Vector Integral Calculus - AAIT CIVIL ENG LECTUR …



Unit III

Vector Integral Calculus

3.1 Line integral of Scalar fields

Instead of integrating a function over an interval [a, b], we will integrate over a curve C in space. Such an integral is called a line integral.

|[pic] |Let [pic] be any partition of C obtained by choosing points [pic], |

| |[pic],…,[pic] of subdivisions along C. For each integer k between l and n|

| |let ([pic],[pic],[pic]) be any point on the curve between [pic] and [pic]|

| |and let [pic]be the length of the portion of C between [pic] and[pic]. |

Now consider the sum

[pic]

Let ║[pic]║ = maximum of the lengths {[pic], [pic], . . ., [pic]} called the norm of the partition [pic].

If f is continuous at every point on C, then

[pic]

exists.

| |

|Definition 3.1 Let f be continuous on a piecewise smooth curve C with finite |

|length. Then the line integral [pic]of f over C is defined by |

|[pic][pic] |

If f is the mass density of a wire, then the mass m of the wire is:

[pic] (1)

Let us assume that C is parameterized by a smooth vector valued function

[pic]

on an interval [a, b]. Then

[pic], since ds =[pic]. (2)

Example 1 Evaluate [pic] where C is the circle [pic]= 4.

Solution We parameterize C by

[pic]= 2 cos t [pic].

Then since x (t) = 2 cos t, y (t) = 2 sin t, z (t) = 0, [pic]

and [pic]=[pic] .

Hence, [pic].

Therefore, [pic]= 0.

Example 2 Let C be the twisted cubic curve parameterized by

[pic]

Evaluate [pic].

Solution Now x (t) = t, y (t) = [pic], z (t) = [pic], [pic]and [pic].

Thus, [pic]

= [pic] = [pic].

Therefore, [pic]= [pic].

The line segment C from (a, 0, 0) to (b, 0, 0) on the x-axis is parameterized by

[pic].

If f is continuous on C, then

[pic][pic]

By identifying f with[pic]where

[pic](t) = f (t, 0, 0) for a ( t ( b

We get:

[pic] (3)

If r parameterizes a curve C on [a, b], then

[pic] (the length of the curve C). (4)

If f is continuous on a piecewise smooth curve C composed of a finite number of smooth curves [pic], [pic], . . ., [pic], then

[pic] (5)

Example 3 Let C be composed of the line segments from (0, 0, 0) to (0, 1, 1) and from (0, 1, 1) to

(1, 1, 2). Evaluate[pic].

Solution Let C be composed of the curves [pic] and [pic] parameterized by

[pic] and [pic].

Consequently;

[pic] and [pic].

Thus, [pic][pic]

Therefore, [pic] = [pic].

3.2 Line Integrals of Vector Fields

| |

|Definition 3.2 Let F be a continuous vector field defined on a smooth oriented curve C. |

|Then the line integral of F over C, denoted and defined by |

|[pic]= [pic] |

|where T(x, y, z) is the unit tangent vector at (x, y, z) for the given |

|orientation of C. |

If F represents the force on an object moving along a curve C that is oriented in the direction of motion, then the work ( done by the force on the object as it traverses C is given by

[pic] (6)

Let[pic]be a parameterization of C with domain [a, b], and assume that the parameterization induces the given orientation on C. Then by definition

T (x (t), y (t), z (t)) = [pic]

Hence, [pic]

Therefore,[pic] (7)

Example 4 A particle moves upward along the circular helix [pic]parameterized by:

[pic]

under a force given by

[pic]

Find the work ω done on the particle by the force.

Solution Now x (t) = cost, y (t) = sin t, z (t) = t and [pic].

Thus, [pic].

[pic][pic][pic].

Therefore, ω [pic].

Note that: The orientation of the curve – C is opposite to the orientation of the curve C. Since the tangent

vector at any point on – C is the negative of the tangent vector at the same point on c. Thus,

[pic]. (8)

3.3 Alternative Form of the Line Integral of Vector Fields

Let [pic] be a continuous vector field defined on a smooth oriented curve C parameterized by:

[pic]

Then[pic]

[pic]

It is common to write the final integral above in the abbreviated form:

[pic] (9)

Or even more briefly as

[pic]

(9) is just another notation for [pic] and is usually evaluated by means of the formula.

[pic]

(10)

Example 5 Evaluate[pic], where C is parameterized by

[pic].

Solution [pic].

