Vector Analysis and Geometry

Vector Analysis and Geometry

E- Content in Mathematics

Vector Integration: Line INtegral of vector valued functions

Objectives

From this unit a learner is expected to achieve the following:

Learn that vector integration is the reverse process of vector differentiation.
Learn the definition of line integral of a vector valued function along a curve and understand that line integral is an integral evaluated along a smooth curve.
Familiarize, by means of examples,the method of finding line integral of vector valued function.
Learns that line integral along a closed smooth curve is the circulation.
Learns that the line integral of a force (vector valued function) represents the work done by the force in moving the particle along the curve.
Learn the definition of vector line integral and understand it through an example

Sections

Introduction

Welcome friends! In this session we first describe integration of vector function f(t) of a scalar variable t. We will see that vector integration is the reverse process of vector differentiation. We recall that line integral (sometimes called a path integral) is the integral of some function along a curve. One can integrate ascalar-valued functionalong a curve, obtaining for example, the mass of a wire from its density. In this session we will see the integration of certain type of vector-valued functions along smooth curves. We will see that the line integral of a force (vector valued function) represents the work done by the force in moving the particle along the curve. We conclude the session by defining vector line integral and illustrating through an example.

Integral of a Vector Function

Let f(t) be a vector valued function in the scalar variable t. If there exists a vector valued function F(t) of the scalar variable t such that

then F(t) is called the indefinite integral of f(t) with respect to t and symbolically we writeas follows:

… (1)

The vector valued function f(t) is called the integrand.

If c is any arbitrary constant vector independent of t, thenwe have the following:

Hence, we have the following:

… (2)

From (1) and (2) it is obvious that the integral F(t) of f(t) is indefinite to the extent of an additive arbitrary constant vector c. Therefore F(t) is called indefinite integral of f(t). The constant vector cis called the constant of integration.

The following result can be obtained easily.

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Definite Integral

If for all t in the interval [a, b], then the definite integral between the limits t = a and t = b can be written asfollows:

Example 1 If F(t) =(t  t 2) i + 2 t 3j  3 k, find

Solution

Example2 The acceleration of a particle at time t is given by a = 12 cos 2ti  8 sin 2tj + 16 tk. If the velocity v and the displacement r be zero at t = 0. Find ( i ) v and (ii) r at time t.

Solution

Given a = 12 cos 2ti  8 sin 2tj + 16 tk.

(i ) On integration,

v = a dt =  (12 cos 2ti  8 sin 2tj + 16 tk) dt

= { 12 cos 2t dt} i + {  (  8 sin 2t ) dt} j + {  16 t dt} k

= 6 sin 2ti + 4 cos 2tj + 8 t2k + c1.

Putting v = 0 , when t = 0, we get

0 = 0 i + 4 j + 0 k + c1,

so that c1 = 4 j; hence v = 6 sin 2ti + 4 (cos 2t 1) j + 8 t2k .

(ii) Since , we have

= 6 sin 2ti + 4 (cos 2t 1) j + 8 t2k .

Integrating, we get

r =  3 cos 2ti + (2 sin 2 t 4t) j + 8/3t3k + c2.

Putting r = 0, when t = 0, we get

0 = 3 i + 0 j + 0 k + c2,

so that c2= 3 i; hence r = 3(1  cos 2t) i + (2 sin 2 t 4t) j + 8/3t3k.

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Line Integrals

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

…(3)

known from calculus. In the expression (3) we integrate the integrandfrom along the x- axis to . In a line integral we shall integrate a given function, called the integrand, along a curve C in space (or in a plane). Hence curve integral would be a better term, but line integral is standard.

Definition Any integral which is to be evaluated along a curve is called a line integral.

We represent the curve C by a parametric representationof the following form:

…(4)

The curve C is called the path of integration. with position vector is its initial point, and with position vector is its terminal point. C is now oriented; the direction from A to B, in which t increases, is called the positive direction on C. We can indicate the direction by an arrow as in Fig. 1. The points A and B may coincide as in Fig. 2, then C is called a closed path.

