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Integral Vector

Theorems

29.3Introduction

Various theorems exist equating integrals involving vectors. Often, use of these theorems canmake certain vector integrals easier. This section introduces the theorems known as GaussTheorem, Stokes Theorem and Greens Theorem.

5

4

2

3Prerequisites

Before starting this Section you should . . .

be able to find the gradient of a scalar fieldand the divergence and curl of a vector field

be familiar with the integration of vectorfunctions

Learning Outcomes

After completing this Section you should beable to . . .

be able to use vector integral theorems tofacilitate vector integration.

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1. Stokes Theorem

This is a theorem that equates a line integral to a surface integral. For any vector field F anda contour C which bounds an area S,

S

( F) dS = C

F drNotes

1. dS is a vector perpendicular to the surface and dr is the line element along the contour C.

2. Both sides of the equation are scalars

3. The theorem is often a useful way of calculating a line integral along a contour composedof several distinct parts (e.g. a square or other figure).

4. F is a vector field representing the curl of the vector field F and may, alternatively,be written as curl F.

Justification

Imagine that the surface is divided into a set of infinitesmally small rectangles ABCD wherethe axes are adjusted so that AB and CD lie parallel to the new xaxis i.e. AB = x and BCand AD lie parallel to the new yaxis i.e. BC = y.Now, CFdr is calculated. The contributions along AB, BC, CD and DA are F(x,y, 0) x =Fx(x,y,z)x, F(x+ x,y, 0) y = Fy(x+ x,y,z)y, F(x, y + y, 0) (x) = Fx(x, y + y,z)xand F(x,y, 0) (x) = Fy(x, y, z )y. Thus,

C

F dr (Fx(x, y, z ) Fx(x, y + y,z))x + (Fy(x + x,y,z) Fy(x,y,z))y

Fyx

xy +Fxy

xy

( F)zS ( F) dS

as S is perpendicular to the x and y axes.

Thus, for each small rectangle,

C

F dr ( F) dS

When the contributions over all the rectangles are summed, the line integrals for the inner partsof the rectangles cancel and all that remains is the line integral around the outside of the shape.The surface integrals sum. Hence, the theorem applies for the area S bounded by the contourC.While the above does not comprise a formal proof of Stokes Theorem, it gives an appreciation

of where the theorem comes from.

HELM (VERSION 1: March 18, 2004): Workbook Level 129.3: Integral Vector Theorems

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Key Point

S

(

F)

dS = C

F

dr

The closed contour integral of the scalar product of a vector function with the vector along thecontour is equal to the integral of the scalar product of the curl of that vector function and theunit normal, over the corresponding surface.

Example Verify Stokes Theorem for the vector function F = y2i (x + z)j + yzk andthe unit square 0 x 1, 0 y 1 for z = 0.

Solution

If F = y2i (x + z)j + yzk then F = (z+ 1)i + (1 2y)k = i + (1 2y)k (as z = 0).Note that dS = dx dyk so that ( F) dS = (1 2y) dy dxThus,

S

( F) dS =

1

x=0

1

y=0

(1 2y) dy dx

=

1

x=0

(y y2)1

y=0dx =

1

x=0

(2) dx

= [2x]10

= 2 + 0 = 2

To evaluateC

F dr, consider it separately on the four sides.When y = 0, F = xj and dr = dxi so F dr = 0 so the contribution to the integral is zero.When x = 1, F = y2i j and dr = dyj so F dr = dy so the contribution to the integral is1

y=0(dy) = [y]1

0= 1.

When y = 1, F = i xj and dr = dxi so F dr = dx so the contribution to the integral is1x=0(dx) = [x]

1

0 = 1.When x = 0, F = y2i and dr = dyj so F dr = 0 so the contribution to the integral is zero.The integral

C

F dr is the sum of the contributions i.e. 0 1 1 + 0 = 2.Thus

S

( F) dS =C

F dr = 2

Example Using cylindrical polar coordinates (effectively plane-polar coordinates as thisexample just considers the plane z = 0), verify Stokes theorem for the function

F = 2 and the circle = a.

