Chapter 10 Vector Integral Calculus. Integral Theorems

**1 Line Inegrals**

The concept of a line integral is a simple and natural generalization of a definite integral.

known from calculus. In a line integral we shall integrate a given function, also called the integrand, along a curve C in space. Hence curve integral would be a better name, but line integral is standard

We represent the curve C by a parametric representation **i**

The curve C is called the **path of integration**, A: its initial point, and B: its terminal point. C is now oriented. For figure b, the curve C is called a closed path.

C is called a **smooth curve** if it has at each point a unique tangent whose direction varies continuosly as we move along C.

in (2) is differentiable and the derivative is continous and different from the zero vector at every point of C.'

**1.1 Definition and Evaluation of Line Integrals**

A line integral of a vector function **F(r)** over a curve C: [as in (2)] is defined by

(See also chapter 9 for dot produnct). In terms of components, with and , formula 3 becomes

If the path of integration of C in (3) is a closed curve, then instead of , we also write

Note: The integrand in (3) is a scalar because we take dot product.

**Example 1** Find the value of the line integral (3) when and C is the circular arc as shown in the following figure form A to B.

**Solution** We may represent C by where. Then , and,

By differentiation,, , so that by (3), we have,

**Exercise 1 **Find the value of (3) when and C is the helix.

(The evaluation of the line integrals in space is practically the same as it is in the plane.)

Simple general properties of the line integral (3) follow directly form corresponding properties of the definition integral in calculus, namely,

**1.2 Motivation of the Line Integral (3): Work Done by the Force**

The work W done by a constance force **F** in the displacement along a straight segment d is . This suggests that we define the work W done by a variable force **F** in the displacement along a curve C: **r(t)** as the limit of sums of tworks done in displacements along small chords of C.

**Example 2** In example 1, if F is a force, the work done by F in the displacement along the quarter-circle will be 0.4521, measured in newton-meters.

**Exercise 2 - (PS 10.1 q1) **Calculate . If **F** is a force, this gives the work done in the displacement along C.

(Hint: As in example 1, we should represent **F** in parametric form first.)

, C the parabola from A(0,0) to B (2, 20).

**1.3 Other Forms of Line Integals**

The line integrals

are special cases of (3) when respectively.

Furthermore, without taking a dot product as in (3), we can obtain a line integral whose value is a vector rather than a scalar, namely,

If we take , . Then

with C as in (2).

**Example 3 **Integrate along the helix in Exercise 2.

**Solution ** inegrated with respect to *t* form *0* to* 2π *gives

**Exercise 3 -PS10.1 q4 **Integrate

**1.4 Path Dependence**

**Theorem 2 - Path Dependence **The line integral (3) generally depends not only on F and on the endpoints A and B of the path, but also on the path itself along which the integral is taken.

**Example 4 - Proof of theorem 2** Almost any example will show this. Take, for instance, the straight segment

and the parabola with and integrate

Then

so that the integration gives and respectively.

**2 Path Independence of Line Integrals**

In this section we consider line integrals

as before, and we shall now find conditions under which (8) is path independent in a domain D in space.

By definition this means that for ever pair of endpoints A, B in D, the integral (8) has the same value for all paths in D that begin at A and end at B.

We shall follow up three ideas that will give path independence of (8) in a domain D if and only if:

- (Theorem 1)
- (Theorem 2) Integration around closed curves C in D always gives 0.
- (Theorem 3) curl
**F** = **0** (privided D is simply connected)

**2.1 Theorem 1**

A line integral (8) with continuous in a domain D in space is path independent in D if and only if is the gradient of some function *f* in D.

**Proof** (a) We assume that (9) holds for some function f in D and show that this implies path independence.

Let C be any path in D from any point A to any point B in D, given by ,where . Then from (9), the chain rule and (3) we have

The last formula in part (a) of the proof,

is the analog of the usual formula for definite integrals in calculus,

Formula (10) should be applied when a line integral is independent of the path.

**Example 6** Show that the integral is path independent in any domain in space and

find its value in the integration from A: to B: .

**Solution **= grad , where because

Hence the integral is independent of path according to theorem 1 and (10) gives

If you want to check this, use the most convenient path C:

, on which

so that , and integration from 0 to 2 gives .

**Exercise 6** Evaluate the integral from A: to B: by showing that has a potential (that is to find ) and applying (10).

(Hint: let , and )

**2.2 Theorem 2 - Path Independence and Integration Around Closed Curves**

The integral (10) is path independent in a domain D if and only if its value around every closed path in D is zero.

The idea is that two paths with common end-points make up a single closed curve.

**Proof**

- If we have path independence, then integration from A to B along in the figure above gives the same value.

