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Using integration, find the area in the first quadrant bounded by the curve y = x|x|, the circle x 2 + y 2 = 2 and the y-axis

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Find area of x|x|,  circle x^2 + y^2 = 2 and y-axis using integration

Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9


Transcript

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–ˆ(π‘₯Γ—π‘₯, π‘₯β‰₯0@π‘₯ Γ—(βˆ’π‘₯), π‘₯<0)─ y = {β–ˆ(π‘₯^2, π‘₯β‰₯0@βˆ’π‘₯^2, π‘₯<0)─ And x2 + y2 = 2 (x – 0)2 + (y – 0)2 = (√2)^2 So, it is a circle with center (0, 0) and Radius = √2 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–ˆ(π‘₯Γ—π‘₯, π‘₯β‰₯0@π‘₯ Γ—(βˆ’π‘₯), π‘₯<0)─ y = {β–ˆ(π‘₯^2, π‘₯β‰₯0@βˆ’π‘₯^2, π‘₯<0)─ And x2 + y2 = 2 (x – 0)2 + (y – 0)2 = (√2)^2 So, it is a circle with center (0, 0) and Radius = √2 π’š=γ€–βˆ’π’™γ€—^𝟐 for x < 0 π’š=γ€–βˆ’π’™γ€—^𝟐 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+𝑦^2=2 Putting x2 = y 𝑦+𝑦^2=2 𝑦^2+π‘¦βˆ’2=0 𝑦^2+2π‘¦βˆ’π‘¦βˆ’2=0 𝑦(𝑦+2)βˆ’1(𝑦+2)=0 (yβˆ’1)(𝑦+2)=0 So, y = 1, y = –2 So, y = 1, y = –2 Since y is in 1st quadrant, it is positive ∴ y = 1 Now, y = x2 1 = x2 x2 = 1 x = Β± √1 x = Β± 1 ∴ x = 1, –1 Since x is in 1st quadrant x = 1 ∴ x = 1, y = 1 So, point B = (1, 1) Area Required = Area AODB – Area ODB Area AODB Area AODB = ∫_0^1▒〖𝑦 𝑑π‘₯γ€— For circle x2 + y2 = 2 y2 = 2 – x2 y = Β± √(2βˆ’π‘₯^2 ) Since AODB is in 1st quadrant y =√(2βˆ’π‘₯^2 ) ∴ Area AODB = ∫_0^1β–’βˆš(2βˆ’π‘₯^2 ) 𝑑π‘₯ = ∫_0^1β–’βˆš((√2)^2βˆ’π‘₯^2 ) 𝑑π‘₯ It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 Here, a = √2 = [π‘₯/2 √((√2)^2βˆ’π‘₯^2 )+(√2)^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/√2γ€— " " ]_0^1 = [π‘₯/2 √(2βˆ’π‘₯^2 )+2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/√2γ€— " " ]_0^1 = [π‘₯/2 √(2βˆ’π‘₯^2 )+〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/√2γ€— " " ]_0^1 = [1/2 √(2βˆ’1^2 )+〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1/√2γ€— " " ] – [0/2 √(2βˆ’0^2 )+〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 0/√2γ€— " " ] = [1/2 √(2βˆ’1)+〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1/√2γ€— " " ] – [0+〖𝑠𝑖𝑛〗^(βˆ’1)⁑0 " " ] = [1/2+〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1/√2γ€— " " ] – [0+0" " ] = 1/2+πœ‹/4 Area ODB Area ODB = ∫_0^1▒〖𝑦 𝑑π‘₯γ€— For parabola y = x2 ∴ Area ODB = ∫_0^1β–’π‘₯^2 𝑑π‘₯ = [π‘₯^3/3]_0^1 =1^3/3βˆ’0^3/3 = 1/3 Thus, Area Required = Area AODB – Area ODB = 1/2+πœ‹/4 – 1/3 = 1/2βˆ’1/3+πœ‹/4 = 𝟏/πŸ”+𝝅/πŸ’ square units

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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.