**
Question 26 (OR 2
**
**
nd
**
**
question)
**

Find the area of the region.

{(x, y) : x
^{
2
}
+ y
^{
2
}
≤ 8, x
^{
2
}
≤ 2y}

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd) You are here

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 1, 2019 by Teachoo

**
Question 26 (OR 2
**
**
nd
**
**
question)
**

Find the area of the region.

{(x, y) : x
^{
2
}
+ y
^{
2
}
≤ 8, x
^{
2
}
≤ 2y}

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