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Similar to Applied Mathematics Multiple Integration by Mrs. Geetanjali P.Kale.pdf
Applied Mathematics Multiple Integration by Mrs. Geetanjali P.Kale.pdf
- 1. 1 By Mrs. Geetanjali P.Kale Guru Nanak Institute of Technology, Nagpur
- 2. Contain : ❑ Introduction ❑Double Integrals in Cartesian Coordinates ❑ Double Integrals in Polar Coordinates ❑ Change of Order of Integration ❑ Triple Integration ❑ Application 2
- 3. Why to studyMultiple Integral ? ▪ To find the area of the region bounded by curves ,we require double integrals. ▪ To find the volume of a solid , we require triple integrals. ▪ To find the surface area of the solid obtained by revolving the curve. 3
- 4. ▪ To findmass , Center of gravity , Center of Pressure , Moment of Inertia we require Multiple integral. ▪In vector calculus we have the relationship between these multiple integrals (Green’s Theorem , Stoke’s Theorem and Gauss Divergence Theorem) which have applications in the subject E.M.F. of Engineering field. ▪ Multiple Integrals are very much required to understand many concepts of Engineering and Technology. 4
- 5. The definite integralis defined as the limit of the sum b a dx x f ) ( n n x x f x x f x x f ) ( . .......... ) ( ) ( 2 2 1 1 + + + When and each of the lengths tends to zero . Here are n sub- intervals into which the range b - a has been divided and are values of x lying respectively in the first , second , . . . . . . . ., nth subinterval . → n n x x x x ,........, , , 3 2 1 n x x x x ,........, , , 3 2 1 5
- 6. Our goal isto find the volume of S. } ) , ( ), , ( 0 ) , , {( 3 R y x y x f z R z y x s = 6
- 7.
- 8.
- 9. TYPE – I: When are function of y and are constants 2 1 , x x 2 , 1 y y A Q P C D B O axis Y − axis X − ) ( 2 2 y x = ) ( 1 1 y x = 2 y y = 1 y y = = 2 1 2 1 ) ( ) ( ) , ( ) , ( y y y y R dy dx y x f dA y x f 9
- 10. TYPE – I: When are function of x and are constants 2 1 , y y 2 , 1 x x Y-axis X – axis A Q C D B O ) ( 2 2 x y = ) ( 1 1 x y = 2 x x = 1 x x = = 2 1 2 1 ) ( ) ( ) , ( ) , ( y y x x R dy dx y x f dA y x f 10
- 11. TYPE – I: When are constants 2 1 2 , 1 , , y y x x R axis Y − axis X − 2 x x = 1 x x = 2 y y = 1 y y = = = 2 1 2 1 2 1 2 1 ) , ( ) , ( ) , ( then R y y x x x x y y dy dx y x f dx dy y x f dA y x f 11
- 12. Evaluate the following integral − + 2 1 1 1 2 ) 2 3 ( (i) dx dy xy x 12
- 13. 14 ) 1 ( 2 ) 2 ( 2 2 6 ] ) 1 ( ) 1 ( 3 [ ] ) 1 ( ) 1 ( 3 [ 3 ) 2 3 ( ) 2 3 ( have we , constants are n integratio of limits the Since 3 3 2 1 3 2 1 2 2 1 2 2 2 2 2 1 1 1 2 2 2 1 1 1 2 2 1 1 1 2 = − = = = − + − − + = + = + = + = − = = = = − = − x dx x dx x x x x dx xy y x dx dy xy x dydx xy x y y x x y y 13
- 14. 2 0 , 2 : ) , ( ; 4 ) , ( where ) , ( Evaluate 2 = − = y y x y y x R y x y x f dy dx y x f R x y 2 = y 0 = y y x 2 = 2 y x= as d illustrate is R region The (i) 14
- 15.
- 16.
- 17.
- 18. 2 0 1 2 2 (ii) 0 sin (i) Evaluate y dxdy x e x dydx y y (i)The Region of Integration R is : y : x to x : 0 to x = y = x y 0 = x x y = R 18
- 19.
- 20. as described is R n integratio of region The (i) 2 4 x y − = 0 = y 0 = x2 = x 2 0 2 2 − X Y 20
- 21. 4 4 sin 4 sin sin ) (cos ) (cos ) cos( Therefore 2 / 0 2 1 2 / 0 4 0 2 1 2 / 0 4 0 2 1 2 / 0 2 0 2 2 0 4 0 2 2 2 = = = = = + − d d u dud u rdrd r dydx y x x 0 = r 2 = r 0 = 2 / = 21
- 22.