- Now and together make up a closed curve C and if we integratefrom A along to B as before, but then in the opposite sense along back to A, the sum of the two integrals is zero, but this is the integral around the crosed curve C.
- Conversely, assume that the integral around any closed path C in D is zero.

- Given any points A and B and any two curves and from A to B in D, we see that with the orientation reversed and together form a closed path C.

- By assumpltion, the integral over C is zero.

- Hence the integrals overand , both taken from A to B, must be equal.

**2.2.1 Work Conservative and Noconservative Physical System **

- From the last section, we know that the integral (8) gives the work done by a forcein the displacement of a body along curve C.

- Then Theorem 2 states tahat work is path independent in D if and only if its value is zero for diplacement around every closed path in D.

- Furthermore, theorem 1 tells us that this happens if and only if is the fradient of a potential in D.

- In this case, and the vector field defined by are called
**conservative** in D because in this case mechanical energy is conserbved;

- that is, now work is done in the displacement from a point A and back to A.

**2.3 Theorem 3* - Path Independence and Exactness of Differential Forms**

Theorem 1 relates path independence of the line integral (1) to the gradient and Theorem 2 to integration around closed curves. A third idea relates path independence to the exactness of the **differential form**

under the integtral sign in (8). This form (11) is called exact in a domain D in space if it is differential

of a differentiable function f(x,y,z) evertwhere in D, that is, if we have

Comparing these two formulae, we see that the form (4) is exact if and only if there is differentiable funcion f(x,y,z) in D such that everywhere in D,

Hence theorem 1 implies theorem 3.

The integral (8) is path independent in a domian D in space if and only if the differential from (11) has continous coefficient functions , , and is exact in D.

A domain D is called **simply connected** if every clsoed curve in D can be continuously strucnk to any point in D without leaving D.

**2.4 Theorem 3 - Criterion for Exactness and Path Independence**

Let , , in the line integral (8),

be continuos and have continuos first partial derivatives in a domain D in space. Then:

- If the differential form (11) is exact in D - and thus (8) is path independent by theorem 3* - , then in D,

in components (6') , refer to chapter 9

- If (12) holds in D and D is
**simply connected**, then (11) is exact in D - and thus (8) is path independent by theorem 3*.

**Proof**

- If (11) is exact in D, then in D by theorem 3*, and, futhermore, curl by theorem of Grad, Div, Curl in chapter 9.

Linde Integral in the Plane.

For the curl has only one component (z-component), so that (6') can be reduced to the single rotation

**Example 6 - Exactness and Independence if Path. Determination of a Potential **Using (6'), show that the differential form under the integral sign of

is exact, so taht we have independence of path in any domain, and find the value of I from A: to B: .

**Solution** Eactness follows from (6'), which gives

To find we integrate (which is ``long'' ,so that we save work) and then diferentiate to compare with and .

implies and we can take , so that in the first line. This gives, by (10)

The assumption in Theorem 3 that D is simply connected is essential and cannot be omitted.

**Exercise 7 PS10.2 q5** Show that the form under the integral sign is exact in the space.

**3 Double Integrals**

In a definite integral (1), we integrate a function f(x) iver an interval (a segment) of the x-axis.

- In a double integral we integrate a function f(x,y), called the integrand, over a closed bounded region R in the x-y plane, whose boundary curve has a unique tangent at each point, but may have finitely many cusps.

- This double integral can be considered as a limit of the following sum whereis the area of theth rectangle as shown in the figure below.

The Double Integral of f(x,y) over the region R is denoted by

**3.1 Properties of the double integrals**

Double integrals have properties quite similar to those of definite integrals.

Furthermore if R is simply connected, then there exists at least one point in R such that we have

where A is the area of R. This is called the **mean value theorem** for double integrals.

**3.2 Evaluation of Double Integrals by Two Successive Integrations**

Double integrals over a region R may be evaluated by two successive integrations. We may integrate first over y and then over x. Then the formula is

- Here and represent the boundary curve of ZR and,

- keeping x constant, we integrate over from to The result will be a funciton of x.

- Then we integrate the result from to .

Similarly, for integrating first over x and then over y the formula is

- The boundary cuve of R is now represented by and

- Treating y as constant, we first integrate over x from to and then

- we integrate the resulting function of y from to .

**3.3 Application of Double Integrals**

Double Integrals have various phyiscal and geometric application. For instance, the area **A** of a reguin R in the xy-plane is given by the double integral.

The volume **V** beneath the surface z=f(x,y) (>0) and above a region R in xy-plane is

because the term in at the beginning of this section represents the volume of a rectangular box with base of areaand altitude

As another application, let f(x,y) be the density (mass per unit area) of a distribution of mass in the xy-plane. Then the **total mass M** in R is

the **center of gravity** of the mass in R has the coordinates , , where

the **moments of inertia** andof the mass in R about the x- and y- axes, respectively, are

**3.4 Change of Variables in Double Integrals. Jacobian**

- Here we assume that is continous and

- has a continuous derivative in some interval such that

- , , or [ , ] and varies between a and b when u varies between and .