We call C a smooth curve (regular arc) if C has a unique tangent at each of its points whose direction varies continuously as we move along C. Mathematically this is equivalent to saying that C has a representation (4) such that is differentiable and the derivative is continuous and different from the zero vector at every point of C.

Line Integrals of Vector Valued Function over a Smooth Curve

Let

F (x, y, z) = F1 i + F2 j + F3 k

be a vector valued function of x, y, z defined and continuous along a smooth curve C. Let

r =r(x, y, z)= xi + yj + zk

be the position vector of a point P(x, y, z) on the curve. If s denotes the arc-length of the curve C from a fixed point on it to the point (x, y, z), then is the unit tangent vector at the point P (x, y, z). The component of F along the tangent at P is The integral of along C from the fixed point A to the point B, written asfollows:

This is known as line integral of the vector valued function F along C (or more precisely known as the tangent line integral or scalar integral of F along C) .

Remarks

The line integral from the pointA to B is negative of that from the point B to A. i.e.,
Let F = F1i + F2j + F3k, r = xi + yj + zk, then we have the following:

If the curve C is given by parametric equations x = x(t), y = y(t), z = z(t), and if t = t1 at A and t = t2 at B, we havethe following:

Example 3 If F = (3x2 + 6y)i 14 yzj + 20 xz2k, evaluate , where:

(i) C is a curve from (0, 0, 0) to (1, 1, 1) with parametric form x = t, y = t 2, z = t 3.

(ii)C is the straight line joining (0, 0, 0) and (1, 1, 1).

Solution

Here F = (3x2 + 6y)i 14 yzj + 20 xz2k.

Hence we have the following.

…(5)

(i) C is in the parametric form x = t, y = t2, z = t3. Substituting x = t, y = t2, z = t3 in (5) , we get the following:

At the point (0, 0, 0), t = 0 and at (1, 1, 1), t = 1. Moving from the point (0, 0, 0) to (1, 1, 1) means t varies from t = 0 to t =1. Therefore, we have the following:

(ii) The straight line joining (0, 0, 0) and (1, 1, 1) is given in the parametric form by x = t, y = t , z = t and t varies from 0 to 1. Substituting x = t, y = t , z = t in (1), we obtain the following:

Hence, we have the following:

Example 4 If F = (3x2 + 6y)i 14 yzj + 20 xz2k, evaluate the line integrals over the line segments from (0, 0, 0) to (1, 0, 0), then to (1, 1, 0) and then to (1, 1, 1).

Solution

Here F = (3x2 + 6y)i 14 yzj + 20 xz2k. Hence, we have the following:

…(6)

Along straight line C1 from (0, 0, 0) to (1, 0, 0): x varies from 0 to 1; y = 0, and z = 0 implies dy = 0, dz = 0. Thus using (6), we obtain the following:

Hence, we obtain the following:

Along the straight line C2from (1, 0, 0) to (1, 1, 0): y varies from 0 to 1; x = 1, z = 0 implies dx = 0, dz =0. Thus using (6), we have F.dr = 0 and hence

Along straight line C3 from (1, 1, 0) to (1, 1, 1): z varies from 0 to 1; x = 1, y =1 implies dx = 0, dy =0. Thus,

F.dr = 20 z2dz

and hence

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Circulation

Before defining circulation of a vector field we recall some definitions.

Definition A curve given by the parametric equation is a simple curve if it does not cross itself; that is, is simple if when

Definition When a curve is simple except for the fact that we say that is a simple closed curve.

Now we define circulation of a vector field.

Definition If C is a simple closed curve ( i.e., a curve which does not intersect itself anywhere), then the line integral of F around C is called the circulation of F about C and is denoted as follows:

Example5 Evaluate , where F = (x2 + y2)i 2 xyj, the curve C is the rectangle in the xy plane bounded by x = 0, x = a, y = 0, y = b.

