Topic 3. Integral calculus

Transcription

1 Integral calculus Topic 3 Line, surface and volume integrals Fundamental theorems of calculus Fundamental theorems for gradients Fundamental theorems for divergences! Green s theorem Fundamental theorems for curls! Stokes theorem

2 Line, surface and volume integrals (1) Line integral: b A line integral is an expression of the form b ap v dl where v is a vector function, dl is the infinitesimal displacement vector, and the integral is to be carried along the path P from point a to point b. If the path is a closed loop (i.e. if b = a), put a circle in the integral sign: v dl At each point on the path we take the dot product of v with the displacement dl to the next point. a x dl

3 Problem: Calculate the line integral of the function v = x ˆx + ŷ + ẑ from the origin to the point (1, 1, 1) b two different routes: (a) (,, ) " (1,, ) " (1, 1, ) " (1, 1, 1) (b) The direct straight line

4 Problem: Calculate the line integral of the function v = x ˆx + ŷ + ẑ from the origin to the point (1, 1, 1) b two different routes: (a) (,, ) " (1,, ) " (1, 1, ) " (1, 1, 1) (,, ) ( 1,, ). x : 1, = = dl = dxˆx, v dl = x dx v dl = x dx 1 1 = x3 3 = 1 3 ( 1,, ) ( 1,1, ). x =1, : 1, = dl = dŷ, v dl = d = v dl = ( 1,1, ) ( 1,1,1 ). x = =1, : 1 dl = dẑ, v dl = 1 1 d = d v dl = d = =1 v dl = # % 1& $ 3 ' ( + +1= 4 3

5 The direct straight line x = = : 1;dx = d = d v dl = x dx + d + d = x dx + x dx + x dx = 4x dx 1 v dl = 4x dx # & = % 4x3 ( $ 3 ' 1 = 4 3

6 () Surface integral: A surface integral is an expression of the form da S v da where v is a vector function, da is the infinitesimal area, with direction perpendicular to the surface. x If the surface is closed such as a balloon, put a circle in the integral sign: v da The outward is positive but for an open area it is arbitrar.

7 Example: Calculate the surface integral of the function v = xˆx + ( x + )ŷ + ( 3)ẑ over the 5 sides (excluding the bottom) of the cubical box (side ). (v) (ii) x (iv) (i) (iii)

8 Example: Calculate the surface integral of the function v = xˆx + ( x + )ŷ + ( 3)ẑ over the 5 sides (excluding the bottom) of the cubical box (side ). (v) (ii) x (iv) (i) (iii) ( i) x =, da = ddˆx, v da = xdd = 4dd v da = 4 d d =16

9 ( ii) x =, da = ddˆx, v da = xdd = v da = ( iii) =, da = dxdŷ, v da = ( x + )dxd v da = ( x + )dx d =1 ( iv) =, da = dxdŷ, v da = ( x + )dxd v da = ( x + )dx d = 1 ( v) =, da = dxdẑ, v da = ( 3)dxd = dxd v da = dx d = 4 v da = = surface

10 (3) Volume integral: A volume integral is an expression of the form T dτ where T is a scalar function, dτ is the infinitesimal volume element. V In Cartesian coordinates dτ = dx d d If T is the densit of a substance, then the volume integral would give the total mass. Ma encounter volume integrals of vector functions but because the unit vectors are constants, the come outside the integral. v dτ = ( v x ˆx + v ŷ + v ẑ)dτ = ˆx v x dτ + ŷ v dτ + ẑ v dτ

11 Example: Calculate the volume integral of T = x over the prism. Do the 3 integrals in an order: x-first: it runs from to (1 ); then (from to 1); and finall ( to 3) 3 { } d = 3 T dτ = 1 # 1 x dx & $% '( d x d 1 ( 1 ) d = 1 9 ( ) # 1 & % ( = 3 $ 1 ' 8

