RF/Microwave Engineering and Applications in Energy Systems. Abdullah Eroglu

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Coordinate System

      In a spherical coordinate system, the differential length is given by

      (1.41)d ModifyingAbove l With right-arrow equals italic d r dot ModifyingAbove r With ampersand c period circ semicolon plus italic r d theta dot ModifyingAbove theta With ampersand c period circ semicolon plus r sine italic theta d phi dot ModifyingAbove phi With ampersand c period circ semicolon

      Its magnitude is defined by

      (1.42)italic d l equals StartRoot italic d r squared plus r squared d theta squared plus r squared sine squared italic theta d phi squared EndRoot

Schematic illustration of differential length, area, and volume in a cylindrical coordinate system.

      (1.43)StartLayout 1st Row d ModifyingAbove s With right-arrow Subscript upper R Baseline equals ModifyingAbove upper R With ampersand c period circ semicolon upper R squared sine italic theta d theta d phi 2nd Row d ModifyingAbove s With right-arrow Subscript theta Baseline equals ModifyingAbove theta With ampersand c period circ semicolon upper R sine italic theta dRd phi 3rd Row d ModifyingAbove s With right-arrow Subscript phi Baseline equals ModifyingAbove phi With ampersand c period circ semicolon italic RdRd theta EndLayout

      The infinitesimal change for the volume is defined by

      (1.44)italic d v equals upper R squared sine italic theta dRd theta d phi

      The illustration of the changes in length, area, and volume for a spherical coordinate system is given in Figure 1.11.

      1.2.4 Line Integral

Schematic illustration of differential length, area, and volume in a spherical coordinate system.

      One of the practical examples for the line integral application is finding work that is required to move a charge from point a to point b, similar to the one shown in Figure 1.12. This can be expressed in terms of the line integral of the electric field, ModifyingAbove upper E With right-arrow.

Schematic illustration of vector F implies along the curve.
along the curve.

      Solution

      We need to find

integral Underscript upper P Endscripts ModifyingAbove upper E With right-arrow dot d ModifyingAbove l With right-arrow equals integral Subscript upper P 1 Superscript upper P 2 Baseline ModifyingAbove upper E With right-arrow dot d ModifyingAbove l With right-arrow Schematic illustration of path for line integral Example 1.1. ModifyingAbove upper E With right-arrow equals ModifyingAbove r With ampersand c period circ semicolon 2 left-parenthesis cosine squared phi minus sine squared phi right-parenthesis plus ModifyingAbove phi With ampersand c period circ semicolon 2 r cosine phi sine phi

      In a cylindrical coordinate system, the differential length along the path is defined as

d ModifyingAbove l With right-arrow equals ModifyingAbove phi With ampersand c period circ semicolon italic r d phi

      where r = 2 along the path. So

StartLayout 1st Row integral Subscript upper P 1 Superscript upper P 2 Baseline ModifyingAbove upper E With right-arrow dot d ModifyingAbove l With right-arrow equals integral Subscript phi equals 180 Superscript degree Baseline Superscript phi equals 90 Superscript degree Baseline Baseline left-parenthesis ModifyingAbove r With ampersand c period circ semicolon 2 left-parenthesis cosine squared phi minus sine squared phi right-parenthesis plus ModifyingAbove phi With ampersand c period circ semicolon 2 r cosine phi sine phi right-parenthesis dot ModifyingAbove phi With ampersand c period circ semicolon italic r d phi 2nd Row equals integral Subscript phi equals 180 Superscript degree Baseline Superscript phi equals 90 Superscript degree Baseline Baseline left-parenthesis 2 r squared cosine phi sine phi right-parenthesis italic d phi 3rd Row equals 2 r squared StartFraction sine squared phi Over 2 EndFraction vertical-bar Subscript phi equals 180 Sub Superscript degree Superscript phi equals 90 Super Superscript degree Baseline equals 4 EndLayout

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