This document discusses vector calculus concepts and operations involving the del operator (∇). It provides proofs and examples of: 1. Curl of the gradient of a scalar field is equal to zero. 2. The curl of the curl of a vector field is equal to the gradient of the divergence minus the Laplace operator. 3. The curl of the product of a scalar field and a vector field is equal to the scalar field times the curl of the vector field plus the cross product of the gradient of the scalar field and the vector field. It also defines the Laplace operator and discusses its representation in spherical coordinates.

1. Product Rules
MSc Mathematics 3rd

2. Contents
1. Prove that curl of gradient of 𝛗 𝐢𝐬 𝟎.
2. 𝐏𝐭𝐨𝐯𝐞 𝐭𝐡𝐚𝐭 𝛁 × 𝛁 × 𝑽 = 𝛁 𝛁. 𝑽 − 𝛁 𝟐
𝑽 .
3. Prove that 𝛁 × 𝐕 × 𝐁 = 𝐁. 𝛁 𝐕 − 𝐁 𝛁. 𝑽 − 𝐕. 𝛁 𝐁 + 𝑽 𝛁. 𝐁
4. Prove that 𝛁 × (𝝋𝑽)=φ(𝛁×𝐕) + ( 𝛁φ)×V
5. Examples
6. Del Operator
7. Laplacian
8. MCQs

3. Del Operator
𝛁 = (
𝜕
𝜕x
𝐢 +
𝜕
𝜕x
𝐣+
𝜕
𝜕x
𝐤)
Del operator = differential operator.
By itself it has no specific use.
like
𝑑
dx
It can be used to find the gradient of a scalar. → (vector)
It can be used to find the divergence of a vector . → (scalar)
It can be used to find the curl of a vector field . → (vector)

4. Prove that 𝛁×(φV)=φ(𝛁×V) + ( 𝛁φ)×V)
Proof: let V=𝑽1i+𝑽2j+𝑽3k is a vector & 𝜑(x, y, z) is a scalar .
Then
𝛁×(φV)=
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜑𝑽1 𝜑𝑽2 𝜑𝑽3
=[φ
𝜕𝑽3
𝜕y
+
𝜕φ
𝜕𝑦
𝑽3 − φ
𝜕𝑽2
𝜕z
− (
𝜕φ
𝜕𝑧
)𝑽2]i+[φ
𝜕𝑽1
𝜕z
+ (
𝜕φ
𝜕𝑧
)𝑽1 − φ
𝜕𝑽3
𝜕x
−
(
𝜕φ
𝜕𝑥
)𝑽3]j +[φ
𝜕𝑽2
𝜕x
+
𝜕φ
𝜕𝑥
𝑽2 − φ
𝜕𝑽1
𝜕y
− (
𝜕φ
𝜕𝑦
)𝑽1]k

5. .=[φ
𝜕𝑽3
𝜕y
+
𝜕φ
𝜕𝑦
𝑽3 − φ
𝜕𝑽2
𝜕z
− (
𝜕φ
𝜕𝑧
)𝑽2]i+[φ
𝜕𝑽1
𝜕z
+ (
𝜕φ
𝜕𝑧
)𝑽1 − φ
𝜕𝑽3
𝜕x
−
(
𝜕φ
𝜕𝑥
)𝑽3]j +[φ
𝜕𝑽2
𝜕x
+
𝜕φ
𝜕𝑥
𝑽2 − φ
𝜕𝑽1
𝜕y
− (
𝜕φ
𝜕𝑦
)𝑽1]k
By taking common phi from the terms which contains phi .
= 𝜑 [(
𝜕𝑽3
𝜕y
−
𝜕𝑽2
𝜕y
)i+(
𝜕𝑽1
𝜕z
−
𝜕𝑽3
𝜕x
)j+(
𝜕𝑽2
𝜕x
−
𝜕𝑽1
𝜕y
)k] +[
𝜕𝜑
𝜕𝑦
𝑽3 −
𝜕𝜑
𝜕𝑧
𝑽2 i +
𝜕𝜑
𝜕𝑧
𝑽1 −
𝜕𝜑
𝜕𝑥
𝑽3 +
𝜕𝜑
𝜕𝑥
𝑽2 −
𝜕𝜑
𝜕𝑦
𝑽1 k]
=φ 𝛁 × 𝑽 +
i j k
𝜕φ
𝜕x
𝜕φ
𝜕y
𝜕φ
𝜕z
𝑽1 𝑽2 𝑽3
𝛁×(φV)= φ(𝛁×𝐕) + ( 𝛁φ)×V

6. Example: If V=y𝑧2i-3x𝑧2j+2xyzk & 𝜑(x, y, z)=xyz then find
𝛁×(φV)=φ(𝛁×V) + ( 𝛁φ)×V)
Solution: given that V=y𝑧2i-3x𝑧2j+2xyzk & 𝜑(x, y, z)=xyz . Then
φV =x𝑦2
𝑧3
i-3𝑥2
𝑦𝑧3
j+2𝑥2
𝑦2
𝑧2
k
𝛁×(φV)=
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
x𝑦2 𝑧3 −3𝑥2 𝑦𝑧3 2𝑥2 𝑦2 𝑧2
𝛁×(φV)= 13𝑥2 𝑦𝑧2I -x𝑦2 𝑧2j -8x𝑦𝑧3k … (1)
φ(𝛁×V)=xyz
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
y𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧

