Question 26 (OR 2 nd question)

Find the area of the region.

{(x, y) : x 2 + y 2 ≤ 8, x 2 ≤ 2y}

Find the area of the region {(x, y) : x^2 + y^2 ≤ 8, x^2 ≤ 2y}

Question 26 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
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Transcript

Question 26 (OR 2nd question) Find the area of the region. {(x, y) : x2 + y2 ≤ 8, x2 ≤ 2y} "x2 + y2 ≤ 8" Let’s draw "x2 + y2 = 8" Taking Point (0, 0) "02 + 02 ≤ 8" 0 ≤ 8 Is True So, We Shade inside of graph x2 ≤ 2y Let’s draw x2 = 2y Taking Point (0, 1) 02 ≤ 2 × 1 0 ≤ 2 Is True So, We Shade center of graph Now, our equations are x2 + y2 ≤ 8 x2 ≤ 2y Drawing circle x2 + y2 ≤ 8 x2 + y2 ≤ 4 × 2 𝑥2+ 𝑦2 ≤(2√2)^2 ∴ Radius of circle = 2√2 So, Point C (2√2, 0) Finding point of intersection A & B Finding point of intersection of circle and parallelogram Equation of circle is x2 + y2 = 8 Putting x2 = 2y from parabola 2y + y2 = 8 y2 + 2y – 8 = 0 y2 + 4y – 2y – 8 = 0 y(y + 4) – 2(y + 4) = 0 (y – 2) (y + 4) = 0 So, y = 2, –4 Since point is in positive y-axis y = 2 Putting y = 2 in equation of parabola x2 = 2y x2 = 2(2) x2 = 4 x = ± 2 So, A = (2, 2) and B = (–2, 2) Now, Since figure is symmetric Area AOC = Area BOC ∴ Required Area = Area AOBC = 2 × Area AOC Area AOC Here, point D is (0, 2) Area AOC = Area AOD + Area ADC Let’s find Area AOD and Area ADC separately Area AOD Area AOD =∫_(1/2)^(3/2)▒〖𝑥 𝑑𝑦〗 𝑥→ Equation of parabola 𝑥^2=2𝑦 𝑥=√(2𝑦" " ) Therefore Area AOD =∫_0^2▒〖 √2𝑦 𝑑𝑦〗 =√2 ∫_0^2▒〖 √(𝑦" " ) 𝑑𝑦〗 =√2 ∫_0^2▒〖 𝑦^(1/2) 𝑑𝑥〗 =√2 [𝑦^(3/2)/(3/2)]_0^2 =(2√2)/3 〖[𝑦^(3/2)]〗_0^2 =(2√2)/3 [(2)^(3/2)−0^(3/2) ] =(2√2)/3 [(2)^(3/2) ] =(2√2)/3 [(2^3 )^(1/2) ] =(2√2)/3 [(2^2×2)^(1/2) ] =(2√2)/3 [2×√2] =(4 × 2)/3 =8/3 Area ACD Area ACD =∫_2^(2√2)▒〖𝑥 𝑑𝑦〗 𝑥→ Equation of circle 𝑥^2+𝑦^2=8 𝑥^2=8−𝑦^2 𝑥=√(8−𝑦^2 ) Therefore Area ACD =∫_2^(2√2)▒〖√(8−𝑦^2 ) 𝑑𝑦〗 =∫_2^(2√2)▒〖 √((2√2)^2−𝑦^2 ) 𝑑𝑦〗 It is of form ∫1▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )〗+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1)⁡〖𝑥/𝑎+𝑐〗 Replacing a by 2√2, x with y, we get = [𝑦/2 √((2√2)^2−𝑦^2 )+(2√2)^2/2 sin^(−1)⁡〖𝑦/(2√2)〗 ]_2^(2√2) = [𝑦/2 √(8−𝑦^2 )+8/2 sin^(−1)⁡〖𝑦/(2√2)〗 ]_2^(2√2) = [𝑦/2 √(8−𝑦^2 )+4 sin^(−1)⁡〖𝑦/(2√2)〗 ]_2^(2√2) = [(2√2)/2 √(8−〖(2√2)〗^2 )+4 sin^(−1)⁡〖(2√2)/(2√2)〗 ] – [2/2 √(8−2^2 )+4 sin^(−1)⁡〖2/(2√2)〗 ] = [√2 √(8−8)+4 sin^(−1)⁡1 ] – [√(8−4)+4 sin^(−1)⁡〖1/√2〗 ] = [0+4×𝜋/2] – [2+4×𝜋/4] = 2π – 2 – π = π – 2 Now, Area AOC = Area AOD + Area ADC = 8/3 + π – 2 = π + 2/3 And Required Area = 2 × Area AOC = 2(π + 2/3) = 2π + 𝟒/𝟑 square units

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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.