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