Abdullah Eroglu

RF/Microwave Engineering and Applications in Energy Systems


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(1.69)

contour-integral Underscript upper S Endscripts upper T overbar dot d s overbar equals integral Subscript x equals 0 Superscript a Baseline integral Subscript y equals 0 Superscript b left-parenthesis 1 minus StartFraction x Over a EndFraction right-parenthesis Baseline left-parenthesis x ModifyingAbove x With ampersand c period circ semicolon plus y ModifyingAbove y With ampersand c period circ semicolon plus z ModifyingAbove z With ampersand c period circ semicolon right-parenthesis dot left-parenthesis italic dxdy StartFraction c Over a EndFraction left-parenthesis ModifyingAbove x With ampersand c period circ semicolon plus StartFraction a Over b EndFraction ModifyingAbove y With ampersand c period circ semicolon plus StartFraction a Over c EndFraction ModifyingAbove z With ampersand c period circ semicolon right-parenthesis right-parenthesis

      or

      From (1.76)

      (1.83)a equals x plus StartFraction a Over b EndFraction y plus StartFraction a Over c EndFraction z

      (1.85)contour-integral Underscript upper S Endscripts upper T overbar dot d s overbar equals integral Subscript x equals 0 Superscript a Baseline italic c b left-parenthesis 1 minus StartFraction x Over a EndFraction right-parenthesis italic d x equals StartFraction italic a b c Over 2 EndFraction

      The left‐hand side of the equation now can be found as

      (1.86)integral Underscript upper V Endscripts nabla dot upper T overbar italic d v equals 3 left-parenthesis ModifyingBelow one third StartFraction italic a b c Over 2 EndFraction With presentation form for vertical right-brace Underscript volume of pyramid Endscripts right-parenthesis equals StartFraction italic a b c Over 2 EndFraction

      1.3.6 Stokes' Theorem

      Stokes' theorem states that the line integral of a vector around a closed contour is equal to the surface integral of the curl of a vector over an open surface. It is expressed by

      Solution

      We first calculate the left‐hand side of Eq. (1.87) as

nabla times upper T overbar equals left-parenthesis StartFraction partial-differential Over partial-differential x EndFraction ModifyingAbove x With ampersand c period circ semicolon plus StartFraction partial-differential Over partial-differential y EndFraction ModifyingAbove y With ampersand c period circ </p>
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