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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.