3 HELM (VERSION 1: March 18, 2004): Workbook Level 129.3: Integral Vector Theorems

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Solution

Firstly, findC

F dr. This can be done by integrating along the contour = a from = 0to = 2. Here F = a2 (as = a) and dr = a d (remembering the scale factor) so

F dr = a3

d and hence C

F dr =

2

0

a3 d = 2a3

As F = 2, F = 3z and ( F) dS = 3 as dS = z.Thus

S

( F) dS =

2

=0

1

=0

3 d d =

2

=0

a=0

32 d d

=

2

=0

3a=0 d =

2

0 a

3

d = 2a

3

Hence C

F dr =

S

( F) dS = 2a3

Example Find the closed line integralC

F dr for the vector field F = y2i + (x2 z)j +2xyk and for the contour ABCDEFGHA in Figure 6.


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Solution

To find the line integral directly would require eight line integrals i.e. along AB, BC, CD,DE, EF, F G, GH and HA. It is easier to carry out a surface integral to find

S

( F) dSwhich is equal to the required line integral

C

F

dr by Stokes theorem.

As F = y2i + (x2 z)j + 2xyk , F =

i j k

x

y

z

y2 x2 z 2xy

= (2x + 1)i 2yj + (2x 2y)k

As the contour lies in the x y plane, the unit normal is k and dS = dx dy kHence ( F) dS = (2x 2y) dx dy.To work out

S

( F) dS, it is necessary to divide the area inside the contour into twosmaller areas i.e. the rectangle ABCDGHand the trapezium DEFG. On the ABCDGH, the

integral is 4

y=0

6

x=0

(2x 2y) dx dy =

4

y=0

x2 2xy6

x=0dy =

4

y=0

(36 12y) dy

=

36y 6y240

= 36 4 6 16 0 = 48

On DEFG, the integral is

7

y=4

y2x=1

(2x 2y) dx dy =

7

y=4

x2 2xyy2

x=1dy =

7

y=4

(y2 + 2y + 3) dy

=1

3y3 + y2 + 3y

74

= 3433

+ 49 + 21 + 643

16 12 = 51

So the full integral,

S( F) dS = 48 51 = 3. By Stokes Theorem,

S

( F) dS =C

F dr = 3

From Stokes theorem, it can be seen that surface integrals of the form

S( F) dS depend

only on the contour bounding the surface and not on the internal part of the surface.


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1. Verify Stokes Theorem for the vector field F = x2i + 2xyj + zk and thetriangle with vertices at (0, 0, 0), (3, 0, 0) and (3, 1, 0).

(a) Find the normal vector dS

(b) Find the vector F

(c) Evaluate the double integral

3

x=0

x/3y=0

( F dS

(d) Find the integral

F dr along the 3 sides of the triangle

(e) Verify that the two sides of the equation in the theorem are equal.

2. Using plane-polar coordinates (or cylindrical polar coordinates with z =0), verify Stokes Theorem for the vector field F = + cos

2 and

the semi-circle 1, 2 2 .3. Verify Stokes theorem for the vector field F = 2xi + (y2 z)j + xzk and

the contour around the rectangle with vertices at (0, 2, 0),(2, 2, 0),(2, 0, 1) and (0, 0, 1).

4. Verify Stokes Theorem for the vector field F = yi +xj +zk and for thecontour starting from the origin and going to (1, 0, 0), (0, 0, 0), (1, 1, 0)and (1, 1, 1) before returning to the origin.

(a) Find the surface integral over the triangle (0,0,0), (1,0,0), (1,1,0)

(b) Find the surface integral over the triangle (1,0,0), (1,1,0), (1,1,1)

(c) Find the line integrals along the four parts of the contour

(d) Show that the two sides of the equation of the theorem are equal

5. Use Stokes theorem to evaluate the integralC

F dr where F =sin( 1

x+ 1) + 5y

i + (2x ey2)j and C is the contour starting at (0, 0)

and going to (5, 0), (5, 2), (6, 2), (6, 5), (3, 5), (3, 2), (0, 2) and returningto (0, 0).

Your solution

1.)

(a)dxdyk,(b)2yk,(c)1,(d)9,3,11,(e)Bothsidesare1.


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Your solution

2.)

