1. Chapter 8 Class 12 Application of Integrals
2. Serial order wise

Transcript

Example 8 In Fig AOBA is the part of the ellipse 9𝑥2+𝑦2=36 in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB Area Required = Area ABC Area ABC = Area OACB – Area ∆ OAB Area OACB Area OACB = 0﷮2﷮𝑦𝑑𝑥﷯ Here, 𝑦 → ellipse equation 9𝑥2+𝑦2=36 𝑦2=36−9𝑥2 𝑦=± ﷮36−9 𝑥﷮2﷯﷯ As OACB is in 1st quadrant, 𝑦= ﷮36−9 𝑥﷮2﷯﷯ Area OACB = 0﷮2﷮𝑦𝑑𝑥﷯ = 0﷮2﷮ ﷮36−9 𝑥﷮2﷯﷯﷯ = 0﷮2﷮ ﷮9( 36﷮9﷯− 𝑥﷮2﷯)﷯﷯ = 0﷮2﷮ ﷮9(4− 𝑥﷮2﷯)﷯﷯ = 0﷮2﷮ ﷮ 3﷮2﷯( 2﷮2﷯− 𝑥﷮2﷯)﷯﷯ = 3 0﷮2﷮ ﷮ 2﷮2﷯− 𝑥﷮2﷯﷯𝑑𝑥 ﷯ = 3 𝑥﷮2﷯ ﷮ 2﷮2﷯− 𝑥﷮2﷯﷯+ 2﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮2﷯﷯﷯﷮0﷮2﷯ = 3 2﷮2﷯ ﷮ 2﷮2﷯− 2﷮2﷯﷯+2 sin﷮−1﷯﷮ 2﷮2﷯− 0﷮2﷯﷯ ﷮ 2﷮2﷯− 0﷮2﷯﷯−2 sin﷮−1﷯﷮0﷯﷯ = 3 1 ﷮0﷯+2 sin﷮−1﷯﷮1−0−0﷯﷯ = 3 0+2 sin﷮−1﷯﷮1﷯﷯ = 3 × 2 × 𝜋﷮2﷯ = 3𝜋 Area OAB Area OAB = 0﷮2﷮𝑦𝑑𝑥﷯ Here, 𝑦 → equation of line AB Equation of line between A(2, 0) & B(0, 6) is 𝑦 − 0﷮𝑥 − 2﷯= 6 − 0﷮0 − 2﷯ 𝑦﷮𝑥−2﷯= 6﷮−2﷯ 𝑦﷮𝑥−2﷯=−3 𝑦=−3 𝑥−2﷯ Area OBC = 0﷮2﷮𝑦 𝑑𝑥﷯ = 0﷮2﷮−3 𝑥−2﷯ 𝑑𝑥﷯ = – 3 0﷮2﷮ 𝑥−2﷯ 𝑑𝑥﷯ = – 3 𝑥﷮2﷯﷮2﷯ −2𝑥 ﷯﷮0﷮2﷯ = – 3 2﷮2﷯﷮2﷯−2 ×2− 0﷮2﷯﷮2﷯−2 ×0﷯ = – 3 −2﷯ = 6 Area Required = Area OACB – Area ∆ OAB =3𝜋−6