Chapter 2 Vector Calculus Elementary Vector Product Differentiation of Vectors Integration of Vectors Del Operator or Nabla (Symbol ) Polar Coordinates

Chapter 2 Continued Line Integral Volume Integral Surface Integral Green’s Theorem Divergence Theorem (Gauss’ Theorem) Stokes’ Theorem

2.1 Elementary Vector Analysis Definition 2.1 (Scalar and vector) Scalar is a quantity that has magnitude but not direction. For instance mass, volume, distance Vector is a directed quantity, one with both magnitude and direction. For instance acceleration, velocity, force

We represent a vector as an arrow from the origin O to a point A. The length of the arrow is the magnitude of the vector written as or . A A or O O

2.1.1 Basic Vector System Perpendicular to each other In the positive directions of the axes have magnitude (length) 1 Unit vectors , ,

Define a basic vector system and form a right-handed set, i.e

2.1.2 Magnitude of vectors Let P = (x, y, z). Vector is defined by with magnitude (length)

2.1.3 Calculation of Vectors 1. Vector Equation Two vectors are equal if and only if the corresponding components are equals

2. Addition and Subtraction of Vectors 3. Multiplication of Vectors by Scalars

Example 2.1

2.2 Vector Products 1) Scalar Product (Dot product) 2) Vector Product (Cross product)

3) Application of Multiplication of Vectors Given 2 vectors and , projection onto is defined by compb a b) The area of triangle

A = c) The area of parallelogram d) The volume of tetrahedrone e) The volume of parallelepiped A =

Example 2.3

2.4 Vector Differential Calculus Let A be a vector depending on parameter u, The derivative of A(u) is obtained by differentiating each component separately,

The nth derivative of vector is given by The magnitude of is

Example 2.4

Example 2.5 The position of a moving particle at time t is given by x = 4t + 3, y = t2 + 3t, z = t3 + 5t2. Obtain The velocity and acceleration of the particle. The magnitude of both velocity and acceleration at t = 1.

Solution The parameter is t, and the position vector is The velocity is given by The acceleration is

At t = 1, the velocity of the particle is and the magnitude of the velocity is

At t = 1, the acceleration of the particle is and the magnitude of the acceleration is

2.4.1 Differentiation of Two Vectors If both and are vectors, then

2.4.2 Partial Derivatives of a Vector If vector depends on more than one parameter, i.e

Partial derivative of with respect to is given by e.t.c.

Example 2.6

Exercise 2.1

2.5 Vector Integral Calculus The concept of vector integral is the same as the integral of real-valued functions except that the result of vector integral is a vector.

Example 2.7

Exercise 2.2

2.6 Del Operator Or Nabla (Symbol ) Operator  is called vector differential operator, defined as

2.6.1 Grad (Gradient of Scalar Functions) If  x,y,z is a scalar function of three variables and f is differentiable, the gradient of f is defined as

Example 2.8

Exercise 2.3

Solution

2.6.1.1 Grad Properties If A and B are two scalars, then

2.6.2 Directional Derivative

Example 2.9

Solution Directional derivative of  in the direction of

2.6.3 Unit Normal Vector Equation  (x, y, z) = constant is a surface equation. Since  (x, y, z) = constant, the derivative of  is zero; i.e.

This shows that when  (x, y, z) = constant, Vector grad  =   is called normal vector to the surface  (x, y, z) = constant y ds grad  z x

Unit normal vector is denoted by Example 2.10 Calculate the unit normal vector at (-1,1,1) for 2yz + xz + xy = 0.

Solution Given 2yz + xz + xy = 0. Thus

2.6.4 Divergence of a Vector

Example 2.11

Exercise 2.4

Remarks

2.6.5 Curl of a Vector

Example 2.12

Solution

Exercise 2.5

Answer Remark

2.7 Polar Coordinates Polar coordinate is used in calculus to calculate an area and volume of small elements in easy way. Lets look at 3 situations where des Cartes Coordinate can be rewritten in the form of Polar coordinate.

