Asst. Prof. Mehmet Çayören
Electromagnetic Waves
2014-2015 Fall Semester Exercise 1
1. a) 𝑉 = 𝑥 2 + 𝑦 2 + 𝑧 2 . Find 𝐸⃗ if 𝐸⃗ = −∇𝑉.
𝜕
𝜕
𝜕
𝐸⃗ = − ( ⃗⃗⃗⃗
𝑒𝑥 +
𝑒𝑦 + ⃗⃗⃗
⃗⃗⃗⃗
𝑒 ) (𝑥 2 + 𝑦 2 + 𝑧 2 )
𝜕𝑥
𝜕𝑦
𝜕𝑧 𝑧
𝐸⃗ = −2𝑥𝑒⃗⃗⃗⃗𝑥 − 2𝑦𝑒⃗⃗⃗⃗𝑦 − 2𝑧𝑒
⃗⃗⃗𝑧
b) 𝑉 = 𝜌2 + 𝛷2 + 𝑧 2 . Find 𝐸⃗ if 𝐸⃗ = −∇𝑉.
𝜕
1 𝜕
𝜕
𝐸⃗ = − ( ⃗⃗⃗
𝑒𝜌 +
𝑒𝛷 + ⃗⃗⃗
⃗⃗⃗⃗
𝑒 ) (𝜌2 + 𝛷2 + 𝑧 2 )
𝜕𝜌
𝜌 𝜕𝛷
𝜕𝑧 𝑧
𝐸⃗ = −2𝜌𝑒⃗⃗⃗𝜌 − 2
𝛷
𝑒 − 2𝑧𝑒
⃗⃗⃗⃗
⃗⃗⃗𝑧
𝜌 𝛷
2. (P2.19 In Cheng’s Book) For vector function 𝐴 = 𝜌2 ⃗⃗⃗
𝑒𝜌 + 2𝑧𝑒⃗⃗⃗𝑧 , verify the divergence theorem
for the circular cylindrical region enclosed by 𝜌 = 5, 𝑧 = 0 and 𝑧 = 4.
Divergence theorem:
⃗⃗⃗⃗
∫ ∇. 𝐴𝑑𝑣 = ∮ 𝐴. 𝑑𝑠
𝑉
∇. 𝐴 =
𝑆
1 𝜕
1 𝜕
𝜕𝐴𝑧
(𝐴𝛷 ) +
(𝜌𝐴𝜌 ) +
𝜌 𝜕𝜌
𝜌 𝜕𝛷
𝜕𝑧
⏟
=0
1 𝜕 3
𝜕
(𝜌 ) + (2𝑧) = 3𝜌 + 2
∇. 𝐴 =
𝜌 𝜕𝜌
𝜕𝑧
4
2𝜋
5
∫ ∇. 𝐴𝑑𝑣 = ∫ ∫ ∫ (3𝜌 + 2)𝜌 𝑑𝜌 𝑑𝛷 𝑑𝑧
𝑉
𝑧=0 𝛷=0 𝜌=0
4
2𝜋
∫ ∇. 𝐴𝑑𝑣 = ∫ ∫ (𝜌3 + 𝜌2 )|5𝜌=0 𝑑𝛷 𝑑𝑧 = 8𝜋(150) = 1200𝜋
𝑉
𝑧=0 𝛷=0
⃗⃗⃗⃗ = ( ∫ + ∫ + ∫ ) (𝐴. 𝑑𝑠
⃗⃗⃗⃗ )
∮ 𝐴. 𝑑𝑠
𝑆
𝑆1
𝑆2
𝑆3
𝑆2
𝑆1
On 𝑆1 (𝜌 = 5)
𝑆3
⃗⃗⃗⃗ = 𝜌 𝑑𝛷 𝑑𝑧 ⃗⃗⃗
𝑑𝑠
𝑒𝜌
1
Asst. Prof. Mehmet Çayören
2𝜋
4
3
∫ 𝜌 𝑑𝛷 𝑑𝑧 = 125 ∫ ∫ 𝑑𝛷 𝑑𝑧 = 8𝜋(125) = 1000𝜋
𝑆1
𝑧=0 𝛷=0
On 𝑆2 (𝑧 = 4)
⃗⃗⃗⃗
𝑑𝑠 = 𝜌 𝑑𝜌 𝑑𝛷 ⃗⃗⃗
𝑒𝑧
2𝜋
5
∫ 2𝑧𝜌 𝑑𝜌 𝑑𝛷 = 4 × ∫ ∫ 2𝜌 𝑑𝜌 𝑑𝛷 = 2𝜋 × 4 × 𝜌2 |5𝜌=0 = 8𝜋 × 25 = 200𝜋
𝑆2
𝛷=0 𝜌=0
On 𝑆2 (𝑧 = 0)