12 Fundamental theorem of calculus Suppose f(x) is a function of one variable. The fundamental theorem of calculus states: a b df dx dx = f ( b) f ( a) Geometrical interpretation: a b F ( x)dx = f ( b) f ( a) Two was to determine the total change of a function: subtract the values at the ends or go step-b-step, adding up all the tin increments as ou go along. f(x) f(b) f(a) df dx = F x ( ) a dx b x

13 Fundamental theorem of gradients A scalar function of three variables T(x,, ). Starting at point a, we move a small distance dl 1. According to the gradient dt = T dl = T dl cosθ b the function T will change b an amount dt = ( T ) dl 1 a x dl Move b an additional displacement dl ; the incremental change in T will be ( T ) dl The total change in T in going from a to b along the selected path is b a P ( T ) dl = T ( b) T ( a)

14 Problem: Check the fundamental theorem for gradients, using T = x + 4x + 3, the points a = (,, ), b = (1, 1, 1), and the path in the figurs below. x

15 (a) (,, ) " (1,, ) " (1, 1, ) " (1, 1, 1) T(b) = = 7; T(a) = " T(b) T(a) = 7 x T = ( x + 4) ˆx + ( 4x + 3 )ŷ + ( 6 )ẑ T dl = ( x + 4)dx + ( 4x + 3 )d + ( 6 )d Segment 1: x: "1, = = d = d = Segment : : "1, x = 1, =, dx = d = Segment 3: : "1, x = = 1, dx = d = 1 T dl = ( x ) dx = x 1 =1 1 T dl = ( 4 ) 1 d = 4 = 4 1 T dl = ( 6 )d = 3 1 = b T dl = 7 a

16 Fundamental theorem of divergences ( v)dτ = v da 3 names for this special theorem: Gauss s theorem V S Green s theorem The divergence theorem The integral of a derivative (divergence) over a region (volume) is equal to the value of the function at the boundar (surface).

17 Problem: Test the divergence theorem for the function v = ( x) ˆx + ( )ŷ + ( 3x)ẑ Take as our volume the cube shown below with sides of length. (v) (ii) x (iv) (i) (vi) (iii) ( v)dτ = v da V S

18 v = + + 3x ( v)dτ = ( + + 3x)dx dd = { ( + + 3x)dx} dd " { } = ( + ) x + 3 x # $ % & ' = ( + ) + 6 { } d = ( )d { } = " + ( 4 + 6) # $ % = ( ) = ( v)dτ = ( 8 +16)d = ( ) = = 48

19 Now for the surfaces: ( i) da = ddˆx, x =, v da = dd, v da = dd = = 8 ( ii) da = ddˆx, x =, v da =, v da = ( iii) da = dxdŷ, =, v da = 4dxd, v da = 4 dx d =16 ( iv) da = dxdŷ, =, v da =, v da = ( v) da = dxdẑ, =, v da = 6xdxd, v da = 4 ( vi) da = dxdẑ, =, v da =, v da = v da = = 48

20 Fundamental theorem of curls Stoke s theorem ( v) da = v dl S P The integral of a derivative (curl) over a region (surface) is equal to the value of the function at the boundar (perimeter).

21 Example: Test Stoke s theorem for the function v = ( x) ˆx + ( )ŷ + ( 3x)ẑ using the triangular shaded area below. x ( v) da = v dl S P

22 Example: Test Stoke s theorem for the function v = ( x) ˆx + ( )ŷ + ( 3x)ẑ using the triangular shaded area below. x v = ˆx ( ) + ŷ( 3) + ẑ( x) = ˆx 3ŷ xẑ da = ddˆx

23 ( v) da = dd { } d ( v) da = ( )d ( v) da = ( )d & = ) ( + = 8 ' 3* 3 { } = ( ) = & ) = ( ' 3 * v dl = ( x)dx + ( )d + ( 3x)d ( i) x = = ; dx = d =. :, v dl = (3) () ( ii) x = ; = ; dx =, d = d, :, v dl = d (1) v dl = ( )d = ( 4 )d = $ ' $ & % 3 3 ) = & 8 ( % 3.8 ' ) = 8 ( 3 ( iii) x = = ; dx = d = ; :, v dl =. v dl = v dl = 8 3