The formula for a change of variables in double integrals form to is

that is, the integrand is expressed in terms of u and v, and dx dy is replaced by du dv times the absolute value of the **Jacobian**.(where represents tohe corresponding region of R on the u-v plane.

**Example 7 -Change if Variables in Double Integral **Evaluate the following doyble integral over the squre R in the following figure.

**Solution** The shape of R suggests the transformation , Then , , the Jacobia is

R corresponds to the square, . Therefore,

If polar coordinates andare to be used, we can set , .Then

and

**Exercise 8** Let be the mass density in the region as shown in the figure. Find the total mass, the center of gravity, and the moment of inertia , .

**4 Green's Theorem in the plane**

Double iintegrals over a plane region may be transformed into line integrals over the boundary of the region and conversely. This is of practical interest because it may simplify the evaluation of an integral.

**4.1 Theorem 4 - Green's Theorem in the Plane (Transformation between Double Integrals and Line Integrals**

Let R be a closed bounded region in the xy-plane whose boundary C consists of finitely many smooth curves. Let and be functions taht are continuous and have continuous partial derivativesand everywhere in some domain containing R. Then

Here we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration.

Region R whose boundary C consists of two parts: is traversed counterclockwise, while is traversed clockwise in such a way that R is on the left for both curves.

Setting and using the curl theorem in chapter 9, we have

**Example 9 -Verification of Green's Theorem in the plane** Green's theorem in the plane will be quite important in our further work. Before proving it, let us get used to it by verifying it for and C the circle

**Solution** In (22) on the left we get

since the circular disk R area is.

Now, we have to show that the line integral in (22) on the right gives the same value. We must orient C counterclockwise, say,

Then ,and on C

Hence the line integral in (22) becomes, verifying Green's theorem,

**4.2 Some Applications of Green's theorem**

**Example 10 - Area of a Plane Region as a Line Integral Over the Boundary** In (22), we first chooseand then .This gives

respectively. The double integral is the area A of R. By addition we have

where we integrate as indicated in Green's theorem. This interesting formula expresses the area of R in therms of line integral over the boundary. It is used, for instance, in the theory of certain planimeters.

For an ellipse or we get , . Thus from (25) we obtain the familiar formula for the area of region bounded by an ellipse.

**Example 11 - Area if Plane Region in Polar Coordinates** Let and be polar coordinates defined by ,. Then

and (25) becomes a formula that is well known from calculus, namely,

As an application of (26), we consider the cardioid ,where We find

**5 Surface for surface Integral **

**5.1 Representation of surfaces**

Represebtatuibs if a syrface S in xyz-space are

For example, represents a hemisphere of radius a and center 0.

Now, for curve C in line integrals, it was more practical and gave greater flexibility yo use a parametric representation , where This is a mapping of the interval ,located on the t-axis, onto the curve C. It maps every t in that interval onto the point of C with position vector

Similarly, for surfaces S in surface integrals, it will often be more practical to use a parametric representation. Surfaces are two dimensional. Hence we need two parameters, which we call u and v. Thus a of a surface S in space is of the form

**Example 12 -Parametric Representation of a Cylinder** The circle cylinder ,, has radius a, height 2 and the z-axis as axis. A parametric representation is

- The components of and

- The parametrics u,v vary in the rectangle R: in the uv-plane.

- The curves u=const are vertical straight lines.

- The curves v=const are parallel circles. The point P in the following figure corresponds to

**Example 13 - Parametric Representation of a Sphere** A sphere can be represented in the form

where the parameters u, v vasry in the rectangle R in the uv-plane and give by the inequalities, , . The components of are

The curves u=const and v - const are the ``meridians'' and parallel on S. This representation is used in geography for measuring the latitude and longitudinal of points on the globe.

Another parametric representation of the sphere also used in mathematic is

where

**Example 14- Parametric Representation of a Cone** A circular cone, can be represented by

in components. THe parameters vary in the rectangle R: . Check that , as it should be.

**5.2 Tangent Plane and Surface Normal**

- Recall from chapter 9 that the tangent vectors of all the curves on a surface S through a point P oof S from a plane, called the
**tangent plane** of S at P.

- Excepteions are points where S has an edge or a cusp (like a cone),so that S cannot have a tangent plane at such a point.

- Furthermore, a vector perpendicular to the tangent plane is called a
**normal vector** of S at P.