Bothsidesare0.

Your solution

3.)

Bothsidesare2.

Your solution

4.)

(a)1(b)0(asFisperpendiculartodS),(c)0,1,1,-1(d)Bothsidesare1.

Your solution

5.)

57.


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Your solution

6.)

Bothsidesarei+2j.

2. Gauss TheoremThis is sometimes also known as the divergence theorem and is similar to Stokes theorem butequates a surface integral to a volume integral. Gauss Theorem states that for a volume V,bounded by a closed surface S, any vector field F satisfies

S

F dS =

V

F dV

Notes

1. dS is a unit normal pointing outwards.

2. Both sides of the equation are scalars

3. The theorem is often a useful way of calculating a surface integral over a surface composedof several distinct parts (e.g. a cube).

4. F is a scalar field representing the divergence of the vector field F and may, alternatively,be written as div F.

5. Gauss theorem can be justified in a manner similar to that used for Stokes theorem.

Key Point S

F dS =

V

F dV

The closed surface integral of the scalar product of a vector function with the unit normal isequal to the integral of the divergence of that vector function over the corresponding volume.


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Example Verify Gauss Theorem for the unit cube 0 x 1, 0 y 1, 0 z 1 andthe function F = xi + zj

Solution

To find

SF dS, the integral must be evaluated for all six faces of the cube and the results

summed.On the left face, x = 0, F = zj and dS = idydzso FdS = 0 and

SFdS = 1

0

1

00dydz = 0

On the right face, x = 1, F = i + zj and dS = idydz so F dS = 1dydz and

S

F

dS = 10 10 1dydz = 1On the front face, y = 0, F = xi + zj and dS = jdxdz so F dS = zdxdz and

SF dS = 1

0

1

0zdxdz = 1

2

On the back face, y = 1, F = xi + zj and dS = jdxdz so F dS = zdxdz and S

F dS = 10

1

0zdxdz = 1

2

On the bottom face, z = 0, F = xi and dS = kdydz so F dS = 0dxdy and S

F dS = 10

1

00dxdy = 0

On the top face, z = 1, F = xi + j and dS = kdydz so F dS = 0dxdy and S

F dS = 10

1

00dxdy = 0

Thus, summing over all six faces,

SF dS = 0 + 1 1

2 +1

2 + 0 + 0 = 1

To find

V F dV note that F =

xx +

yz = 1 + 0 = 1.

So

V F dV = 1

0

1

0

1

01 dx dy dz = 1.

So

SF dS =

V F dV = 1

Note that in the above example, the volume integral was accomplished by one triple integralwhile the surface integral required six double integrals. This is often the motivation for using

Gauss Theorem i.e. to carry out one integral rather than several.

Example Use Gauss theorem to evaluate the surface integral

SF dS where F is the

vector field x2yi + 2xyj + z3k and S is the surface of the unit cube 0 x 1,0 y 1, 0 z 1.


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Solution

Note that to carry out the surface integral directly will involve the evaluation of six doubleintegrals (one for each face of the cube). However, by Gauss theorem, the same result comesfrom the surface integral

V F dV. As

F = 2xy + 2x + 3z2, the surface integral

becomes the triple integral

1

0

1

0

1

0

(2xy + 2x + 3z2) dx dy dz

=

1

0

1

0

x2y + x2 + 3xz2

1

x=0dy dz =

1

0

1

0

(y + 1 + 3z2) dy dz

=

1

0

1

2y2 + y + 3yz2

1

y=0

dz =

1

0

(1

2+ 1 + 3z2) dz =

1

0

(3

2+ 3z2) dz

=3

2z+ z

31

0 =

11

6

The six double integrals also sum to 116

but this approach requires a greater amount of work.

Example Verify Gauss Theorem for the vector field F = y2j xzk and the triangularprism with vertices at (0, 0, 0), (2, 0, 0), (0, 0, 1), (0, 4, 0), (2, 4, 0) and (0, 4, 1)(see figure 7 ).


