Misc 15 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Misc 15 Find the area of the region { , : 2 4 , 4 2+ 4 2 9} Drawing figures Now, our figure will be Circle is 4 2+ 4 2 9 2+ 2 9 4 2+ 2 3 2 2 Radius of circle = 3 2 So, Point C ( 3 2 , 0) Finding point of intersection A & B Solving 2 =4 (1) & 4 2+ 4 2= 9 (2) Putting (1) in (2) 4 2+ 4 2= 9 4 2+ 4 4 =9 4 2+16 9=0 4 2+18 2 9=0 2 (2 +9) 1(2 +9)=0 2 1 2 +9 =0 So, = 1 2 & = 9 2 Area OAC Area OAC = Area ACD + Area OAD Area ACD Area ACD = 1 2 3 2 Equation of circle 2 + 2 = 9 4 2 = 9 4 2 = 9 4 2 Therefore Area ACD = 1 2 3 2 9 4 2 = 1 2 3 2 3 2 2 2 = 2 3 2 2 2 + 3 2 2 2 sin 1 3 2 1 2 3 2 = 2 9 4 2 + 9 8 sin 1 2 3 1 2 3 2 = 3 2 2 9 4 3 2 2 + 9 8 sin 1 2 3 2 3 1 2 2 9 4 1 2 2 + 9 8 sin 1 2 1 2 3 = 3 4 9 4 9 4 + 9 8 sin 1 1 1 4 9 4 1 4 + 9 8 sin 1 1 3 = 9 8 sin 1 1 2 4 9 8 sin 1 1 3 = 9 16 2 4 9 8 sin 1 1 3 Area OAD Area OAD = 0 1 2 Equation of parabola 2 =4 = 4 Therefore Area OAD = 0 1 2 4 =2 0 1 2 =2 0 1 2 1 2 =2 3 2 3 2 0 1 2 = 4 3 [ 3 2 ] 0 1 2 = 4 3 1 2 3 2 0 3 2 = 4 3 1 2 2 = 4 3 1 2 2 2 2 = 2 3 Area OAC = Area ACD + Area OAD = 9 16 2 4 9 8 sin 1 1 3 + 2 3 = 9 16 9 8 sin 1 1 3 + 2 12 Required Area = 2 Area OAC = 2 9 16 9 8 sin 1 1 3 + 2 12 = 9 8 9 4 sin 1 1 3 + 2 6 = 9 8 9 4 sin 1 1 3 + 2 6 2 2 = 9 8 9 4 sin 1 1 3 + 2 6 2 = +
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Misc 4 Important
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Misc 13
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