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Ex 8.1, 5 - Find area by ellipse x2/4 + y2/9 =1 - Class 12

Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 3 Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals - Part 4

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Ex 8.1, 5 Find the area of the region bounded by the ellipse π‘₯^2/4+𝑦^2/9=1 Given Equation of Ellipse π‘₯^2/4+𝑦^2/9=1 π‘₯^2/(2)^2 +𝑦^2/(3)^2 =1 Area of Ellipse = Area of ABCD = 2 Γ— [Area of BCD] = 2 Γ— ∫_(βˆ’2)^2▒〖𝑦 𝑑π‘₯γ€— We know that π‘₯^2/4+𝑦^2/9=1 𝑦^2/9=1βˆ’π‘₯^2/4 𝑦^2/9=(4 βˆ’ π‘₯^2)/4 𝑦^2=9/4 (4βˆ’π‘₯^2 ) Taking Square Root on Both Sides 𝑦="Β±" √(9/4 (4βˆ’π‘₯^2 ) ) 𝑦 ="Β±" 3/2 √(4βˆ’π‘₯^2 ) Since Area BCD is above the x-axis, The value of y will be positive ∴ 𝑦=3/2 √(4βˆ’π‘₯^2 ) Now, Area of Ellipse = 2 Γ— ∫_(βˆ’2)^2▒〖𝑦 𝑑π‘₯γ€— = 3 ∫_(βˆ’2)^2β–’γ€– √((2)^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— = 3 [π‘₯/2 √((2)^2βˆ’π‘₯^2 )+(2)^2/2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]_(βˆ’2)^2 It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 Replacing a by 2 we get = 3 [π‘₯/2 √(4βˆ’π‘₯^2 )+2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]_(βˆ’2)^2 = 3[(2/2 √(4βˆ’2^2 )+2 sin^(βˆ’1)⁑〖2/2γ€— )βˆ’((βˆ’2)/2 √(4βˆ’γ€–(βˆ’2)γ€—^2 )+2 sin^(βˆ’1)⁑〖(βˆ’2)/2γ€— )] = 3[(0+2 sin^(βˆ’1)⁑1 )βˆ’(0+2 sin^(βˆ’1)⁑〖(βˆ’1)γ€— )] = 3[2 sin^(βˆ’1)⁑1βˆ’2 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖(βˆ’πŸ)γ€— ] = 3[2 sin^(βˆ’1)⁑1βˆ’2Γ—βˆ’γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑𝟏 ] = 3[2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑1+2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑1 ] = 3 [4 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑(𝟏) ] = 3 Γ— 4 Γ— 𝝅/𝟐 = 6Ο€ ∴ Area of Ellipse = 6Ο€ square units (As sin-1 (–θ) = –sin–1 ΞΈ )

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