# Find area enclosed by circle x2 + y2 = a2

Derive the Area of a Circle Using Integration (x^2+y^2=r^2)
Derive the Area of a Circle Using Integration (x^2+y^2=r^2)

Examples

Example 2 Important

Example 3

Example 4 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Important Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Important Deleted for CBSE Board 2024 Exams

Question 9 Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams

Last updated at Aug. 11, 2023 by Teachoo

Example 1 Find the area enclosed by the circle 𝑥2 + 𝑦2 = 𝑎2Given 𝑥^2 + 𝑦^2= 𝑎^2 This is a circle with Center = (0, 0) Radius = 𝑎 Since radius is a, OA = OB = 𝑎 A = (𝑎, 0) B = (0, 𝑎) Now, Area of circle = 4 × Area of Region OBAO = 4 × ∫1_𝟎^𝒂▒〖𝒚 𝒅𝒙〗 Here, y → Equation of Circle We know that 𝑥^2 + 𝑦^2 = 𝑎^2 𝑦^2 = 𝑎^2− 𝑥^2 y = ± √(𝑎^2−𝑥^2 ) Since AOBA lies in 1st Quadrant y = √(𝒂^𝟐−𝒙^𝟐 ) Now, Area of circle = 4 × ∫1_0^𝑎▒〖𝑦 𝑑𝑥〗 = 4 × ∫1_0^𝑎▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥〗 Using: √(𝑎^2−𝑥^2 )dx = 1/2 √(𝑎^2−𝑥^2 ) + 𝑎^2/2 〖”sin” 〗^(−1) 𝑥/4 + c = 4[𝒙/𝟐 √(𝒂^𝟐−𝒙^𝟐 )+𝒂^𝟐/𝟐 〖”sin” 〗^(−𝟏) 𝒙/𝒂]_𝟎^𝒂 = 4[𝑎/2 √(𝑎^2−𝑎^2 )+𝑎^2/2 〖”sin” 〗^(−1) 𝑎/𝑎−0/2 √(𝑎^2−0)−0^2/2 〖”sin” 〗^(−1) (0)] = 4[0+𝑎^2/2 〖”sin” 〗^(−1) (1)−0−0] = 4.𝑎^2/2. 𝜋/2 = 𝝅𝒂^𝟐

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