Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
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Chapter 1, Problem 1.23P

The surfaces p = 3 , p = 5 , φ = 130 o , z = 3 a n d z = 4.5 define a closed surface. Find (a) the enclosed volume; (b) the total area of the enclosing surface; (c) the total length of the 12 edges of the surfaces; (a) the Length of the longest straight line that lies entirely within the volume.

Expert Solution
Check Mark
To determine

(a)

The enclosed volume.

Answer to Problem 1.23P

The enclosed volume is 6.28

Explanation of Solution

Given information:

For the closed surface,

ρ=3ρ=5ϕ=100°ϕ=130°z=3z=4.5

Concept used:

Volume= ρρ ϕ ϕ z z ρdρdϕdz ......... (1)

Calculation:

Substituting the values in equation (1), we get

   Volume= 3 4.5 100 130 3 5 ρdρdϕdz

   = 3 4.5 1.74 2.26 3 5 ρdρdϕdz

   = 3 4.5 1.74 2.26 [ ρ 2 2 ] 3 5 dϕdz

   = 3 4.5 1.74 2.26 [ ρ 2 2 ] 3 5 dϕdz

=3 4.5 1.74 2.26 [ 16 2 ] dϕdz= 3 4.5 [ 8ϕ]1.742.26dz=3 4.5[ 8×2.268×1.74]dz=3 4.54.16dz

=[4.16z]34.5=4.5×4.163×4.16=6.28

Conclusion:

Therefore, the enclosed volume is 6.28.

Expert Solution
Check Mark
To determine

(b)

Total area of the enclosing surface.

Answer to Problem 1.23P

Total area of the enclosing surface is 20.7.

Explanation of Solution

Given information:

For the closed surface,

ρ=3ρ=5ϕ=100°ϕ=130°z=3z=4.5

Concept used:

Area= 2ϕϕ ρ ρ ρdρdϕ+zz ϕ ϕ 3dϕdz+zz ϕ ϕ 5dϕdz+2zz ϕ ϕ dϕdz ........... (2)

Calculation:

Substituting the values in equation (2), we get

   Area=2 100 130 3 5 ρdρdϕ + 3 4.5 100 130 3dϕdz+ 3 4.5 100 130 5dϕdz+ 2 3 4.5 3 5 dρdz

   =2 1.74 2.26 3 5 ρdρdϕ + 3 4.5 1.74 2.26 3dϕdz + 3 4.5 1.74 2.26 5dϕdz +2 3 4.5 3 5 dρdz

   [ converting degree to radian for ϕ ]

   =2 1.74 2.26 [ ρ 2 2 ] 3 5 dϕ+ 3 4.5 [ 3ϕ ] 1.74 2.26 dz+ 3 4.5 [ 5ϕ ] 1.74 2.26 dz+2 3 4.5 [ ρ ] 3 5 dz

   =2 1.74 2.26 [ 25 2 9 2 ] dϕ+ 3 4.5 [ 3×2.263×1.74 ] dz+ 3 4.5 [ 5×2.265×1.74 ] dz+2 3 4.5 [ 53 ] dz

Further integrating we have,

   Area=2 1.74 2.268dϕ+ 3 4.5 1.56dz+ 2.26 4.52.6dz+23 4.52dz=2[8ϕ]1.742.26+[1.56]34.5+[2.6z]34.5+2[2z]34.5=2[8×2.268×1.74]+[1.56×4.51.56×3]+[2.6×4.52.6×3]+2[2×4.52×3]=2×4.16+2.34+3.9+6=20.7

Conclusion:

Therefore, the total area of the enclosing surface is 20.7

Expert Solution
Check Mark
To determine

(c)

Total length of the twelve edges of the surfaces.

Answer to Problem 1.23P

Total length of the twelve edges of the surfaces is 22.4.

Explanation of Solution

Given information:

For the closed surface,

ρ=3ρ=5ϕ=100°ϕ=130°z=3z=4.5

Concept used:

Length=4×1.5+4×2+2[303602π×3+303602π×5]    ..........(3)

Calculation:

From equation (3), we get

Length=4×1.5+4×2+2[ 30 360 2π×3+ 30 360 2π×5]=6+8+2[0.083×6π+0.083×10π]=14+2×4.17=22.4

Conclusion:

Therefore, the total length is 22.4.

Expert Solution
Check Mark
To determine

(d)

The length of the longest straight line that lies entirely within the volume.

Answer to Problem 1.23P

The length of the longest straight line that lies entirely within the volume is 3.21.

Explanation of Solution

Given information:

For the closed surface,

ρ=3ρ=5ϕ=100°ϕ=130°z=3z=4.5

Concept used:

Length= |BA|    ..........(4)

Calculation:

The longest straight line will be between the points A(ρ=3,ϕ=100,z=3) and

B(ρ=5,ϕ=130,z=4.5)

Now converting A and B to Cartesian co-ordinates, x=ρcosϕ,y=ρsinϕ,z=z

For A ,

x=3cos100°=0.519y=3sin100°=2.952z=3

Therefore, A(x=0.52,y=2.95,z=3)

For B ,

x=5cos130=3.21y=5sin130=3.83z=4.5

Therefore, B(x=3.21,y=3.83,z=4.5)

So, from equation (4),

Length=|BA|=|(3.21+0.51,3.832.95,4.53)|= 2.692+ 0.882+ 1.52=9.8888=3.21

Conclusion:

Therefore, the total length is 3.21.

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