𝐴𝑧 = 0 ∫𝑆 (𝐴. ⃗⃗⃗⃗
𝑑𝑠) = 0
3
∮ 𝐴. ⃗⃗⃗⃗
𝑑𝑠 = 1000𝜋 + 200𝜋 + 0 = 1200𝜋
𝑆
∫ ∇. 𝐴𝑑𝑣 = ∮ 𝐴. ⃗⃗⃗⃗
𝑑𝑠 = 1200𝜋
𝑉
𝑆
⃗ =
3. (P2.19 In Cheng’s Book) A vector field 𝐷
cos2 𝛷
𝑒𝑟
⃗⃗⃗
𝑟3
exists in the region between two
spherical shells defined by 𝑟 = 1 and 𝑟 = 2. Evaluate
⃗ . ⃗⃗⃗⃗
a) ∮ 𝐷
𝑑𝑠
⃗ 𝑑𝑣
b) ∫ ∇. 𝐷
⃗ . ⃗⃗⃗⃗
⃗ . ⃗⃗⃗⃗
a) ∮ 𝐷
𝑑𝑠 = (∫𝑆 + ∫𝑆 ) (𝐷
𝑑𝑠) where 𝑆1 and 𝑆2 represents 𝑟 = 1 and 𝑟 = 2 respectively.
1
2
On 𝑆1 (𝑟 = 1)
⃗⃗⃗⃗ = −𝑟 2 sinθ 𝑑θ 𝑑𝛷 ⃗⃗⃗
𝑑𝑠
𝑒𝑟
2𝜋
𝜋
2𝜋
𝜋
cos 2 𝛷 2
1
⃗ . ⃗⃗⃗⃗
∫𝐷
𝑑𝑠 = ∫ ∫ −
𝑟 sinθ 𝑑θ 𝑑𝛷 = − ∫ ∫ cos2 𝛷 sinθ 𝑑θ 𝑑𝛷
3
𝑟
1
𝑆1
𝛷=0 θ=0
𝛷=0 θ=0
2𝜋
2𝜋
⃗⃗⃗⃗ = ( ∫ cos
⃗ . 𝑑𝑠
∫𝐷
𝑆1
2
𝛷 𝑑𝛷) (𝑐𝑜𝑠θ)|𝜋θ=0
= −2 ∫ cos2 𝛷 𝑑𝛷
𝛷=0
𝛷=0
1
cos2 𝑥 = (cos(2𝑥) + 1)
2
2𝜋
⃗⃗⃗⃗ = −2 ∫
⃗ . 𝑑𝑠
∫𝐷
𝑆1
𝛷=0
2𝜋
1
1
(cos 2𝛷 + 1)𝑑𝛷 = ( sin2𝛷 + 𝛷)|
= −2𝜋
2
2
𝛷=0
On 𝑆2 (𝑟 = 2)
2
Asst. Prof. Mehmet Çayören
⃗⃗⃗⃗ = 𝑟 2 sinθ 𝑑θ 𝑑𝛷 ⃗⃗⃗
𝑑𝑠
𝑒𝑟
2𝜋
𝜋
2𝜋
𝜋
cos2 𝛷 2
1
𝑟 sinθ 𝑑θ 𝑑𝛷 =
∫ ∫ cos2 𝛷 sinθ 𝑑θ 𝑑𝛷 = 𝜋
3
𝑟
2
⏟ θ=0
𝛷=0 θ=0
𝛷=0
⃗ . ⃗⃗⃗⃗
∫𝐷
𝑑𝑠 = ∫ ∫
𝑆2
2𝜋
⃗ . ⃗⃗⃗⃗
( ∫ + ∫ ) (𝐷
𝑑𝑠) = −2𝜋 + 𝜋 = −𝜋
𝑆1
1 𝜕
1
𝑆2
𝜕
1
𝜕
⃗ = 2 (𝑟 2 𝐷𝑟 ) +
(𝐷θ 𝑠𝑖𝑛θ) +
(𝐷𝛷 )
b) ∇. 𝐷
⏟
𝑟 𝜕𝑟
𝑟𝑠𝑖𝑛θ 𝜕θ
𝑟𝑠𝑖𝑛θ 𝜕𝛷
=0
⃗ =
∇. 𝐷
1 𝜕
cos2 𝛷
1 𝜕 cos2 𝛷
cos2 𝛷
2
(𝑟
)= 2 (
)=
𝑟 2 𝜕𝑟
𝑟3
𝑟 𝜕𝑟
𝑟
𝑟4
𝑑𝑣 = 𝑟 2 𝑠𝑖𝑛θ 𝑑r𝑑𝛷𝑑θ
𝜋
2𝜋
2
