Question 2 - Area between curve and line - Chapter 8 Class 12 Application of Integrals
Last updated at April 16, 2024 by Teachoo
Area between curve and line
Area between curve and line
Last updated at April 16, 2024 by Teachoo
Question 2 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