^2+π¦β2=0
π¦^2+2π¦βπ¦β2=0
π¦(π¦+2)β1(π¦+2)=0
(yβ1)(π¦+2)=0
So, y = 1, y = β2
So, y = 1, y = β2
Since y is in 1st quadrant, it is positive
β΄ y = 1
Now,
y = x2
1 = x2
x2 = 1
x = Β± β1
x = Β± 1
β΄ x = 1, β1
Since x is in 1st quadrant
x = 1
β΄ x = 1, y = 1
So, point B = (1, 1)
Area Required = Area AODB β Area ODB
Area AODB
Area AODB = β«_0^1βγπ¦ ππ₯γ
For circle
x2 + y2 = 2
y2 = 2 β x2
y = Β± β(2βπ₯^2 )
Since AODB is in 1st quadrant
y =β(2βπ₯^2 )
β΄ Area AODB = β«_0^1ββ(2βπ₯^2 ) ππ₯
= β«_0^1ββ((β2)^2βπ₯^2 ) ππ₯ It is of form
β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ
Here, a = β2
= [π₯/2 β((β2)^2βπ₯^2 )+(β2)^2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [π₯/2 β(2βπ₯^2 )+2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [π₯/2 β(2βπ₯^2 )+γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [1/2 β(2β1^2 )+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0/2 β(2β0^2 )+γπ ππγ^(β1)β‘γ 0/β2γ " " ] = [1/2 β(2β1)+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+γπ ππγ^(β1)β‘0 " " ] = [1/2+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+0" " ]
= 1/2+π/4
Area ODB
Area ODB = β«_0^1βγπ¦ ππ₯γ
For parabola
y = x2
β΄ Area ODB = β«_0^1βπ₯^2 ππ₯
= [π₯^3/3]_0^1
=1^3/3β0^3/3 = 1/3
Thus,
Area Required = Area AODB β Area ODB
= 1/2+π/4 β 1/3
= 1/2β1/3+π/4
= π/π+π /π square units

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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