⃗ 𝑑𝑣 = ∫ ∫ ∫
∫ ∇. 𝐷
θ=0 𝛷=0 𝑟=1
𝜋
cos 2 𝛷
𝑠𝑖𝑛θ𝑑r𝑑𝛷𝑑θ
𝑟2
2𝜋
𝜋
2𝜋
1 2
1
2
⃗
∫ ∇. 𝐷 𝑑𝑣 = ∫ ∫ cos 𝛷 𝑠𝑖𝑛θ𝑑𝛷𝑑θ ( 2 )|
= − ∫ ∫ cos 2 𝛷 𝑠𝑖𝑛θ𝑑𝛷𝑑θ = −𝜋
𝑟 𝑟=1
2
⏟ 𝛷=0
θ=0 𝛷=0
θ=0
=2𝜋
4. (P2.24 In Cheng’s Book) Assume the vector function 𝐴 = 3𝑥 2 𝑦 2 ⃗⃗⃗⃗
𝑒𝑥 − 𝑥 3 𝑦 2 ⃗⃗⃗⃗
𝑒𝑦
a) Find ∮ 𝐴. ⃗⃗⃗
𝑑𝑙 around the triangular contour given in figure.
⃗⃗⃗⃗ over the triangular area.
b) Evaluate ∫(∇ × 𝐴). 𝑑𝑠
c) Can 𝐴 be expressed as the gradiant of a scalar? Explain.
a) ∮ 𝐴. ⃗⃗⃗
𝑑𝑙 = (∫𝐿 + ∫𝐿 + ∫𝐿 ) 𝐴. ⃗⃗⃗
𝑑𝑙
1
2
3
On 𝐿1 (𝑥 = 𝑦), ⃗⃗⃗
𝑑𝑙 = 𝑑𝑥𝑒⃗⃗⃗⃗𝑥 + 𝑑𝑦𝑒⃗⃗⃗⃗𝑦
2
∫ 𝐴. ⃗⃗⃗
𝑑𝑙 = ∫ 2𝑥 5 𝑑𝑥 = 21
𝐿1
1
On 𝐿2 (𝑥 = 2), ⃗⃗⃗
𝑑𝑙 = 𝑑𝑦𝑒⃗⃗⃗⃗𝑦
3
Asst. Prof. Mehmet Çayören
1
1
⃗⃗⃗ = ∫(−8𝑦 2 )𝑑𝑦 = −
∫ 𝐴. 𝑑𝑙
𝐿2
2
8𝑦 3
8 64 56
| =− +
=
3 2
3 3
3
On 𝐿3 (𝑦 = 1), ⃗⃗⃗
𝑑𝑙 = 𝑑𝑥𝑒⃗⃗⃗⃗𝑥
1
⃗⃗⃗ = ∫(3𝑥 2 )𝑑𝑥 = 𝑥 3 |12 = 1 − 8 = −7
∫ 𝐴. 𝑑𝑙
𝐿3
2
∮ 𝐴. ⃗⃗⃗
𝑑𝑙 = 21 +
b) (∇ × 𝐴) = |
𝑒𝑥
⃗⃗⃗⃗
𝑒𝑦
⃗⃗⃗⃗
⃗⃗⃗
𝑒𝑧
𝜕
𝜕𝑥
2 2
𝜕
𝜕𝑦
3 2
𝜕
|
𝜕𝑧
3𝑥 𝑦
−𝑥 𝑦
56
98
−7=
3
3
= ⃗⃗⃗⃗
𝑒𝑥 (… ) + ⃗⃗⃗⃗
𝑒𝑦 (… ) + ⃗⃗⃗
𝑒𝑧 (
𝜕
𝜕
(−𝑥 3 𝑦 2 ) − (3𝑥 2 𝑦 2 ))
𝜕𝑥
𝜕𝑦
0
𝑒𝑧 (−3𝑥 2 𝑦 2 − 9𝑥 2 𝑦 2 ) = −12𝑥 2 𝑦 2 ⃗⃗⃗
𝑒𝑧
(∇ × 𝐴) = ⃗⃗⃗
⃗⃗⃗⃗ = −𝑑𝑥𝑑𝑦𝑒
𝑑𝑠
⃗⃗⃗𝑧
𝑦=2 𝑥=2
⃗⃗⃗⃗ = ∫
∫ (∇ × 𝐴). 𝑑𝑠
𝑆
𝑦=2
𝑥=2
𝑦=2
𝑥3
∫ 12𝑥 𝑦 𝑑𝑥𝑑𝑦 = 12 ∫ 𝑦 ( )|
𝑑𝑦 = 4 ∫ (8 − 𝑦 3 ) 𝑦 2 𝑑𝑦
3 𝑥=𝑦
2 2
2