- 23. 8 4 2 2 1 2 1 cos 2 sec cos 2 cos ) cos ( Therefore 2 / 4 / 2 / 4 / 2 2 2 / 4 / 2 sec 0 2 2 / 4 / sec 0 2 2 / 4 / sec 0 2 2 1 0 0 2 2 2 = − = = = = = = + = = d d d r drd r rdrd r r dxdy y x y r r y 4 / = y sec = r x 2 / = 23
- 24. R dA y x f ) , ( Evaluate (-2,1) and (3,1) (0,0), vertices h the regionwit r triangula closed the is ; ) , ( , 2 R xy y x f where = (i) The Region of Integration R is illustrated below : 0 = y y x 2 − = X 1 = y y x 3 = 1 2 − 3 R Y ) 0 , 0 ( ) 1 , 2 (− ) 1 , 3 ( 24
- 25. = = = = = = = = = = = = = b z a z (z) y (z) y ) z , (y f x ) z , y ( f x 2 1 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 2 1 1 dx ) z , y , x ( f ) z , y x, f( followes as evaluated is integral triple then the , b z , a z and (z) y , (z) y , ) z , (y f x , ) z , y ( f x Let dz dy 25
- 26. − 1 0 11 0 2 6 x y dz dy dx z Evaluate dx dy dz z dz dy dx z x x y y z x y = = − = − = 1 0 1 1 0 1 0 1 1 0 2 2 6 6 dx dy y dx dy z x x y x x y y z z = = = = − = = − = = 1 0 1 2 1 0 1 1 0 2 2 2 ) 1 ( 3 2 6 26
- 27.
- 28. Area by Double Integration a)Cartesian Co-ordinates : The area A of the region bounded by the curves and the lines x = a , x= b is given by , ) ( , ) ( 2 1 x f y x f y = = = b a x f x f dx dy A ) ( ) ( 2 1 The area A of the region bounded by the curves and the lines y = c , y = d is given by, ) ( , ) ( 2 1 y f x y f x = = = d c y f y f dy dx A ) ( ) ( 2 1 28
- 29. b) Polar Co-ordinates: The area of region bounded by the curve and the lines is given by , = ) ( ) ( 2 1 f f d dr r A ) ( , ) ( 2 1 f r f r = = = = , 29
- 30. Find the smallerof the areas bounded by the ellipse and the straight line 36 9 4 2 2 = + y x 6 3 2 = + y x 1 4 9 2 2 = + y x The equation of the ellipse is And the line is 1 2 3 = + y x The point of intersection of both curves are A(3 , 0 ) and B(0 , 2) 30
- 31. R A(3 , 0) B(0,2) Y-axis X-axis 2 4 2 3 y x − = ) 2 ( 2 3 y x − = (0 , 0 ) From figure we have : 2 4 2 3 ) 2 ( 2 3 : y to y x − − Q P 2 0 : to y 31
- 32. Thus the requiredarea is : = − − = = 2 0 4 2 3 ) 2 ( 2 3 2 y y y x dy dx A 2 0 2 1 2 2 2 0 4 2 3 ) 2 ( 2 3 2 0 2 2 2 sin 2 4 2 4 2 3 ) 2 ( 2 3 4 2 3 2 + − + − = − − − = = − = − − = y y y y y dy y y dy x y y y y 32
- 33. ( ) 2 2 3 2 2 2 2 3 2 4 1 sin 2 2 31 − = − • = + − = − A 33
- 34.
- 35. The total electriccharge on a solid object occupying a region E and having charge density σ(x, y, z) is: ( ) , , E Q x y z dV = TOTAL ELECTRIC CHARGE 35
- 36. Find the centerof mass of a solid of constant density that is bounded by the parabolic cylinder x = y2 and the planes x = z, z = 0, and x = 1. Example 5 APPLICATIONS 36
- 37. The solid Eand its projection onto the xy-plane are shown. Example 5 APPLICATIONS Fig. 16.6.14a, p. 1033 Fig. 16.6.14b, p. 1033 37
- 38. The lower andupper surfaces of E are the planes z = 0 and z = x. So, we describe E as a type 1 region: ( ) 2 , , 1 1, 1,0 E x y z y y x z x = − Example 5 APPLICATIONS Fig. 16.6.14a, p. 1033 38
- 39. Then, if thedensity is ρ(x, y, z), the mass is: Example 5 APPLICATIONS 2 2 1 1 1 0 1 1 1 E x y y m dV dz dxdy xdxdy − − = = = 39
- 40. APPLICATIONS ( ) ( ) 2 1 2 1 1 1 4 1 1 4 0 1 5 0 2 1 2 1 4 55 x x y x dy y dy y dy y y = − = − = = − = − = − = Example 5 40
- 41. Due to thesymmetry of E and ρ about the xz-plane, we can immediately say that Mxz = 0, and therefore . The other moments are calculated as follows. 0 y = Example 5 APPLICATIONS 41
- 42. 2 2 1 1 1 0 11 2 1 yz E x y y M x dV x dz dxdy x dxdy − − = = = Example 5 APPLICATIONS 42
- 43. APPLICATIONS ( ) 2 1 3 1 1 1 6 0 1 7 0 3 2 1 3 2 3 7 4 7 x xy x dy y dy y y = − = = = − = − = Example 5 43
- 44. ( ) 2 2 2 1 1 10 2 1 1 1 0 1 1 2 1 1 6 0 2 2 2 1 3 7 x xy y E z x y z y M z dV z dz dxdy z dxdy x dxdy y dy − = − = − = = = = = − = Example 5 APPLICATIONS 44
- 45. Therefore, the centerof mass is: ( ) ( ) 5 5 7 14 , , , , ,0, yz xy xz M M M x y z m m m = = Example 5 APPLICATIONS 45