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Solution

As F = y2j xzk , F = 0 + 2y x = 2y x.Thus

V

F dV =2x=0

4y=0

1x/2z=0

(2y x) dz dy dx

=

2

x=0

4

y=0

[2yz xz]1x/2z=0 dy dx =

2

x=0

4

y=0

(2y xy x + 12

x2) dy dx

=

2

x=0

y2 1

2xy2 xy + 1

2x2y

4

y=0

dx =

2

x=0

(16 12x + 2x2) dx

=

16x 6x2 + 2

3x320

=40

3

To work out

SF dS, it is necessary to consider the contributions from the five facesseparately.On the front face, y = 0, F = xzk and dS = j thus F dS = 0 and the contribution to theintegral is zero.On the back face, y = 4, F = 16j xzk and dS = j thus F dS = 16 and the contribution tothe integral is

2

x=0

1x/2z=0

16 dz dx =2

x=0[16z]1x/2z=0 dx =

2

x=016(1x/2) dx = [16x 4x2]2

0=

32 16 = 16On the left face, x = 0, F = y2j and dS = i thus F dS = 0 and the contribution to theintegral is zero.On the bottom face, z = 0, F = y2j and dS =

k thus F

dS = 0 and the contribution to the

integral is zero.On the top right face, z = 1 x/2, F = y2j + ( 1

2x2 x)k and the unit normal n = 1

5i + 2

5k

Thus dS =

15

i + 25

k

dy dw where dw measures the distance along the slope for a constant

y. As dw =5

2dx, dS =

1

2i + k

dy dx thus F dS = 16 and the contribution to the integral

is2

x=0

4

y=0(12

x2 x) dy dx = 2x=0

(2x2 4x) dx = 23

x3 2x220

= 83

.

Adding together the contributions,

SF dS = 0 + 16 + 0 + 0 8

3= 40

3

Thus

SF dS =

V F dV = 40

3

Gauss Theorem also applies using orthogonal curvilinear coordinates.


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1. Verify Gauss Theorem for the vector field F = xi yj + zk and the unitcube 0 x 1, 0 y 1, 0 z 1.

(a) Find the vector

F

(b) Evaluate the triple integral

1

z=0

1

y=0

1

x=0

F dxdydz

(c) For each side, evaluate the normal vector dSand the surface integral S

F dS

(d) Show that the two sides of the equation of the theorem are equal

2. Verify Gauss Theorem for the vector field F = 4xzi y2j + yzk and thecuboid 0 x 2, 0 y 3, 0 z 4.

3. Verify Gauss Theorem, using cylindrical polar coordinates, for the vectorfield F = 2 over the cylinder 0 r0, 1 z 1 for

(a) r0 = 1

(b) r0 = 2

4. For S being the surface of the tetrahedron with vertices at(0, 0, 0),(1, 0, 0),(0, 1, 0) and (0, 0, 1), find the surface integral

S(xi + yzj) dS

(a) directly

(b) by using Gauss Theorem

Hint :- When evaluating directly, show that the unit normal on the slop-ing face is 1

3(x + y + z) but that dS = x + y + z

Your solution

1.)

(a)1(b)1(c)-dxdyk,0;dxdyk,1;-dxdyj,0;dxdzj,-1;-dydzi,0;dydzi,1(d)Bothsidesare1.


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Your solution

2.)

Bothsidesare156.

Your solution

3.)

Bothsidesequal4,2.


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Your solution

4.)

Bothsidesequal5

24.

3. Greens TheoremLike Gauss Theorem, Greens Theorem equates a surface integral to a volume integral. However,Greens Theorem is concerned with two scalar fields u(x, y, z ) and w(x,y,z). Two statementsof Greens Theorem are as follows

S

(u w) ds =

V

u w + u 2w dVand

S

[u

w

v

u]

ds =

Vu

2w

w

2u dV

Proof of Greens Theorem

Greens Theorem can be derived from Gauss Theorem and a vector derivative identity.

Vector identity (1) from subsection 6 of 37.2 states that (A) = () A + ( A)

Letting = u and A = w,

(uw) = (u) (w) + u( (w))= (u) (w) + u2w


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Gauss Theorem states S

F dS =

V

F dV

Now, letting F = uw, S

(uw) dS =

V

(uw) dV

=

V

(u) (w) + u2w dV

This is the first statement of Greens Theorem. Reversing the roles of u and w,

S(wu) dS =

V

(w) (u) + w

2

u

dV

Subtracting the last two equations yields the second statement of Greens Theorem.