𝑦=1 𝑥=𝑦
𝑦=1
𝑦=1
𝑦=2
∫ (∇ × 𝐴). ⃗⃗⃗⃗
𝑑𝑠 = 4 (
𝑆
8𝑦 3 𝑦 6
128 256 2 32
98
− )|
=−
+
+ −
=
3
6 𝑦=1
3
3
3 3
3
c) Since ∮ 𝐴. ⃗⃗⃗
𝑑𝑙 = ∫(∇ × 𝐴). ⃗⃗⃗⃗
𝑑𝑠 ≠ 0, 𝐴 cannot be expressed as gradient of a scalar.
𝛷
2
5. (P2.25 In Cheng’s Book) Given the vector function 𝐴 = 𝑠𝑖𝑛 ( ) ⃗⃗⃗⃗
𝑒𝛷 , verify Stokes’s theorem
over the hemispherical surface and its circular contour that are shown in figure.
∫ (∇ × 𝐴). ⃗⃗⃗⃗
𝑑𝑠 = ∮ 𝐴. ⃗⃗⃗
𝑑𝑙
𝑆
𝐶
Integral on surface;
⃗⃗⃗⃗
𝑑𝑠 = 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜃𝑑𝛷𝑒⃗⃗⃗𝑟
4
Asst. Prof. Mehmet Çayören
𝑒⃗⃗⃗𝑟
1 |𝜕
×
𝐴
=
(∇
)
𝑟 2 𝑠𝑖𝑛𝜃 |𝜕𝑟
𝑟𝑒⃗⃗⃗⃗𝜃
𝜕
𝜕𝜃
0
(∇ × 𝐴) =
1
0
𝑟𝑠𝑖𝑛𝜃𝑒⃗⃗⃗⃗𝛷
𝜕
|
𝜕𝛷
|
𝛷
𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛 ( )
2
𝜕
𝛷
𝜕
𝛷
⃗⃗⃗ ( (𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛 ( ))) − 𝑟𝑒⃗⃗⃗⃗𝜃 ( (𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛 ( )))]
𝜕𝜃
2
𝜕𝑟
2
[𝑒
𝑟 2 𝑠𝑖𝑛𝜃 𝑟
𝛷
𝛷
𝑐𝑜𝑠𝜃𝑠𝑖𝑛 ( 2 )
𝑠𝑖𝑛𝜃𝑠𝑖𝑛 ( 2 )
) ⃗⃗⃗
𝑒𝑟 − (
) ⃗⃗⃗⃗
𝑒𝜃
(∇ × 𝐴) = (
𝑟𝑠𝑖𝑛𝜃
𝑟𝑠𝑖𝑛𝜃
2𝜋 𝜋/2
𝛷
∫ (∇ × 𝐴). ⃗⃗⃗⃗
𝑑𝑠 = ∫ ∫ 𝑐𝑜𝑠𝜃𝑠𝑖𝑛 ( ) 𝑟𝑑𝜃𝑑𝛷
2
𝑆
𝛷=0 𝜃=0
Using 𝑟 = 𝑏,
2𝜋
2𝜋
𝛷=0
𝛷=0
𝛷
𝛷
⃗⃗⃗⃗ = 𝑏 ∫ 𝑠𝑖𝑛 ( ) 𝑑𝛷 ((𝑠𝑖𝑛𝜃)|𝜋/2 ) = 𝑏 ∫ 𝑠𝑖𝑛 ( ) 𝑑𝛷
∫ (∇ × 𝐴). 𝑑𝑠
𝜃=0
2
2
𝑆
2𝜋
𝛷
⃗⃗⃗⃗ = −2𝑏 (𝑐𝑜𝑠 ( ))|
∫ (∇ × 𝐴). 𝑑𝑠
= −2𝑏(−1 − 1) = 4𝑏