Then [pic].

Thus, [pic].

Therefore, [pic].

Note that: [pic]

[pic]

and[pic]. [pic]

Example 6 Let C be the unit circle in the xz-plane, oriented by the parameterization

[pic]

Then evaluate[pic].

Solution [pic].

Then [pic].

[pic][pic].

[pic]

Therefore, [pic].

Remark:

1. If an oriented curve C is piecewise smooth, composed of smooth curves [pic], [pic], . . ., [pic], then

[pic][pic]

which is the same as

[pic]

2. If C is a plane curve with smooth parameterization

[pic]

is continuous at every point on C, then

[pic]

and [pic]. (11)

Example 7 Evaluate [pic], where C is the portion of the circle [pic]from (2, 0)

to (0, 2) oriented in a counterclockwise direction.

Solution The curve C is parameterized by

[pic]

Consequently;

[pic]

[pic]= [pic] (verify!)

Therefore, [pic].

Example 8 Evaluate[pic] where C is the quarter circle from (1, 0) to (0, 1).

Solution [pic].

Thus [pic]

[pic] (verify!)

Therefore, [pic].

3.4 The fundamental Theorem of line Integrals

This section is devoted to the generalization of the fundamental theorem of calculus that is applicable

to integrate integral of the form [pic] for certain vector field F on a curve C.

| |

|Theorem 3.1 (Fundamental theorem of line integrals) |

|Let C be an oriented curve with initial point [pic]and |

|terminal point [pic]. Let f be a function of three variables |

|that is differentiable at every point on C, and assume that grad f |

|is continuous on C. Then |

|[pic] = [pic]–[pic]. |

Note that: To evaluate[pic], where F is a continuous vector field on C such that F = grad f for some

differentiable function f we need to find a potential function f for F.

Example 9 Let C be the curve from (1, ( 1, ( 1) to (1, 1, 1) parameterized by

[pic]

and [pic].

[pic] Evaluate [pic].

Solution Note that: In unit II we found out that, F = grad f, where [pic].

Thus, by the fundamental theorem of line integrals

[pic]= f (1, 1, 1) ( f (1, ( 1, ( 1) = 2.

Therefore, [pic]= 2.

The fundamental theorem of line integrals can be formulated in two dimensions (a Plane) as follows. Let f be a differentiable function of two variables on an oriented curve C with initial point ([pic], [pic]) and terminal point ([pic], [pic]) and let gradient of f be continuous on C. Then

[pic] (12)

More over; if F is a continuous vector field such that F = grad f, then

[pic]

Example 10 Let C be any piecewise smooth curve in the xy-plane from (( 1, 2) to (2, 3) and let

[pic]

Then evaluate[pic]

Solution Now [pic]. Since [pic], F is the

gradient of some differentiable function f.

Hence, [pic] (*)

Integrating both sides of the second equation in (*) with respect to y we get:

[pic] (**)

Taking the partial derivative of both sides of (**) with respect to x and comparing with the first

equation in (*) we get:

[pic]

Therefore, [pic].

Consequently; [pic].

Therefore,[pic]= 35.

3.5 Independence of path

| |

|Definition 3.3 Let F be a continuous vector field with domain D. [pic] |

|is said to be independent of path if for any oriented curve [pic]in |

|D having the same end points as curve C. |

|[pic] |

Note that: The following are equivalent.

1. F = grad f for some differentiable function f of three variables. i. e., F is conservative.

2. [pic] is independent of path.

3. [pic]= 0 for every oriented closed curve C in the domain of F.

If F = grad f for some differentiable function f, then

4. curl F = 0.

More over; if the domain of F is [pic]or any region with no "holes", then condition 4 implies condition 1, so that condition 1-4 are equivalent.

Note that: curl F = 0 need not imply F is conservative.

Example 11 Let [pic].

Then [pic].

Hence, [pic]

Thus, curl F = 0.

To show that F is not conservative we need to show that[pic]( 0 for some closed oriented curve C.

Let C be the unit circle [pic]with parameterization

[pic]

Then [pic].

Therefore, F is not conservative.

3.6 Green’s Theorem

| |

|Theorem 3.2 Green’s theorem |

| |

|Let R be a simple region in the xy-plane with a piecewise smooth |

|boundary C oriented counterclockwise. Let M and N be functions |

|of two variables having continuous partial derivatives on R. Then |

|[pic] |

Remark: By using Green’s theorem we can some times evaluate a line integral

[pic] without using a parameterization of C.

