Question 34 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Oct. 26, 2020 by Teachoo
Find the area of the region bounded by the curves π₯
2
+ π¦
2
= 4, y = β3π₯ πππ π₯ β ππ₯ππ ππ π‘βπ ππππ π‘ ππ’ππππππ‘
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Question 34 (Choice 1) Find the area of the region bounded by the curves π₯^2+π¦^2=4, π¦=β3 π₯ πππ π₯ β ππ₯ππ ππ π‘βπ ππππ π‘ ππ’ππππππ‘
Given Equation of Circle
π₯2+π¦2=4
π₯2+π¦2=2^2
So, Radius = 2
β΄ Point A (2, 0) and B is (0, 2)
Let point where line and circle intersect be point M
Required Area = Area of shaded region
= Area OMA
Finding Point M
Point M is intersection of line and circle
Putting π=βπ π in Equation of Circle
π₯2+π¦2=4
π₯2+(β3 π₯)2=4
4π₯2=4
π₯2=1
β΄ π₯=Β±1
When π=π
π¦=β3 π₯=β3
So, point is (1, β3)
When π=βπ
π¦=β3 π₯=ββ3
So, point is (β1, ββ3)
As Point M is in 1st Quadrant
β΄ M = (1, β3)
β΄ Required Area = Area OMP + Area PMA
= β«1_0^1βπ¦1ππ₯+β«1_1^2βπ¦2ππ₯
Finding y1 and y2
For y1 β Equation of Line
y = β3x
So, y1 = β3x
For y2 β Equation of Circle
π₯^2+π¦^2=4
π¦^2=4βπ₯^2
β΄ π¦=Β±β(2^2βπ₯^2 )
As Required Area is in first Quadrant
β΄ π¦2=β(2^2βπ₯^2 )
Thus,
Required Area = β«1_0^1βγβ3 π₯ ππ₯γ+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ
=β3 β«1_0^1βγπ₯ ππ₯γ+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ
=β3 [π₯^2/2]_0^1+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ
=β3 [1^2/2β0^2/2]_0^1+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ
=β3/2+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ
It is of form
β«1βγβ(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )γ+π^2/2 γπ ππγ^(β1)β‘γπ₯/π+πγ
Replacing a by 2 , we get
γ=β3/2+[π₯/2 β((2)^2βπ₯^2 )+(2)^2/2 sin^(β1)β‘γπ₯/2γ ]γ_1^2
=β3/2+2/2 β((2)^2β(2)^2 )+2 sin^(β1)β‘γ2/2γβ1/2 β((2)^2β("1" )^2 )β2 sin^(β1)β‘γ1/2γ
=β3/2+0+2 sin^(β1)β‘γ(1)β1/2 β(4β1)βγ 2 sin^(β1)β‘(1/2)
=β3/2+2 Γπ/2ββ3/2β2Γπ/6
=π βπ/3
=ππ /π square units
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
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