- Now, since S can be given by in the parametric representation, the new idea is that we get a curve C on S by taking a pair of differentiable functions

- whose derivatives u;, v; are continuous. Then C has the position vector . By differentation and the use of the chain rule, we ontain a tangent vector of C on S

- Hence, the partial derivatives at P are tangential to S at P. We assume that they are linearly independent, which geometrically means that the curves u=const and v=const on S intersect at P at a nonzero angle. Then span the tangent plane of S at P.

- Hence their cross product gives a
**normal vector N** of S at P

- The corresponding unit
**normal vector n** of S at P is

Also, if S is represented by g(x,y,z)=0, then the unit normal vector will be

A surface S is called a **smooth surface** if its surface normal depends continuously on the ponet of S.

**Example 15 - Unit Normal Vector of a Sphere **From (30) we find that the sphere has the unit normal vector

**6 Surface Integral**

- For a given vector function
** F** we can now define the **surface integral** over S by

- Here and is the area of the parallelogram with sides

- By definition of cross product. Hence

- And we see that dA= is the element of area of S.

- Also is the normal component of .This integral arises naturally in flow problem where it gives the flux(mass of fluid acrossing S per unit time) across S.

- We can express (30) in components, using and

- Here, are the angles betweenand the coordinate are indeed, for the angel between and , as and so on. Therefore,

- In (33) we can writeThen (33) becomes the following integral for the flux:

**Example 16 - Surface Integral** Evaluate (30) when and S is the portion of the plane x+y+z=1 in the first octant as shown.

**Solution **Writing x=u and y=v, we have z=1-x-y=1-u-v.

Hence we can represent theplane x+y+z=1 in the form . We obtain the first-octant portion S of this plane by restricting x=u and y=v to the projection R of S in the xy-plane.

R is the triangle bounded by the two coordinate axes and the straight line x+y=1, obtained from x+y+z=1 by setting z=0.

Thus,

By differentiation

Hence. By (30),

**Example 16 -Surface Integral of a cylinder (PS10.6 q4) **Given , and S:

Find the surface integral.

**Solution** This is a circular cylinder of radius 3 and height 2 in the first octant with the z-axis as axis.

A parametric representation is

From this we have

Integration over u from *0* to gives . Integration of this over v from *0* to *2* gives the answer:

**Exercise 17 - Surface Integral of a sphere (PS10.6 q 9)** Given ,

Find the surface integral.

**7 Triple intergals.**

** Divergence Theorem of gauss**

**7.1 Triple intergal**

A triple integral is an integral of a function taken over a closed bounded region *T* in space.It is denoted by

Triple integerals can be evaluated by three successive integrations.

**7.2 Divergence Theorem of gauss**

** (Transformation between triple and surface integrals)**

Let *T* be a closed bounded region in space whose boundary is a piecewise smooth orientable surface *S*. Let be a vector function that is continuous and has continuous first partial derivatives in some domain containing *T*. Then

In components of and of the outer unit normal vector of *S*, (35) becomes

**Example 18 - Evalution of a surface integral by the Divergence Theorem** Evaluate

where *S* is the closed surface in (36) consisting of the cylinder and the circular disks and

**Solution.**, , Hence div . The from of the surface suggests that we introduce polar coordinates defined by , (thus cylindrical coordinates ). Then the volume element is , and we obtain

**Example 19 - Verification of the Divergence Theorem** Evaluate over the sphere (a) by (35), (b) directly.

**Solution.**

- div

the answer is .

2. Wirte *S* in parameter form of and using ,

Then

Now on *S* we have , , so that become on *S*

and

On *S* we have to integrate over *u* form *0* to . This gives

Then integrating over *v* form to , we get

**7.3 Invariance of the Divergence**

The divergence of a vector function with continuous first partial derivatives in a region is independent of the particular choice of Cartesian coordinates. For any point in it is given by

**8 Stokes's Theorem**

**(Transformation between surface an line integrals)**

Let *S* be a piecewise smooth oriented surface in space and let the boundary of *S* be a piecewise smooth simple close curve *C*. Let be a countions vector function that has continuous first partial derivatices in a domain in space containing *S*. Then

Here is a unit normal vector of *S* and, depending on , the integration around *C* is taken in the sense shown in (38). Furthermore, is the unit tangent vector and *s* the arc length of *C*.

In components, (38) becomes

Here, and *R* is the region with boundary curve in the -plane corresponding to *S* represented by .

**Example 20 - Verifying Stokes's Theorem for and ***S* the paraboloid

**Solution.** The curve *C*, oriented as in (), is the circle . Its unit tangent vector is .The function on *C* is .Hence

We now consider the surface integral. We have , so that in (38) we obtain

A normal vector of *S* is .Hece Now Using polar coordinates defined by and denoting the projection of *S* into *xy*-plane by *R*, we thus obtain

Note. Green's Theorem in the plane as a special case of Stokes's Theorem

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