Solution We know that in the xy plane z = 0,

r = x i + yj , and dr = dx i + dyj .

Now C is the rectangle OABC indicated in the figure.

Fig.3

Hence the required integral is given as follows:

Along OA : x varies from0 to a and y = 0 and hence dy = 0. Hence, we have the following:

Along AB : y varies from0 to b and x = a implies dy = 0. Hence, we have the following:

Along BC: x varies from a to 0 and y = b implies dy = 0. Hence, we have the following:

Along CO: y varies from b to 0 and x = 0 implies dx = 0. Hence, we have the following:

F.dr = 0 and

Adding the above, we obtain the following:

7. Work done by a Force

If is the representation of a smooth curve C, and aforceF actson a particle moving along the curve C,then the line integral

represents the work done by the force in moving the particle from A to B along the curve C. Hence the line integral is also called work integral.

Example 6 Find the total work done in moving a particle in a force field given by

F = 3xy i 5 zj + 10 x k along the curve x = t 2 +1, y = 2 t 2, z = t 3from t = 1 to t = 2.

Solution

Work done is given by , where C is the arc of the given curve from t = 1 to t = 2.

Given F = 3xy i 5 zj + 10 x k. Hence we have the following:

C is in the parametric form x = t 2 + 1, y = 2 t 2, z = t 3.

Substituting x = t 2 + 1, dx = 2 t dt etc., in the above equations, we obtain the following:

In the line integral t varies from t = 1 to t = 2. Hence we have the following:

Example7 Find the work done in moving a particle once round a circle C in the xy plane: the circle has centre at the origin at radius 3 and the force field is given by

F = (2x  y + z)i + (x + y  z2)j + (3x 2y + 4z)k.

Solution

In the xy plane z = 0,

r = x i + yj , and

dr = dx i + dyj .

Now from the figure, we have the following:

fig.4

Hence we can choose the parametric equations of the circle C with center at the origin and radius 3 as :

x = 3 cos, y = 3 sin,

where varies from 0 to 2 . Hence, we have the following:

= 3 sin

 dx = 3 sin d. Similarly, dy = 3 cos  d.

 dr = 3 sin d i + 3 cos  d j

Hence, we have the following:

F.dr ={(2x y+ z)i +(x + y z2)j +(3x2y +4z)k}.{3sin d i +3cos dj}

Hence, we obtain

8. Vector Line Integral

is a representation of a smooth curve C,we define the vector line integral of F over the curve C as follows:

Example 8 If F = xy i  z j + x2k and C is the curve x = t 2, y = 2 t, z = t 3from t = 0 to t = 1. Evaluate the vector line integral.

Solution

Substituting x = t 2, y = 2 t, z = t 3 , we get the following:

F = xy i  z j + x2k = 2t3i  t3 j + t4 k

dr = dxi + dy j + dz k = (2 ti + 2 j + 3t2 k) dt.

Hence, we have

= [ i (3t5 + 2t4)  j 4t5 + k (4t3 + 2t4)] dt.

Hence the vector line integral is given by

Summary

In this session we have described integration of vector function f(t) of a scalar variable t. We have seen that the vector integration is the reverse process of vector differentiation. In this session we have discussed integration of certain type of vector-valued functions along smooth curves. We have learnt that the line integral of a force represents the work done by the force in moving the particle along the curve. Definition and illustration of vector line integral has been made in this session. See you next time. Till then, good bye!

Assignments

1.Show that , where r is a vector function in the variable t and c is a scalar.

2.Prove that , where A is a vector function in the variable t and c is a constant vector function.

3.Evaluate the line integral from (0, 0, 0) to (1, 2, 4) if

F = x2i + yj + (xz  y)k.

(i)along the line segment joining these two points.

(ii)Along the curve given parametrically by x = t2, y = 2 t, z = 4 t3

4.Find the line integral of the tangential component of F = x i + x2j from (1, 0) to (1, 0) in the xy plane

(i)along the x axis

(ii)along the semicircle y = (1 x2)

5.Find , where F = x2i + j + yz k along C, x = t, y =2t2, z= 3 t, 0  t  1.