Example 12 Let M (x, y) = [pic] and N(x, y) =[pic], and let C be the circle

[pic]oriented counterclockwise.

Evaluate[pic]

Solution Now consider R to be the disk [pic].

Since, [pic], we have [pic]

Therefore,[pic].

Example 13 Let C be the closed curve shown in the figure below.

Evaluate [pic].

|Solution Now M = [pic]and N = [pic]. |[pic] |

|Then [pic] | |

|Thus, [pic] | |

|Since R is vertically simple | |

[pic].

Therefore, [pic]

Example 14 Evaluate the line integral[pic], where C is the closed curve shown below.

|Solution Let C be the boundary of the simple region R in |[pic] |

|the xy plane and C be composed the curves [pic]and [pic] | |

|parameterized by | |

|[pic]: [pic] for 0 ≤ t ≤ 2 | |

|[pic]: [pic] for 0 ≤ t ≤ 2 | |

Now M = [pic]and N = 4xy. Then [pic]

Thus, [pic]

= [pic]= [pic]= [pic]= [pic].

Therefore, [pic]= [pic].

Green’s theorem can be used, in the reverse way, to evaluate a double integral by evaluating a line integral.

Suppose we wish to evaluate [pic] where R is a simple region with piecewise smooth boundary and if M and N are chosen so that

[pic] (*)

Then by Green's theorem

[pic]

Where C is the boundary of R oriented counterclockwise.

We can choose M and N satisfying (*) in one of the following three ways.

i) M (x, y) = 0 and N (x, y) = x

ii) M(x, y) = ( y and N(x, y) = 0

iii) M(x, y) = ( [pic]y and N(x, y) = [pic]x.

Hence we get the following formulas for the area A of R.

[pic] (13)

Example 15 Compute the area A of the region R enclosed by the ellipse.

[pic]

Solution The ellipse is oriented counter clockwise by the parameterization

[pic]= a cost [pic]

Then by (13)

[pic], where x (t) = a cos t and y (t) = b sin t.

Thus, [pic].

Therefore, [pic] square units.

Green’s theorem can be applied for a non-simple region R that can be broken up into collections of simple regions.

Example 16 Let R be the semi annular region where y ≥ 0 with[pic]= 1 and [pic]= 2, in the xy plane.

Evaluate[pic].

Solution Let M(x, y) = ℓn ([pic]) and N(x, y) = ℓn ([pic])

Then[pic]

Thus by Green’s theorem we get:

[pic] =[pic]

= [pic]

= [pic]( 4.

Therefore,[pic]= ( 4.

Note that: We have three ways of evaluating line integrals of the form [pic]

i) Parameterize C and then use the formula

[pic]

ii) Use the fundamental theorem of line integrals (Provided that, F is a conservative vector field).

iii) Use Green’s theorem (provided that C is closed) and F is a vector field of two variables.

3.7 Alternative Forms of Green’s Theorem

There are two other forms of Green’s Theorem.

I. Let[pic]be a continuous vector field defined on a simple region R in the

xy plane, and assume that the range of F is contained in the xy plane. Further more let the boundary

C of R have its counterclockwise orientation. Then

[pic]

In addition,[pic], [pic]

Therefore, the formula in Green’s Theorem can be translated into

[pic] (14)

II. To obtain the second alternative form of Green’s Theorem, we assume that the boundary C of a

simple region R is oriented counterclockwise by a smooth parameterization

[pic]

Then the tangent T of C is given by

[pic]

and let [pic] [pic] [pic]

Then [pic]lies in the xy plane and [pic][pic]T, so [pic]is parallel to the normal vector [pic]

If[pic]is a continuous vector field, then

[pic]

[pic]

[pic]

[pic]

Then using Green’s theorem and the fact that div F(x, y) = [pic]

[pic]

Therefore, [pic].

Example 17 Let[pic]and let R be the region bounded by the unit circle[pic].

Evaluate [pic] where C is the boundary of R.

Solution Now M(x, y) = 2x and N (x, y) = 3y. Then[pic].

We can parameterize C by [pic] with counterclockwise orientation

[pic].

Thus, [pic].

Therefore, [pic].

Or using the above result we get:

[pic]= [pic].

