Ex 8.1, 12 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Aug. 11, 2021 by
Last updated at Aug. 11, 2021 by
Transcript
Ex 8.1, 12 Area lying in the first quadrant and bounded by the circle 𝑥2+𝑦2=4 and the lines 𝑥 = 0 and 𝑥 = 2 is (A) π (B) 𝜋2 (C) 𝜋3 (D) 𝜋4 Equation of Given Circle :- 𝑥2+ 𝑦2=4 𝑥2+ 𝑦2= 22 ∴ 𝑟𝑎𝑑𝑖𝑢𝑠 , 𝑟=2 Line 𝑥=0 is y-axis & Line x = 2 passes through point A 2 , 0 So, Required area = Area of shaded region = Area OAB = 02𝑦.𝑑𝑥 We know that, 𝑥2+ 𝑦2=4 𝑦2=4− 𝑥2 ∴ 𝑦=± 4− 𝑥2 As, OBA is in 1st Quadrant ∴ 𝑦= 4− 𝑥2 ∴ Required area = 02𝑦.𝑑𝑥 = 02 4− 𝑥2 𝑑𝑥 = 02 22− 𝑥2 𝑑𝑥 = 𝑥 2 22− 𝑥2+2 sin−1 𝑥2 02 = 22 22− 22+2 sin−1 22 − 02 22− 02−2 sin−1 02 = 0+2 sin−1 1−0 4−2 sin−1 0 = 2 sin−1 1−2 sin−1 0−0 = 2 sin−1 1− sin−1 0 = 2 𝜋2−0 = 2 . 𝜋2 = π ∴ Area Required = π square units Hence, Option (A) is correct
Area bounded by curve and horizontal or vertical line
Area bounded by curve and horizontal or vertical line
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