6.Find from (1, 0, 0) to (1, 0, 4), if F = x i yj + z along the line segment joining (1, 0, 0) and (1, 0, 4).

7.Find the value of  [(3x+4y)dx+(2x+3y2)dy] around the circle x2+y2=4.

8.Find the line integral along the line segment from (1, 0, 2) to (3, 4, 1) where F = 2 xy i + (x2 + z) j + y k.

9.Find the integral around the circumference of the circle x2 2x + y2 = 2, z =1, where F = y i + xj + xyz2k.

10.Evaluate , where F = x2i + y2 j and C is the arc of the parabola y = x2 in the xy plane from (0, 0) to (1, 1).

11. If F = x iyj , compute along the straight line joining the points (0, 0) and (1, 1).

12.If F = (2 y+3)i + xzj + (yz  x)k, evaluate along the following paths C.

(i)x = 2 t2, y = t, z = t3 from t = 0 to t =1.

(ii)the straight line from (0, 0, 0) to (0, 0, 1), then to (0, 1, 1) and then to (2, 1, 1)

(iii)the straight line joining (0, 0, 0) and (2, 1, 1).

13.Evaluate , where F = (x 3 y)i + (y 2 x)j and C is the closed curve from t = 0 to t = 2 .

14.Find the work done in moving a particle in the force field F = 3x2i + (2 xz  y) j + z k along

(i)the straight line from (0, 0, 0) to (2, 1, 3).

(ii)The space curve x = 2 t2, y =t, z = 4 t2t, from t =1 to t=1.

(iii)The curve defined by x 2 = 4y, 3x3 = 8z from x = 0 to x =2.

15.Evaluate along the curve x2+y2=4, z = 1 in the positive direction from (0, 1, 1) to (1, 0, 1) if F = (yz + 2 x)i + xzj + (xy + 2z k.

16.If F = 2 y i zj + x k, evaluate along the curve x = cos t, y = sin t, z = 2 cos t from t = 0 to t = /2.

17.Evaluate where F = (x2 y2)i + 2 xy j and C is the square bounded by the co-ordinate axes and the lines x = a, y = a.

18.Prove the following results:

Quiz

1. Which of the following is not a vector field.

(a) gravitational field

(b) electric field

(d) speed

Ans. (d)

2. Which of the following is not a scalar field.

(a) mass

(b) kinetic energy

(d) electric potential

Ans. (c)

3. If F(t) = (3t2 t) i + (2  6t) j  4tk, the values of

are given, respectively, by ……………

Ans.

(a) (i) (t3 t2/2)i + (2t  3t2)j 2 t2k + c (ii) 50i  32j 24 k

(b) (i) (t3 t2)i + (2t  3t2)j 2 t2k + c (ii) 50i  32j 24 k

(d) (i) (t3 t2/2)i + (2t  3t2)j 2 tk + c (ii) 50i  32j 24 k

Ans. (a)

4. Fill in the blanks:

(a) 3 i + 3j

(b) 4 i + 2j

(d) 3 i + 2j

Ans. (d)

5. If r = t i t 2j + (t 1) k and s = 2 t2 i + 6t k, then

(a) 10

(b) 11

(d) 13

Ans. (c)

6. If r = t i t 2j + (t 1) k and s = 2 t2 i + 6t k, then

(a) 24 i  40/3j + 64/5k

(b) 24 i  40/3j + 64/5k

(d) 24 i +40/3j +64/5k

Ans. (b)

7. The acceleration a of a particle at any time t 0 is given by

a = 18 cos 3ti  8 sin 2tj + 6tk.

If the velocity v and the displacement vector r be zero at t = 0, then the values of v and r at time t are given by ______

(a) v = 6 sin 3ti + (4 cos 2t 4)j + 3 t2k, and r = (2  2 cos 3t)i + (2 sin 2t 4 t)j + t3k.