3.8 Surface Integrals

Let ∑ be the graph of a function having continuous partial derivatives and defined on a region R in the xy plane that is composed of a finite number of vertically or horizontally simple regions. Let g be continuous on ∑. Let R' be a rectangle that circumscribes R and let [pic] be a partition of R' into sub rectangles. Let [pic], [pic], . . . , [pic] be those rectangles in [pic] that are entirely contained in R.

Let [pic], [pic], . . . ,[pic] be the projection of [pic], [pic], . . . , [pic] onto ∑ and let [pic], [pic], . . . , [pic] be their respective surface areas. For each integer k between 1 and n, let ([pic],[pic], [pic]) be an arbitrary point in [pic].

Consider

[pic] (*)

As the maximum [pic] of the dimensions of sub rectangles tends to 0, the sum in (*) tends to a limit!

| |

|Definition 3.4 Let ∑ be the graph of a function having continuous partial derivatives |

|and defined on a region R in the xy plane that is composed of a finite number of |

|vertically or horizontally simple regions. Let g be a function that is continuous |

|on ∑. The surface integral of g over ∑, denoted [pic]is defined by |

|[pic] |

Remarks

1. If ∑ represents a metal plate of negligible thickness and if g(x, y, z) represents the mass density of the path at the point (x, y, z) on ∑, then the mass m of the plate is given by:

[pic]

2. If ∑ is the graph of f on R (i.e. ∑ is the graph of z = f (x, y), Then

[pic] (*)

3. If g (x, y, z) =1 for all (x, y, z) in ∑, then the double integral in (*) represents the surface area of ∑.

[pic]

Example 18 Evaluate [pic], where ∑ is the portion of the cone [pic] between the planes

z = 1 and z = 3.

Solution Let R be the ring[pic], and let f (x, y) =[pic] [pic]R.

Then ∑ is the graph of f on R.

Thus, [pic]

= [pic].

Therefore, [pic]=[pic].

Remark: Some surfaces can not be expressed as graphs of function of x and y but can be expressed as graphs of functions of x and z or of y and z. In such cases we can still apply (*), with the role of x, y and z interchanged.

Example 19 Evaluate [pic], where ∑ is the portion of the plane x + y = 1 in the first

octant for which 0 ≤ z ≤ 1.

Solution Note that there is no function f of x and y whose graph is ∑.

If R is the square in the xz plane consisting of all (x, z) for which 0 ≤ x ≤ 1 and

0 ≤ x ≤ 1 and if f (x, z) = 1 − x, then ∑ is the graph of y = f (x, y) on R.

Thus, [pic] and hence [pic]

= [pic].

Therefore, [pic] = [pic]. [pic][pic]

Remark: If [pic] is the graph of a function f that does not have partial derivatives at every point in R

and if the double integral in (*) is interpreted as an improper integral, then (*) remains valid.

Example 20 Evaluate [pic], where ∑ is the hemisphere [pic].

Solution If f (x, y) =[pic], then ∑ is the graph of f and the domain R of f is the disk bounded

by the circle [pic]. We find that:

[pic] = [pic] for [pic].

These formulas do not hold on the circle [pic], but for 0 < b < 1 they hold on the

disk [pic]bounded by the circle [pic].

Thus, [pic] = [pic]dA

= [pic] = [pic]

= [pic].

Therefore, [pic]= 3(.

Remark: If a surface [pic] is composed of several sub surfaces [pic], [pic], …. , [pic]and if g is continuous

on[pic], then

[pic]

Example 21 Evaluate[pic], where[pic] is the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0)

and (0, 0, 1).

Solution Since [pic], where

[pic] is the graph of f (x, y) = 0[pic](x, y) ([pic], where [pic]is the region in the xy plane bounded

by x + y =1, x = 0 and y = 0.

[pic] is the graph of f (y, z) = 0[pic](y, z) ([pic], where[pic]is the region in the yz plane bounded

by y + z =1, y = 0 and z = 0.

[pic] is the graph of f (x, z) = 0[pic](x, z) ([pic], where [pic]is the region in the xz- plane bounded

by x + z =1, x = 0 and z = 0

and [pic] is the graph of f (x, y) =1– x – y [pic](x, y) ([pic], where [pic]is the region in the xy-plane

bounded by x + y =1, x = 0 and y = 0.

Thus, [pic]

[pic][pic]

[pic]

[pic][pic].

Therefore,[pic].

3.9 Integrals over Oriented Surfaces

In this section we will introduce the notion of orientation of a surface and then define the surface integral of a vector field over an oriented surface.

