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Example 1 Find the area enclosed by the circle π‘₯2 + 𝑦2 = π‘Ž2Given π‘₯^2 + 𝑦^2= π‘Ž^2 This is a circle with Center = (0, 0) Radius = π‘Ž Since radius is a, OA = OB = π‘Ž A = (π‘Ž, 0) B = (0, π‘Ž) Now, Area of circle = 4 Γ— Area of Region OBAO = 4 Γ— ∫1_𝟎^π’‚β–’γ€–π’š 𝒅𝒙〗 Here, y β†’ Equation of Circle We know that π‘₯^2 + 𝑦^2 = π‘Ž^2 𝑦^2 = π‘Ž^2βˆ’ π‘₯^2 y = Β± √(π‘Ž^2βˆ’π‘₯^2 ) Since AOBA lies in 1st Quadrant y = √(𝒂^πŸβˆ’π’™^𝟐 ) Now, Area of circle = 4 Γ— ∫1_0^π‘Žβ–’γ€–π‘¦ 𝑑π‘₯γ€— = 4 Γ— ∫1_0^π‘Žβ–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— = 4[𝒙/𝟐 √(𝒂^πŸβˆ’π’™^𝟐 )+𝒂^𝟐/𝟐 γ€–"sin" γ€—^(βˆ’πŸ) 𝒙/𝒂]_𝟎^𝒂 = 4[π‘Ž/2 √(π‘Ž^2βˆ’π‘Ž^2 )+π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) π‘Ž/π‘Žβˆ’0/2 √(π‘Ž^2βˆ’0)βˆ’0^2/2 γ€–"sin" γ€—^(βˆ’1) (0)] = 4[0+π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) (1)βˆ’0βˆ’0] = 4.π‘Ž^2/2. πœ‹/2 = 𝝅𝒂^𝟐 Using: √(π‘Ž^2βˆ’π‘₯^2 )dx = 1/2 √(π‘Ž^2βˆ’π‘₯^2 ) + π‘Ž^2/2 γ€–"sin" γ€—^(βˆ’1) π‘₯/4 + c

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.