(b) v = 5 sin 3ti + (4 cos 2t 4)j + 3 t2k, and r = (2  2 cos 3t)i + (2 sin 2t 4 t)j + t3k.

(d) v = 6 sin 3ti + (4 cos 2t 4)j + 3 t2k, and r = (2  2 cos 3t)i + (2 sin 2t 4 t)j + 2t3k.

Ans. (a)

FAQ

1. What is the importance of line integrals in vector calculus.

Ans. The line integral of a vector field plays a crucial role in vector calculus. Some of the fundamental theorems of vector calculusinvolve line integrals of vector fields. Green's theorem and Stokes' theorem (that will be discussed in later sessions) relate line integrals around closed curves to double integrals or surface integrals. If you have a conservative vector field (will be discussed in later session), you can relate the line integral over a curve to quantities just at the curve's two boundary points. It’s worth the effort to develop a good understanding of line integrals.

2. Is it possible to define line integral of vector valued function on any curve?

Ans. No. The curve must be smooth.

3. Does the value of a line integral depends on the choice of representation of the path.

Ans. No. That is, the value of the integral

with given F and C is independent on the particular choice of a representation

of C. We note that the notationmakes it explicit that line integrals are independent of the parametrization.

4. Does the value of a line integral depends on the choice of path?

Ans. For certain vector valued functions F, the value of the integral

from the point A to the point Bdepends on the choice of the path. But the value of a line integral of any conservative vector field is independent of the choice of path. (Detailed discussion is given in the session “ Line Integrals that are Independent of Path”.)

5. Is the orientation of the curve has a role in the line integral of vector valued functions?

Ans. Yes. We refer to the choice of the unit tangent vector, or equivalently, the choice of the direction to tranverse C, as the orientation of the curve. Every simple curve has two orientations, one corresponding to one unit tangent vector T and the other corresponding to its opposite −T.

Glossary

Infinite integral (indefinite integral) of a vector valued function f(t) : Let f(t) be a vector valued function in the variable t. If there exists a vector valued function F(t) of the scalar variable t such that

then F(t) is called the infinite integral of f(t) with respect to t and symbolically we write

The vector valued function f(t) is called the integrand.

Definite integral of a vector valued function f(t) : If for all t in the interval [a, b], then the definite integral between the limits t = a and t = bis given by

Line integral: Any integral which is to be evaluated along a curve is called a line integral.

Path of integration: In the line integral , the curve C is called the path of integration.

Smooth curve (regular arc):A curve C is a smooth curve (regular arc) if C has a unique tangent at each of its points whose direction varies continuously as we move along C. Mathematically this is equivalent to saying that C has a representation

such that is differentiable and the derivative is continuous and different from the zero vector at every point of C.

Line integral of the vector valued function F along a smooth curveC: If the smooth curve that C has a representation

Then the integral of along C from the fixed point A to the point B, written as

is known as line integral of the vector valuedfunction F along C (or more precisely known as the tangent line integral or scalar integral of F along C) .

Simple curve:A curve given by the parametric equation is a simple curve if it does not cross itself; that is, is simple if when

Simple closed curve: When a curve is simple except for the fact that we say that is a simple closed curve.

Circulation:If C is a simple closed curve, then the line integral of F around C is called the circulation of F about C and is denoted by .

Work done by a force in moving the particle fromone point to another point along a curve: If

is a representation of a smooth curve C,and aforceF actson a particle moving along the curve C,then the line integral

represents the work done by the force in moving the particle from A to B along the curve C.

Vector Line Integral: If

is a representation of a smooth curve C,we define the vector line integral of F over the curve C by

REFERENCES

Books

Murray R. Spiegel, Vector Analysis, Schaum Publishing Company, New York.
N. Saran and S. N. Nigam, Introduction to Vector Analysis, Pothishala Pvt. Ltd., Allahabad.
Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 1999.
Shanti Narayan, A Text Book of Vector Calculus, S. Chand & Co., New Delhi.

______