Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 3, Problem 44P

(a)

To determine

The value of the line integral LFdl.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The vector (F) is x2yaxyay.

The boundaries for L are as shown in the figure-(1).

Elements Of Electromagnetics, Chapter 3, Problem 44P , additional homework tip  1

Figure-(1)

Calculation:

The line distribution of the given figure is as shown in figure (2).

Elements Of Electromagnetics, Chapter 3, Problem 44P , additional homework tip  2

Figure-(2)

From the figure (2), the equations for lines 1, 2, and 3 are as follows.

For line 1,

  x=y

  dx=dy

The limit of (x) is 0<x<1.

Calculate the value of line integral (1Fdl) using the relation.

  1Fdl=1F(dxax+dyay)

  1Fdl=1(x2yaxyay)(dxax+dyay)=1x2ydxydy=01x2(x)dx(x)(dx)=01(x3x)dx

  1Fdl=[x44x22]01=(144122)(044022)=14

For line 2,

  y=x+2

  dy=dx

The limit of (x) is 1<x<2.

Calculate the value of line integral (2Fdl) using the relation.

  2Fdl=2F(dxax+dyay)

  2Fdl=2(x2yaxyay)(dxax+dyay)=2x2ydxydy=12x2(x+2)dx(x+2)(dx)=12(x3+2x2x+2)dx

  2Fdl=[x44+2x33x22+2x]12=(244+2(2)33222+2(2))(144+2(1)33122+2(1))=(103)(2312)=1712

For line 3,

  y=0

  dy=0

The limit of (x) is 0<x<2.

Calculate the value of line integral (3Fdl) using the relation.

  3Fdl=3F(dxax+dyay)

  3Fdl=3(x2yaxyay)(dxax+dyay)=3x2ydxydy=02x2(0)dx(0)(0)=02(0)dx

  3Fdl=0

Calculate the value of total line integral (LFdl) using the relation.

  LFdl=1Fdl+2Fdl+3Fdl

  LFdl=14+1712+0=1412=1.1667

Thus, the value of line integral LFdl is 1.1667_.

(b)

To determine

The value of surface integral S(×F)dS.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The vector (F) is x2yaxyay.

The area (S) for boundary (L) is as shown in the figure-(3).

Elements Of Electromagnetics, Chapter 3, Problem 44P , additional homework tip  3

Figure-(3)

Calculation:

Calculate the curl of vector (×F) using the relation.

  ×F=|axayazxyzFxFyFz|

  ×F=|axayazxyzx2y(y)0|=ax(y(0)z(y))ay(x(0)zx2y)+az(x(y)yx2y)=ax(00)ay(00)+az(0x2)=x2az

The line distribution of the given figure is as shown in figure-(4).

Elements Of Electromagnetics, Chapter 3, Problem 44P , additional homework tip  4

Figure-(4)

From the figure (2), the equations for line 1 and 2 are as follow.

For line 1,

  x=y

The limit of (x) is 0<x<1.

For line 2,

  y=x+2

The limit of (x) is 1<x<2.

Calculate the value of surface integral (S(×F)dS) using the relation.

  S(×F)dS=S(×F)(dydzax+dxdzaydxdyaz)

  S(×F)dS=S(x2az)(dydzax+dxdzaydxdyaz)=Sx2dxdy=010xx2dxdy+120x+2x2dxdy=01x2dx[y]0x+12x2dx[y]0x+2

  S(×F)dS=01x2(x0)dx+12x2(x+20)dx=01x3dx+12(x3+2x2)dx=[x44]01+[x44+2x33]12=(144044)+[(244+2(2)33)(144+2(1)33)]

  S(×F)dS=14+(4+163+1423)=14+1112=1412=1.667

Thus, the value of surface integral S(×F)dS is 1.1667_.

(c)

To determine

Whether the Stokes’s theorem is satisfied or not.

(c)

Expert Solution
Check Mark

Explanation of Solution

The value of line integral LFdl is 1.667.

The value of surface integral S(×F)dS is 1.667.

The Stokes’s theorem states that if F is a continuously differentiable vector point defined on an open surface bounded by the curve L in a positive direction then,

  LFdl=S(×F)dS

From the above calculations in subpart (a) and subpart (b), these values are same.

Thus, the Stokes’s theorem is satisfied.

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Chapter 3 Solutions

Elements Of Electromagnetics

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