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

Transcript

Example 4 Find the area of the region in the first quadrant enclosed by the 𝑥-axis, the line 𝑦 = 𝑥, and the circle 𝑥2+𝑦2=32 Equation of Given Circle is :- 𝑥2+𝑦2=32 𝑥2+𝑦2=16 ×2 𝑥2+𝑦2= 4﷮2﷯ ×2 𝑥2+𝑦2= 4 ﷮2﷯﷯﷮2﷯ Let point where line and circle intersect be point M Required Area = Area of shaded region = Area OMA First we find Point M Point M is intersection of line and circle We know that 𝑦=𝑥 Putting this in Equation of Circle, we get 𝑥2+𝑦2=32 𝑥2+𝑥2=32 2𝑥2=32 𝑥2=16 ∴ 𝑥=±4 As Point M is in 1st Quadrant ∴ M = 4 , 4﷯ ∴ Required Area = Area OMP + Area PMA = 0﷮4﷮𝑦1𝑑𝑥﷯+ 4﷮4 ﷮2﷯﷮𝑦2𝑑𝑥﷯ ∴ Required Area = 0﷮4﷮𝑥 𝑑𝑥﷯+ 4﷮4 ﷮2﷯﷮ ﷮ 4 ﷮2﷯﷯﷮2﷯− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ Taking I1 i.e. I1= 0﷮4﷮𝑥.𝑑𝑥﷯ = 𝑥﷮2﷯﷮2﷯﷯﷮0﷮4﷯ = 4﷯﷮2﷯ − 0﷮2﷯ = 16﷮2﷯ = 8 Now solving I2 I2= 4﷮4 ﷮2﷯﷮ ﷮ 4 ﷮2﷯﷯﷮2﷯− 𝑥﷮2﷯﷯ 𝑑𝑥﷯ I2= 𝑥﷮2﷯ ﷮ 4 ﷮2﷯﷯﷮2﷯− 𝑥﷮2﷯﷯+ 4 ﷮2﷯﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑥﷮4 ﷮2﷯﷯﷯﷯﷮4﷮4 ﷮2﷯﷯ = 4 ﷮2﷯﷮2﷯ ﷮ 4 ﷮2﷯﷯﷮2﷯− 4 ﷮2﷯﷯﷮2﷯﷯+ 4 ﷮2﷯﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 4 ﷮2﷯﷮4 ﷮2﷯﷯﷯ − 4﷮2﷯ ﷮ 4 ﷮2﷯﷯﷮2﷯− 4﷯﷮2﷯﷯− 4 ﷮2﷯﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 4﷮4 ﷮2﷯﷯﷯ =0+ 16 × 2﷮2﷯ sin﷮−1﷯﷮ 1﷯−2 ﷮32−16﷯−﷯ 16 × 2﷮2﷯ sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯﷯ =16 sin﷮−1﷯﷮(1)﷯−2 ﷮16﷯−16 sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯﷯ =16 sin﷮−1﷯﷮ 1﷯− sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯﷯﷯﷯−8 =16 𝜋﷮2﷯− 𝜋﷮4﷯﷯−8 =16 4𝜋 − 2𝜋﷮4 × 2﷯﷯−8 = 16﷮8﷯ 2𝜋﷯−8 =2 2𝜋﷯−8 =4𝜋−8 Putting the value of I1 & I2 in (1) Area =8+4𝜋−8 =4𝜋 ∴ Required Area =4𝜋 Square units

Serial order wise