Chapter 3 1. Line Integral Volume Integral Surface Integral Green’s Theorem Divergence Theorem (Gauss’ Theorem) Stokes’ Theorem

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

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

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

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

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,