Key Point

1. S

(u w) ds =

V

u w + u 2w dV2.

S

[u w v u] ds =

V

u2w w2u dV

Example Verify Greens Theorem (first statement) for u = (x x2)y, w = xy + z2 and

the unit cube, 0 x 1, 0 y 1, 0 z 1.


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Solution

As w = xy + z2, w = yi + xj + 2zk . Thus uw = (xy x2y)(yi + xj + 2zk) and the surfaceintegral is of this quantity (scalar product with dS integrated over the surface of the unit cube.On the three faces x = 0, x = 1, y = 0, the vector u

w = 0 and so the contribution to the

surface integral is zero.On the face y = 1, uw = (x x2)(i + xj + 2zk) and dS = j so (uw) dS = x2 x3 and the

contribution to the integral is1

x=0

1

z=0(x2 x3) dz dx = 1

0(x2 x3) dx =

x3

3 x

4

4

10

= 112

.

On the face z = 0, uw = (x x2)y(yi + xj) and dS = k so (uw) dS = 0 and thecontribution to the integral is zero.On the face z = 1, uw = (x x2)y(yi + xj + 2k) and dS = k so (uw) dS = 2y(x x2) and the contribution to the integral is

1

x=0

1

y=02y(x x2) dy dx = 1

0(x2 x3) dx =

1x=0

[y2(x

x2)]1

y=0

dx = 10 (x x2) dx = x

2

2 x3

31

0

= 16

.

Thus,

S(u w) ds = 0 + 0 + 0 + 1

12+ 0 + 1

6= 1

4.

Now evaluate

V[u w + u 2w] dV.

Note that u = (1 2x)yi + (x x2)j and 2w = 2 so

u w + u 2w = (1 2x)y2 + (x x2)x + 2(x x2)y= x2 x3 + 2xy 2x2y + y2 2xy2

and the integral

V

u w + u 2w dV = 1

z=0

1y=0

1x=0

(x2 x

3

+ 2xy 2x2

y + y2 2xy

2

) dx dy dz

=

1

z=0

1

y=0

x3

3 x

4

4+ x2y 2

3x3y + xy2 x2y2

1x=0

dy dz

=

1

z=0

1

y=0

(1

12+

y

3) dy dz=

1

z=0

y

12+

y2

6

1y=0

dz

=

1

z=0

(1

4) dz =

z4

1

z=0=

1

4

Hence S

(u w) ds =

V

u w + u 2w dV = 14

Greens Theorem in the Plane

This states that C

(P dx + Q dy) =

S

Q

x P

y

dx dy

Justification


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Greens Theorem in the plane can be derived from Stokes Theorem.

S

( F) dS =C

F dr

Now let F be the vector field P(x, y)i + Q(x, y)j i.e. there is no dependence on z and there areno components in the z direction. Now

F =

i j k

x

y

z

P(x, y) Q(x, y) 0

=

Q

x P

y

k

and dS = dx dy k giving (

F)

dS = Qx

P

y dx dy.

Thus Stokes Theorem becomes

S

Q

x P

y

dx dy =

C

F dr

and Greens Theorem in the plane Follows

Key PointC

(P dx + Q dy) =

S

Q

x P

y

dx dy

Example Evaluate the line integralC

[(4x2 + y 3)dx + (3x2 + 4y2 2)dy] around therectangle 0

x

3, 0

y

1.


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Solution

The integral could be accomplished by four line integrals but it is easier to note that[(4x2 + y 3)dx + (3x2 + 4y2 2)dy] is of the form P dx + Q dy with P = 4x2 + y 3 andQ = 3x2 + 4y2

2. It is thus of a suitable form for Greens Theorem in the plane.

Note that Qx

= 6x and Py

= 1.

Greens Theorem in the plane becomes

C

((4x2 + y 3) dx + (3x2 + 4y2 2) dy) =

1

y=0

3

x=0

(6x 1) dx dy

=

1

y=0

3x2 x3

x=0dy =

1

y=0

24 dy = 24

The same result could be gained by evaluating four line integrals.

