Topic 3 - Edge · A scalar function of three variables T ... gradient the function T will change by...
Transcript of Topic 3 - Edge · A scalar function of three variables T ... gradient the function T will change by...
Topic 3 Integral calculus
• 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
(1) Line integral:
A line integral is an expression of the form
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:
At each point on the path we take the dot product of v with the displacement dl to the next point.
Line, surface and volume integrals
v ⋅dlaP
b∫
v ⋅dl∫
a
b
dl
yx
z
Problem: Calculate the line integral of the function
from the origin to the point (1, 1, 1) by two different routes:
(a) (0, 0, 0) " (1, 0, 0) " (1, 1, 0) " (1, 1, 1)
(b) The direct straight line
v = x2x+ 2yzy+ y2z
Problem: Calculate the line integral of the function
from the origin to the point (1, 1, 1) by two different routes:
(a) (0, 0, 0) " (1, 0, 0) " (1, 1, 0) " (1, 1, 1)
v = x2x+ 2yzy+ y2z
0, 0, 0( )→ 1, 0, 0( ). x : 0→1, y = z = 0
dl = dxx, v ⋅dl = x2dx⇒ v ⋅dl = x2 dx0
1∫∫ =
x3
3 0
1
=13
1, 0, 0( )→ 1,1, 0( ). x =1, y : 0→1, z = 0
dl = dyy, v ⋅dl = 2yzdy = 0 ⇒ v ⋅dl∫ = 0
1,1, 0( )→ 1,1,1( ). x = y =1, z : 0→1
dl = dzz, v ⋅dl = y2dz = dz⇒ v ⋅dl = dz0
1∫∫ = z
0
1=1
v ⋅dl∫ =13#
$%&
'(+ 0+1=
43
The direct straight line
x = y = z : 0→1;dx = dy = dz
v ⋅dl = x2dx + 2yzdy+ y2dz = x2dx + 2x2dx + x2dx = 4x2dx
v ⋅dl∫ = 4x2 dx0
1∫ =
4x3
3#
$%
&
'(0
1
=43
(2) Surface integral:
A surface integral is an expression of the form
where v is a vector function, da is the infinitesimal area, with direction perpendicular to the surface.
If the surface is closed such as a balloon, put a circle in the integral sign:
The outward is positive but for an open area it is arbitrary.
v ⋅daS∫
v ⋅da∫
yx
z da
Example: Calculate the surface integral of the function
over the 5 sides (excluding the bottom) of the cubical box (side 2).
v = 2xzx+ x + 2( ) y + y z2 −3( ) z
y
z
x
(i)(iv)(iii)
(ii)
(v)
2
2
2
Example: Calculate the surface integral of the function
over the 5 sides (excluding the bottom) of the cubical box (side 2).
v = 2xzx+ x + 2( ) y + y z2 −3( ) z
y
z
i( ) x = 2, da = dydzx, v ⋅da = 2xzdydz = 4zdydz
x
(i)(iv)(iii)
(ii)
(v)
2
2
2
v ⋅da∫ = 4 dy0
2∫ zdz
0
2∫ =16
v ⋅da∫ = 0
ii( ) x = 0, da = −dydzx, v ⋅da = −2xzdydz = 0
iii( ) y = 2, da = dxdzy, v ⋅da = x + 2( )dxdz
v ⋅da∫ = x + 2( )dx0
2∫ dz
0
2∫ =12
iv( ) y = 0, da = −dxdzy, v ⋅da = − x + 2( )dxdz
v ⋅da∫ = − x + 2( )dx0
2∫ dz
0
2∫ = −12
v( ) z = 2, da = dxdyz, v ⋅da = y z2 −3( )dxdy = ydxdy
v ⋅da∫ = dx0
2∫ ydy
0
2∫ = 4
v ⋅dasurface∫ =16+ 0+12−12+ 4 = 20
(3) Volume integral:
A volume integral is an expression of the form
where T is a scalar function, dτ is the infinitesimal volume element.
In Cartesian coordinates
dτ = dx dy dz
If T is the density of a substance, then the volume integral would give the total mass.
May encounter volume integrals of vector functions but because the unit vectors are constants, they come outside the integral.
T dτV∫
vdτ∫ = vxx+ vyy+ vzz( )dτ∫ = x vx dτ∫ + y vy dτ∫ + z vz dτ∫
yx
z3
11
T dτ∫ = z20
3∫ y x dx
0
1−y∫#$%
&'(dy0
1∫{ }dz =
Example: Calculate the volume integral of T = xyz2 over the prism.
Do the 3 integrals in any order:
x-first: it runs from 0 to (1 – y); then y (from 0 to 1); and finally z (0 to 3)
12
z2 dz0
3∫ 1− y( )2 ydy = 1
29( )
0
1∫ 1
12#
$%
&
'(=38
Fundamental theorem of calculus
dfdxdx
a
b∫ = f b( )− f a( )
⇒ F x( )dxa
b∫ = f b( )− f a( )
Suppose f(x) is a function of one variable. The fundamental theorem of calculus states:
Geometrical interpretation:
Two ways to determine the total change of a function: subtract the values at the ends or go step-by-step, adding up all the tiny increments as you go along.