2
𝑆
𝛷=0
Integral on contour;
⃗⃗⃗
𝑑𝑙 = 𝑟𝑠𝑖𝑛𝜃𝑑𝛷𝑒⃗⃗⃗⃗𝛷
2𝜋
𝛷
∮ 𝐴. ⃗⃗⃗
𝑑𝑙 = ∫ 𝑠𝑖𝑛 ( ) 𝑟𝑠𝑖𝑛𝜃𝑑𝛷
2
𝐶
Using 𝑟 = 𝑏 and 𝜃 =
𝛷=0
𝜋
2
2𝜋
𝛷
⃗⃗⃗ == −2𝑏 (𝑐𝑜𝑠 ( ))|
∮ 𝐴. 𝑑𝑙
= −2𝑏(−1 − 1) = 4𝑏
2
𝛷=0
𝐶
6. a) If a scalar field is defined as 𝑓 = 𝑥 2 𝑦 2 𝑧 2 , show that the Laplacian ∆𝑓 is equal to ∇. (∇𝑓)
𝜕2𝑓 𝜕2𝑓 𝜕2𝑓
∆𝑓 = ∇2 𝑓 = 2 + 2 + 2 = 2𝑦 2 𝑧 2 + 2𝑥 2 𝑧 2 + 2𝑥 2 𝑦 2
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑓
𝜕𝑓
𝜕𝑓
∇. (∇𝑓) = ∇. ( ⃗⃗⃗⃗
𝑒𝑥 +
𝑒𝑦 +
⃗⃗⃗⃗
⃗⃗⃗
𝑒 ) = ∇. (2𝑥𝑦 2 𝑧 2 ⃗⃗⃗⃗
𝑒𝑥 + 2𝑥 2 𝑦𝑧 2 ⃗⃗⃗⃗
𝑒𝑦 + 2𝑥 2 𝑦 2 𝑧𝑒
⃗⃗⃗𝑧 )
𝜕𝑥
𝜕𝑦
𝜕𝑧 𝑧
∇. (∇𝑓) =
𝜕
𝜕
𝜕
(2𝑥𝑦 2 𝑧 2 ) +
(2𝑥 2 𝑦𝑧 2 ) + (2𝑥 2 𝑦 2 𝑧) = 2𝑦 2 𝑧 2 + 2𝑥 2 𝑧 2 + 2𝑥 2 𝑦 2
𝜕𝑥
𝜕𝑦
𝜕𝑧
5
Asst. Prof. Mehmet Çayören
b) If a vector field is defined as,
𝐴 = 𝑥 2 𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 𝑥𝑦 2 𝑧𝑒⃗⃗⃗⃗𝑦 + 𝑥𝑦𝑧 2 ⃗⃗⃗
𝑒𝑧
show that
∆𝐴 = ∇(∇. 𝐴) − ∇ × ∇ × 𝐴
∆𝐴 = ∇2 𝐴 = (
∆𝐴 = ∇2 𝐴 = ⃗⃗⃗⃗
𝑒𝑥
𝜕2
𝜕2
𝜕2
+
+
𝑒𝑥 + 𝐴𝑦 ⃗⃗⃗⃗
𝑒𝑦 + 𝐴𝑧 ⃗⃗⃗
𝑒𝑧 )
) (𝐴𝑥 ⃗⃗⃗⃗
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
𝜕2 2
𝜕2
𝜕2
2
(𝑥
(𝑥𝑦
(𝑥𝑦𝑧 2 )
𝑦𝑧)
+
𝑒
⃗⃗⃗⃗
𝑧)
+
𝑒
⃗⃗⃗
𝑦
𝑧
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
∆𝐴 = ∇2 𝐴 = 2𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 2𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 2𝑥𝑦𝑒