Let [pic] be a surface and suppose [pic] has a tangent plane at each of its non boundary points. At such a point on [pic] two (unit) normal vectors, that are opposite in direction, exist. If it is possible to select one normal at each non-boundary point in such a way that the chosen normal varies continuously on[pic] , then [pic] is said to be an orientable surface, and the selection of the normal gives an orientation to[pic] . If [pic] is bounded by a curve C, then the orientation of [pic] induces an orientation on C. Such orientation is called induced orientation. If your right-hand thumb points in the direction of the normal, then the remaining fingers naturally curl in a way that defines an orientation on C.

Example 22 (Orientable surfaces)

Spheres, Paraboloids, cylinders, Cones etc are examples of orientable surfaces.

Furthermore; orientable surfaces have two sides.

Example 23 (Non-Orienteable surfaces).

The Mobius band. The mobius band has only one side.

Note that: If [pic] is the boundary of a solid region D in space, then we customarily choose the normal

to[pic] that is directed outward from D.

Flux Integrals

Suppose [pic] is an oriented surface and [pic] is a unit normal. Let F be a continuous vector field on[pic], then the integral

[pic]

is called a flux integral or normal surface integral.

Remarks 1. For the flux integral [pic], [pic]must be oriented.

2. If F represents the velocity ( of a fluid with density (, then the rate of mass flow of the

fluid through the surface [pic] is given by

[pic]

3. If [pic] is the graph of a function f, z = f (x, y), with continuous partial derivatives on a

region R in the xy plane that is composed of either vertically or horizontally simple

regions, then

[pic].

Note that: If [pic] is the graph of z = f (x, y) on R, then the normal to[pic] is given by:

[pic]

and hence the unit normal [pic], depending on the orientation of[pic], is given by:

[pic], if[pic] is directed upward. i.e. its [pic]component is positive.

[pic], if[pic] is directed downward. i.e. its [pic]component is negative.

Now let [pic]be continuous on ∑, then

i) if [pic]is directed upward, then

[pic]

ii) ) if [pic]is directed downward, then

[pic]

Example 24 Evaluate [pic], where [pic]and ∑ is the part of the cone

[pic]above the square with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0) and (0, 1, 0)

and [pic]is directed upward.

Solution Now f (x, y) = [pic]for all (x, y) ( R, where R is the region in the xy-plane

bounded by x = 0, x = 1, y = 0 and y = 1.

Then [pic] and [pic].

But these partial derivatives do not exist at (0, 0) and we need to consider as an improper integral.

|[pic] |[pic] |

| |[pic] |

| |[pic] |

[pic]

Now since, [pic]

and[pic].

We get:

[pic] [pic], [pic],

[pic][pic]

and [pic].

Hence,[pic].

Therefore,[pic].

Example 25 Evaluate [pic], where[pic]and ∑ is the portion

of the plane 2x + y + 2z = 6 in the first octant. [pic] is directed upward.

Solution ∑ is the graph of f (x, y) = [pic] for all (x, y) ( R, where R is the region in the xy plane

bounded by 2x + y = 6, x = 0 and y = 0.

Now [pic].

[pic]

[pic]=[pic]= [pic]

= [pic]

Therefore,[pic].

Remark

If a surface [pic] composed of several oriented surfaces[pic], [pic], ….,[pic], provided that the surfaces induce opposite orientations on the common curves that bind them together, then

[pic]

Example 26 Evaluate[pic], where [pic],[pic] is the sphere

[pic]and [pic]is directed outward from the sphere.

Solution We divide[pic] into the upper hemisphere[pic] and the lower hemisphere[pic].

Now let[pic] is the graph of [pic]on the region R in the xy plane bounded by the

circle [pic].

Thus, [pic] and [pic] ( (x, y) ( [pic] where [pic]

is the region in the xy plane bounded by the circle [pic] where 0 ≤ h ( 2.

Then [pic]= [pic]

[pic]

[pic].

Similarly, [pic] is the graph of [pic]on the region R in the xy plane bounded by

the circle [pic].

Thus, [pic] and [pic] ( (x, y) ( [pic] where [pic]is the

region in the xy plane bounded by the circle [pic], where 0 ≤ h ( 2

Then[pic] = [pic]

= – [pic] = –[pic]

Therefore, [pic]= 0.

