Ex 8.1, 7 Find area of smaller part of circle x2 + y2 = a2 cutoff - Ex 8.1

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Ex 8.1, 7 Find the area of the smaller part of the circle 𝑥2+𝑦2= 𝑎2 cut off by the line 𝑥= 𝑎﷮ ﷮2﷯﷯. Equation of Given Circle is 𝑥﷮2﷯+ 𝑦﷮2﷯= 𝑎﷮2﷯ ∴ Radius , 𝑟 = 𝑎 So, a is positive ∴ 𝑥= 𝑎﷮ ﷮2﷯﷯ will lie on positive side of x-axis Let PQ represent the line 𝑥= 𝑎﷮ ﷮2﷯﷯ We have to find Area of APQ Area APQ = 2 × Area APR = 2 × We know that, 𝑥﷮2﷯+ 𝑦﷮2﷯= 𝑎﷮2﷯ 𝑦﷮2﷯= 𝑎﷮2﷯− 𝑥﷮2﷯ 𝑦=± ﷮ 𝑎﷮2﷯− 𝑥﷮2﷯﷯ Since , APR is in 1st Quadrant 𝑦= ﷮ 𝑎﷮2﷯− 𝑥﷮2﷯﷯ ∴ Area of APQ = 2 × = 2 × = 2 1﷮2﷯𝑥 ﷮ 𝑎﷮2﷯− 𝑥﷮2﷯﷯+ 𝑎﷮2﷯﷮𝑎﷯ sin﷮−1﷯﷮ 𝑥﷮𝑎﷯﷯﷯﷮ 𝑎﷮ ﷮2﷯﷯﷮𝑎﷯ = 2 𝑎﷮2﷯ ﷮ 𝑎﷮2﷯− 𝑎﷮2﷯﷯+ 𝑎﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑎﷮𝑎﷯− 𝑎﷮2 ﷮2﷯﷯﷯ ﷮ 𝑎﷮2﷯− 𝑎﷮ ﷮2﷯﷯﷯﷮2﷯﷯− 𝑎﷯﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑎﷮ ﷮2﷯𝑎﷯ ﷯﷯ = 2 0+ 𝑎﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 1﷯− 𝑎﷮2 ﷮2﷯﷯ ﷮ 𝑎﷮2﷯− 𝑎﷮2﷯﷮2﷯﷯− 𝑎﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯﷯﷯﷯ = 2 𝑎﷮2﷯﷮2﷯𝑠𝑖 𝑛﷮−1﷯ 1﷯− 𝑎﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯− 𝑎﷮2 ﷮2﷯﷯ ﷮ 2𝑎﷮2﷯ − 𝑎﷮2﷯﷮2﷯﷯﷯ ﷯ = 2 𝑎﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 1﷯−﷯ sin﷮−1﷯﷮ 1﷮ ﷮2﷯﷯﷯﷯﷯− 𝑎﷮2 ﷮2﷯﷯ ﷮ 𝑎﷮2﷯ ﷮2﷯﷯﷯ = 2 𝑎﷮2﷯﷮2﷯ 𝜋﷮2﷯− 𝜋﷮4﷯﷯﷯− 𝑎﷮2 ﷮2﷯﷯ × 𝑎﷮ ﷮2﷯﷯﷯ = 2 𝑎﷮2﷯﷮2﷯ 𝜋﷮2﷯− 𝜋﷮4﷯﷯− 𝑎﷮2﷯﷮2.2﷯﷯ = 2 𝑎﷮2﷯﷮2﷯ 𝜋﷮4﷯﷯− 𝑎﷮2﷯﷮4﷯﷯ = 2 𝑎﷮2﷯﷮4﷯ 𝜋﷮2﷯−1﷯ = 𝑎﷮2﷯﷮2﷯ 𝜋﷮2﷯−1﷯ ∴ Required Area = 𝒂﷮𝟐﷯﷮𝟐﷯ 𝝅﷮𝟐﷯−𝟏﷯ Square units

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