x
f(b)
f(a)
f(x)
a bdx
dfdx
= F x( )
A scalar function of three variables T(x, y, z). Starting at point a, we move a small distance dl1. According to the gradient
the function T will change by an amount
Move by an additional displacement dl2; the incremental change in T will be
The total change in T in going from a to b along the selected path is
Fundamental theorem of gradients
dT = ∇T( ) ⋅dl1
a
b
dl
yx
z
∇T( ) ⋅dlaP
b∫ = T b( )−T a( )
dT =∇T ⋅dl = ∇T dl cosθ
∇T( ) ⋅dl2
yx
z
Problem: Check the fundamental theorem for gradients, using T = x2 + 4xy +2yz3, the points a = (0, 0, 0), b = (1, 1, 1), and the path in the figurs below.
yx
z (a) (0, 0, 0) " (1, 0, 0) " (1, 1, 0) " (1, 1, 1)
T(b) = 1 + 4 + 2 = 7; T(a) = 0
" T(b) – T(a) = 7
∇T = 2x + 4y( ) x+ 4x + 2z3( ) y+ 6yz2( ) z
∇T ⋅dl = 2x + 4y( )dx + 4x + 2z3( )dy+ 6yz2( )dzSegment 1: x: 0 "1, y = z = dy = dz = 0
Segment 2: y: 0 "1, x = 1, z = 0, dx = dz = 0
Segment 3: z: 0 "1, x = y = 1, dx = dy = 0
∇T ⋅dl∫ = 2x( )dx0
1∫ = x2
0
1=1
∇T ⋅dl∫ = 4( )dy0
1∫ = 4y
0
1= 4
∇T ⋅dl∫ = 6z2( )dz0
1∫ = 2z3
0
1= 2
∇T ⋅dla
b∫ = 7
Fundamental theorem of divergences
∇⋅v( )dτV∫ = v ⋅d
S∫ a
3 names for this special theorem:
• Gauss’s theorem
• 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 boundary (surface).
Problem: Test the divergence theorem for the function
Take as your volume the cube shown below with sides of length 2.
v = xy( ) x+ 2yz( ) y + 3zx( ) z
y
z
x(i)
(iii)(ii)
(v)
2
2
2
(iv)
(vi)
∇⋅v( )dτV∫ = v ⋅d
S∫ a
∇⋅v( )dτ∫ = y+ 2z+3x( )dxdydz =∫ y+ 2z+3x( )dx0
2∫{ }dydz∫∫
⇒ { }= y+ 2z( ) x + 32x2
"
#$%
&'0
2
= 2 y+ 2z( )+ 6
∇⋅v( )dτ∫ = 8z+16( )dz0
2∫ = 4z2 +16z( )
0
2=16+32 = 48
∇⋅v = y+ 2z+3x
⇒ { }= y2 + 4z+ 6( ) y"# $%02= 4+ 2 4z+ 6( ) = 8z+16
= 2y+ 4z+ 6( )dy0
2∫{ }dz∫
Now for the surfaces:
i( ) da = dydzx, x = 2, v ⋅da = 2ydydz, v ⋅da∫ = 2ydydz∫∫ = 2y2
0
2= 8
ii( ) da = −dydzx, x = 0, v ⋅da = 0, v ⋅da∫ = 0
iii( ) da = dxdzy, y = 2, v ⋅da = 4zdxdz, v ⋅da∫ = 4zdx dz∫∫ =16
iv( ) da = −dxdzy, y = 0, v ⋅da = 0, v ⋅da∫ = 0
v( ) da = dxdyz, z = 2, v ⋅da = 6xdxdy, v ⋅da∫ = 24
vi( ) da = −dxdyz, z = 0, v ⋅da = 0, v ⋅da∫ = 0
⇒ v ⋅da∫ = 8+16+ 24 = 48
Fundamental theorem of curls
∇× v( ) ⋅daS∫ = v ⋅dl
P∫Stoke’s theorem
The integral of a derivative (curl) over a region (surface) is equal to the value of the function at the boundary (perimeter).
Example: Test Stoke’s theorem for the function
using the triangular shaded area below.
v = xy( ) x+ 2yz( ) y + 3zx( ) z
y
z
x 2
2
∇× v( ) ⋅daS∫ = v ⋅dl
P∫
Example: Test Stoke’s theorem for the function
using the triangular shaded area below.
v = xy( ) x+ 2yz( ) y + 3zx( ) z
y
z
x 2
2
∇× v = x 0− 2y( )+ y 0−3z( )+ z 0− x( ) = −2yx−3zy− xz
da = dydzx
∇× v( ) ⋅da = −2ydydz
∇× v( ) ⋅da∫ = −2y( )dy0
2−z∫{ }dz∫
→{ }= y20
2−z= − 2− z( )2
∇× v( ) ⋅da∫ = − 4− 4z+ z2( )dz0
2∫ = − 4z− 2z2 + z
3
3&
'(
)
*+0
2
= − 8−8+ 83
&
'(
)
*+= −
83 y
z
v ⋅dl = xy( )dx + 2yz( )dy+ 3zx( )dz
y
z
(3) (2)
(1)
i( ) x = z = 0; dx = dz = 0. y : 0→ 2, v ⋅dl∫ = 0
ii( ) x = 0; z = 2− y; dx = 0, dz = −dy, y : 2→ 0, v ⋅dl = 2yzdy
v ⋅dl∫ = 2y 2− y( )dy =2
0∫ − 4y− 2y2( )dy = − 2y2 − 23 y
3$
%&
'
()0
2
= − 8− 23.8
$
%&
'
()= −
830
2∫
iii( ) x = y = 0; dx = dy = 0; z : 2→ 0, v ⋅dl = 0. v ⋅dl∫ = 0⇒ v ⋅dl∫ = −
83