3.10 Stokes’s Theorem

Stokes’s Theorem is a three-dimensional version of Green’s theorem, involving three – dimensional surfaces and their boundaries rather- than plane regions and their boundaries.

| |

|Theorem 3.3 (Stokes’s Theorem) |

|Let [pic] be an oriented surface with normal [pic] and finite surface area. Assume |

|that [pic] is bounded by a closed piecewise smooth curve C whose orientation |

|is induced by[pic]. Let F be a continuous vector field defined on [pic] and assume |

|that the component functions of F have continuous partial derivatives at each |

|non boundary point of [pic] . Then |

|[pic] |

If [pic], then[pic].

Example 26 Let C be the oriented triangle shown in the figure, below which lies in the plane [pic].

If [pic], then calculate[pic].

Solution Now [pic] is the graph of f (x, y) = [pic]y on the region in the xy- plane bounded by x = 0, y = 0

and x + y = 2. From the orientation on C, [pic]. Then

|[pic] |[pic] |

|= [pic] | |

|= [pic] [pic] | |

|Therefore, [pic]= 14. | |

Remark

Suppose two oriented surfaces [pic]and [pic] are bounded by the same curve C and induce the same orientation on C. If [pic]and [pic]denote the normals of [pic]and [pic] respectively, then by stokes’s

theorem we infer that

[pic]

Example 27 Let C be the intersection of the paraboloid [pic]and the plane z = y and given C

its counterclockwise orientation as viewed from the positive z axis.

Evaluate [pic]

Solution Let [pic]and let [pic] be the portion of the plane z = y that lies

inside the paraboloid.

Note that: If (x, y, z) is on C, then [pic] and hence thus is an equation of the circular cylinder

r = sin[pic], in cylindrical co-ordinates.

Now if R is the region in the xy plane bounded by the circle r = sin[pic], then [pic] is the graph of

Z = y on R.. If we orient [pic] by the normal directed upward, then the induced orientation on C

is counterclockwise.

Thus, [pic] (show!) and f (x, y) = y on R.

Hence, [pic]

[pic]

[pic].

Therefore,[pic].

Example 28 Let [pic] be the semi-ellipsoid[pic], oriented so that the normal [pic]is directed

up ward, and let [pic].

Evaluate[pic].

Solution Let [pic] be the unit disk centered at (0, 0) in the xy plane, and let [pic] be the unit normal

to [pic]. Then the induced orientations on the common boundary of [pic] and [pic] are identical.

Thus, [pic].

Since, [pic] and[pic].

[pic]

Therefore, [pic].

Remark: Suppose two oriented surfaces [pic] and [pic] are bounded by the same curve C but induce

opposite orientations on C. Then Stokes’ theorem implies that:

[pic] = –[pic].

3.11 The Divergence Theorem

| |

|Definition 3.5 A solid region D is called a simple solid region if D is the |

|solid region between the graphs of two functions [pic]and [pic]on |

|a simple region R in the xy plane and if D has the corresponding |

|properties with respect to the xz plane and the yz – plane. |

Example 29 Region bounded by Spheres, Cylinders, hemispheres, ellipsoids, Cubes and tetra-hedrons

are some examples of simple solid regions.

| |

|Theorem 3.4 (The Divergence Theorem , Gauss’s Theorem) |

|Let D be a simple solid region whose boundary surface [pic] is oriented by |

|the normal n directed outward from D, and let F be a vector field whose |

|component functions have continuous partial derivatives on D. Then |

|[pic] |

Example 30 Evaluate [pic], where[pic]and ∑ is the boundary of the

solid region in side the cylinder [pic] and between the planes z = 0 and z = 2.

Solution [pic].

Now we need to evaluate[pic], where D is the solid region bounded by the

cylinder [pic] and the planes z = 0 and z = 2.

Thus,[pic]

[pic]

[pic]

= [pic]

= [pic].

= – [pic].

Therefore, [pic]= 16 (.

Example 31 Evaluate [pic], where[pic]and ∑ is the boundary of

the solid region bounded by the planes z = 1 and x + y =1 and the coordinate planes.

Solution [pic]= 2xy + 3z.

Thus, by the divergence theorem

[pic]= [pic]

= [pic]= [pic]

= [pic]= – [pic] = [pic].

Therefore, [pic]= [pic].

-----------------------

Figure 3.1

x

y

z

x

[pic]

[pic]

[pic]

y

x4 + y4 = 1

x

y

x

y = 2x

y = x2

y

(0, 1)

(1, 1)

(1, 0)

[pic]

[pic]

y

z

x

C11

Z = [pic]y

C3

x + y = 2

C21

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