Misc 15 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
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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
= +
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
