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Example 5 Important
Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Example 8 Important Deleted for CBSE Board 2023 Exams
Example 9 Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 11
Example 12
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important
Last updated at March 16, 2023 by Teachoo
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