⃗⃗⃗𝑧
∇(∇. 𝐴) = ∇ (
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
+
+
⃗⃗⃗𝑧
) = ∇(2𝑥𝑦𝑧 + 2𝑥𝑦𝑧 + 2𝑥𝑦𝑧) = 6𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 6𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 6𝑥𝑦𝑒
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑒⃗⃗⃗⃗𝑥
𝜕
∇ × ∇ × 𝐴 = ∇ × ||
𝜕𝑥
𝐴𝑥
𝑒⃗⃗⃗⃗𝑦
𝜕
𝜕𝑦
𝐴𝑦
𝑒𝑥
⃗⃗⃗⃗
𝑒⃗⃗⃗𝑧
𝜕
𝜕|
= ∇ × ||
|
𝜕𝑥
𝜕𝑧
𝐴𝑧
𝑥 2 𝑦𝑧
𝑒⃗⃗⃗⃗𝑦
⃗⃗⃗
𝑒𝑧
𝜕
𝜕 |
𝜕𝑦
𝜕𝑧 |
𝑥𝑦 2 𝑧 𝑥𝑦𝑧 2
∇ × ∇ × 𝐴 = ∇ × [𝑒⃗⃗⃗⃗𝑥 (𝑥𝑧 2 − 𝑥𝑦 2 ) − ⃗⃗⃗⃗
𝑒𝑦 (𝑦𝑧 2 − 𝑥 2 𝑦) + ⃗⃗⃗
𝑒𝑧 (𝑦 2 𝑧 − 𝑥 2 𝑧)]
𝑒⃗⃗⃗⃗𝑥
𝜕
∇ × ∇ × 𝐴 = ||
𝜕𝑥
𝑥𝑧 2 − 𝑥𝑦 2
𝑒⃗⃗⃗⃗𝑦
𝑒𝑧
⃗⃗⃗
𝜕
𝜕
||
𝜕𝑦
𝜕𝑧
−𝑦𝑧 2 + 𝑥 2 𝑦 𝑦 2 𝑧 − 𝑥 2 𝑧
∇ × ∇ × 𝐴 = ⃗⃗⃗⃗
𝑒𝑥 (2𝑦𝑧 + 2𝑦𝑧) − ⃗⃗⃗⃗
𝑒𝑦 (−2𝑥𝑧 − 2𝑥𝑧) + ⃗⃗⃗
𝑒𝑧 (2𝑥𝑦 + 2𝑥𝑦)
∇ × ∇ × 𝐴 = 4𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 4𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 4𝑥𝑦𝑒
⃗⃗⃗𝑧
∇(∇. 𝐴) − ∇ × ∇ × 𝐴 = (6𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 6𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 6𝑥𝑦𝑒
⃗⃗⃗𝑧 ) − (4𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 4𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 4𝑥𝑦𝑒
⃗⃗⃗𝑧 )
∇(∇. 𝐴) − ∇ × ∇ × 𝐴 = 2𝑦𝑧𝑒⃗⃗⃗⃗𝑥 + 2𝑥𝑧𝑒⃗⃗⃗⃗𝑦 + 2𝑥𝑦𝑒⃗⃗⃗𝑧
6