2.7.1 Polar Coordinate for Plane (r, θ) y ds d x 

2.7.2 Polar Coordinate for Cylinder (, , z) ds dv z y   x

2.7.3 Polar Coordinate for Sphere (r, q, f) z r  y  x

Example 2.13 (Volume Integral) z  y  4  3

Solution Since it is about a cylinder, it is easier if we use cylindrical polar coordinates, where

2.8 Line Integral Ordinary integral  f (x) dx, we integrate along the x-axis. But for line integral, the integration is along a curve.  f (s) ds =  f (x, y, z) ds A O B

2.8.1 Scalar Field, V Integral If there exists a scalar field V along a curve C, then the line integral of V along C is defined by

Example 2.14

Solution

Exercise 2.6

2.8.2 Vector Field, Integral Let a vector field and The scalar product is written as

Example 2.15

Solution

Exercise 2.7

* Double Integral *

2.9 Volume Integral 2.9.1 Scalar Field, F Integral If V is a closed region and F is a scalar field in region V, volume integral F of V is

Example 2.20 Scalar function F = 2 x defeated in one cubic that has been built by planes x = 0, x = 1, y = 0, y = 3, z = 0 and z = 2. Evaluate volume integral F of the cubic. z x y 3 O 2 1

Solution

2.9.2 Vector Field, Integral If V is a closed region and , vector field in region V, Volume integral of V is

Example 2.21 Evaluate , where V is a region bounded by x = 0, y = 0, z = 0 and 2x + y + z = 2, and also given

Solution If x = y = 0, plane 2x + y + z = 2 intersects z-axis at z = 2. (0,0,2) If x = z = 0, plane 2x + y + z = 2 intersects y-axis at y = 2. (0,2,0) If y = z = 0, plane 2x + y + z = 2 intersects x-axis at x = 1. (1,0,0)

z y x We can generate this integral in 3 steps : 2 O 1 2x + y + z = 2 y = 2 (1  x) We can generate this integral in 3 steps : Line Integral from x = 0 to x = 1. Surface Integral from line y = 0 to line y = 2(1-x). Volume Integral from surface z = 0 to surface 2x + y + z = 2 that is z = 2 (1-x) - y

Therefore,

Example 2.22 Evaluate where and V is region bounded by z = 0, z = 4 and x2 + y2 = 9 x z  y  4  3

Using polar coordinate of cylinder, ; ; ; where

Therefore,

Exercise 2.8

2.10 Surface Integral 2.10.1 Scalar Field, V Integral If scalar field V exists on surface S, surface integral V of S is defined by where

Example 2.23 Scalar field V = x y z defeated on the surface S : x2 + y2 = 4 between z = 0 and z = 3 in the first octant. Evaluate Solution Given S : x2 + y2 = 4 , so grad S is

Also, Therefore, Then,

Surface S : x2 + y2 = 4 is bounded by z = 0 and z = 3 that is a cylinder with z-axis as a cylinder axes and radius, So, we will use polar coordinate of cylinder to find the surface integral. x z  y 2 3 O

Polar Coordinate for Cylinder where (1st octant) and

Using polar coordinate of cylinder, From

Therefore,

Exercise 2.9

2.10.2 Vector Field, Integral If vector field defeated on surface S, surface integral of S is defined as

Example 2.24

Solution x z y 3 O

Using polar coordinate of sphere,

Exercise 2.9

2.11 Green’s Theorem If c is a closed curve in counter-clockwise on plane-xy, and given two functions P(x, y) and Q(x, y), where S is the area of c.

Example 2.25  y 2 x C3 C2 C1 O x2 + y2 = 22 Solution

2.12 Divergence Theorem (Gauss’ Theorem) If S is a closed surface including region V in vector field

Example 2.26

Solution x z y 2 4 O S3 S4 S2 S1 S5

2.13 Stokes’ Theorem If is a vector field on an open surface S and boundary of surface S is a closed curve c, therefore

Example 2.27 Surface S is the combination of

Solution z y x 3 4 O S3 C2 S2 C1 S1

We can also mark the pieces of curve C as C1 : Perimeter of a half circle with radius 3. C2 : Straight line from (-3,0,0) to (3,0,0). Let say, we choose to evaluate first. Given

So,

By integrating each part of the surface,

Then , and

By using polar coordinate of cylinder ( because is a part of the cylinder),

Therefore, Also,

(ii) For surface , normal vector unit to the surface is By using polar coordinate of plane ,

For surface S3 : y = 0, normal vector unit to the surface is dS = dxdz The integration limits : So,