Check sibling questions

Example 4 - Find area enclosed by x-axis, y = x and circle - Examples

Example 4 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 4 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 4 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 4 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 4 - Chapter 8 Class 12 Application of Integrals - Part 6

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Transcript

Example 4 Find the area of the region in the first quadrant enclosed by the -axis, the line = , and the circle 2+ 2=32 Equation of Given Circle is :- 2+ 2=32 2+ 2=16 2 2+ 2= 4 2 2 2+ 2= 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 = Putting this in Equation of Circle, we get 2+ 2=32 2+ 2=32 2 2=32 2=16 = 4 As Point M is in 1st Quadrant M = 4 , 4 Required Area = Area OMP + Area PMA = 0 4 1 + 4 4 2 2 Required Area = 0 4 + 4 4 2 4 2 2 2 Taking I1 i.e. I1= 0 4 . = 2 2 0 4 = 4 2 0 2 = 16 2 = 8 Now solving I2 I2= 4 4 2 4 2 2 2 I2= 2 4 2 2 2 + 4 2 2 2 sin 1 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 2 4 8 =16 4 2 4 2 8 = 16 8 2 8 =2 2 8 =4 8 Putting the value of I1 & I2 in (1) Area =8+4 8 =4 Required Area =4 Square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.