# Ex 8.1, 6 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.1, 6 Find the area of the region in the first quadrant enclosed by 𝑥−axis, line 𝑥 = 3 𝑦 and the circle 𝑥2 + 𝑦2 = 4. Equation of given circle is :- 𝑥2+ 𝑦2=4 𝑥2+ 𝑦2= 22 ∴ radius 𝑟 = 2 Let AB represent the line 𝑥= 3𝑦 We have to find Area of OAC First, we have to find Point C Point C is the intersection of line and circle. So, we solve their equation We know that 𝑥= 3𝑦 Putting value of x in equation of circle 𝑥2+ 𝑦2=4 3𝑦2+ 𝑦2=4 3 𝑦2+ 𝑦2=4 4 𝑦2=4 𝑦2=1 ∴ 𝑦=±1 Now, finding value of x As Point C is in first Quadrant ∴ C is 3 , 1 Now, Area OAC is Area of OAC = Area OCX + Area XCA = 0 3𝑦1𝑑𝑥+ 32𝑦2𝑑𝑥 Area of OAC = 0 3𝑦1𝑑𝑥+ 32𝑦2𝑑𝑥 ∴ Area of OAC = 0 3 𝑥 3 𝑑𝑥+ 32 4− 𝑥2𝑑𝑥 Solving I1 & I2 separately I1 = 0 3 𝑥 3 𝑑𝑥 = 1 3 0 3𝑥 𝑑𝑥 = 1 3 𝑥220 3 = 12 3 𝑥20 3 = 1 2 3 32− 02 = 1 2 3 3 = 32 Now , I2 = 32 4− 𝑥2 𝑑𝑥 I2 = 32 22− 𝑥2 𝑑𝑥 I2 = 12𝑥 22− 𝑥2+ 222 sin−1 𝑥2 32 I2 = 12 2 22− 22− 12 3 22− 32+ 222 sin−1 22− 222 sin−1 32 I2 = 0 – 32 4−3+2 sin−1 1−2 sin−1 32 I2 = − 32+2 sin−1 1−𝑠𝑖 𝑛−1 32 I2 = − 32+2 𝜋2− 𝜋3 I2 = − 32+2 3𝜋 − 2𝜋2 × 3 I2 = − 32+ 𝜋3 Putting value of I1 and I2 in (1) Area of OAC = 32− 32+ 𝜋 3 = 𝜋 3 ∴ Required Area = 𝝅𝟑 square units

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.