Example Verify Greens Theorem in the plane for the integralC

[4z dy + (y2 2)] andthe contour starting at the origin O = (0, 0, 0) and going to A = (0, 2, 0) andB = (0, 0, 1) before returning to the origin.


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Solution

The whole of the contour is in the plane x = 0 and Greens Theorem in the plane becomes

C(P dy + Q dz) =

SQ

y P

z

dy dz

Firstly evaluateC

[4z dy + (y2 2)].On OA, z = 0 and dz = 0. As the integrand is zero, the integral will also be zero.On AB, z = (1 y

2) and dz = 0. The integral is

0

y=2

(4 2y)dy 1

2(y2 2)dy) = 0

2(5 2y 1

2y2)dy =

5y y2 1

6y30

2= 14

3

On BO , y = 0 and dy = 0. The integral is0

1(2)dz = [2z]0

1= 2.

Summing,C

[4z dy + (y2 2)] = 83

In this example, P = 4z and Q = y2 2. Thus Pz

= 4 andQ

y= 2y. Hence,

S

Q

y P

z

dy dz =

2

y=0

1y/2

z=0

(2y 4) dy dz

=

2

y=0

[2yz 4z]1y/2z=0 dz =

2

y=0

y2 + 4y 4 dz=

1

3y3 + 2y2 4y

2

0

= 83

HenceC

(P dy + Q dz) =

S

Qy

P

z

dy dz= 8

3

One very useful, special case of Greens Theorem in the plane is when Q = x and P = y. Thetheorem becomes

C

(y dx + x dy) =

S

(1 (1)) dx dy

The right hand side becomes S2 dx dy i.e. 2A where A is the area inside the contour C.Hence

A =1

2

C

(x dy y dx)

This result is known as the area theorem.

Example Verify the area theorem A = 12 C

(x dy y dx) for the segment of the circle

x2

+ y2

= 4 lying above the line y = 1.


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Solution

Firstly, the area of the segment ADBC can be found by subtracting the area of the triangleOADB from the area of the sector OACB. The triangle has area 1

2 2

3 1 =

3. The

sector has area 3

22 = 43

. Thus segment ADBC has area 43

3.

Now, evaluate the integralC

(x dy y dx) around the segment. Along the line, y = 1, dy = 0so the integral

C

(x dy y dx) becomes 33(x 0 1 dx) = 3

3( dx) = 2

3.

Along the arc of the circle, y =

4 x2 = (4 x2)1/2 so dy = x(4 x2)1/2dx. The integralC

(x dy y dx) becomes33

[x2(4 x2)1/2 (4 x2)1/2]dx =

33 4

4 x2dx

=

/3/3

41

2cos 2cos d

=

/3/3

4d =8

3

So, 12

C

(x dy y dx) = 12

8

3 2

3

= 43

3 Hence both sides of the theorem equal4

3

3

1. Verify the first statement of Greens Theorem for the functions u = xyz,w = y2 and the unit cube 0 x 1, 0 y 1, 0 z 1.

2. Use Greens Theorem in the plane to evaluate the integralC

[(5x + 2y 7)dx + (3x 4y + 5)dy] where Crepresents the perimeterof the trapezium with vertices at (0, 0), (3, 0), (6, 1) and (1, 1).

(a) Find the function (uw)

(b) Evaluate the integral S

(u

w)

dS for the six faces of the cube

(c) Find the function u w + u2w

(d) Evaluate the integral

[u w + u2w] dV

(e) Show that the twow sides in the equation of the theorem are equal.

3. Verify the area theorem for

(a) The area above y = 0 but below y = 1 x2.(b) The segment of the circle x2 + y2 = 1, to the upper left of the line

y = 1 x.


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Your solution

1.)

(a)2xy2

zj(b)0,0,0,1

2,0,0(c)4xyz(d)Bothsidesare

1

2.

Your solution

2.)

4.

Your solution

3.)

4

3,

4

1

2.

21 HELM (VERSION 1 M h 18 2004) W kb k L l 1

29 3 Int Vec Thms

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