7. so
φ(𝛁×V)= 8𝑥2 𝑦𝑧2 𝐢 -4x𝑦𝑧3k &
𝛁φ)×V=
𝑖 𝑗 𝑘
𝑦𝑧 𝑥𝑧 𝑥𝑦
y𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧
𝛁φ)×V = 5𝑥2 𝑦𝑧2I -x𝑦2 𝑧2j -4x𝑦𝑧3k
φ(𝛁×V) + ( 𝛁φ)×V)= 8𝑥2 𝑦𝑧2 𝐢 -4x𝑦𝑧3k + 5𝑥2 𝑦𝑧2I -x𝑦2 𝑧2j -4x𝑦𝑧3k
φ(𝛁×V) + ( 𝛁φ)×V)= 13𝑥2 𝑦𝑧2I -x𝑦2 𝑧2j -8x𝑦𝑧3k …(2)
From equation (1)&(2) prove that
𝛁×(φV)=φ(𝛁×V) + ( 𝛁φ)×V)

8. Prove that curl of gradient of 𝛗 𝐢𝐬 𝟎.or
Prove that 𝛁 × (𝛁𝛗)=𝟎
Proof: let 𝜑(x, y, z) is a scalar function.
𝛁 × (𝛁φ)=𝛁 × (
𝜕φ
𝜕x
𝐢 +
𝜕φ
𝜕x
𝐣+
𝜕φ
𝜕x
𝐤)
𝛁 × (𝛁φ) =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕φ
𝜕𝑥
𝜕φ
𝜕𝑦
𝜕φ
𝜕𝑧
𝛁 × (𝛁φ)= [
𝜕
𝜕𝑦
(
𝜕φ
𝜕𝑧
) −
𝜕
𝜕𝑧
(
𝜕φ
𝜕𝑦
)]𝒊+[
𝜕
𝜕𝑧
(
𝜕φ
𝜕𝑥
) −
𝜕
𝜕𝑥
(
𝜕φ
𝜕𝑧
)]𝒋 +[
𝜕
𝜕𝑥
(
𝜕φ
𝜕𝑦
) −
𝜕
𝜕𝑦
(
𝜕φ
𝜕𝑥
)]𝒌
Hence prove that
𝛁 × (𝛁φ)= 0i + 0j + 0k = 0

9. Example: If 𝜑(x, y, z)=xyz is a scalar function then find 𝛁 × (𝛁φ)=0
Solution: Given that 𝜑(x, y, z)=xyz is a scalar function.
Now we find
𝛁 × (𝛁φ)=𝛁 × (
𝜕φ
𝜕x
𝐢 +
𝜕φ
𝜕x
𝐣+
𝜕φ
𝜕x
𝐤)
𝛁 × (𝛁φ) = 𝛁 × (
𝜕(𝑥𝑦𝑧)
𝜕x
𝐢 +
𝜕(𝑥𝑦𝑧)
𝜕x
𝐣+
𝜕 𝑥𝑦𝑧
𝜕x
𝐤)
𝛁 × (𝛁φ)= 𝛁 × (𝐲𝐳 𝐢 + 𝐱𝐳 𝐣+xy 𝐤)
𝛁 × (𝛁φ) =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑦𝑧 𝑥𝑧 𝑥𝑦
𝛁 × (𝛁φ)= [𝑥 − 𝑥]𝒊+[𝑦 − 𝑦]𝒋 +[z − 𝑧]𝒌
𝛁 × (𝛁φ)= 0i + 0j + 0k = 0

10. 𝐏𝐭𝐨𝐯𝐞 𝐭𝐡𝐚𝐭 𝛁 × 𝛁 × 𝑽 = 𝛁 𝛁. 𝑽 − 𝛁 𝟐
𝑽
Proof:
𝛁 × 𝛁 × 𝑽 = 𝛁 ×
i j k
𝜕
𝜕x
𝜕
𝜕y
𝜕
𝜕z
𝑽1 𝑽2 𝑽3
𝛁 × 𝛁 × 𝑽 = 𝛁 × [(
𝜕𝑽3
𝜕y
−
𝜕𝑽2
𝜕y
)i+(
𝜕𝑽1
𝜕z
−
𝜕𝑽3
𝜕x
)j+(
𝜕𝑽2
𝜕x
−
𝜕𝑽1
𝜕y
)k]
𝛁 × 𝛁 × 𝑽 =
i j k
𝜕
𝜕x
𝜕
𝜕y
𝜕
𝜕z
𝜕𝑽3
𝜕y
−
𝜕𝑽2
𝜕y
𝜕𝑽1
𝜕z
−
𝜕𝑽3
𝜕x
𝜕𝑽2
𝜕x
−
𝜕𝑽1
𝜕y

11. = [
𝜕
𝜕𝑦
𝜕𝑽2
𝜕x
−
𝜕𝑽1
𝜕y
−
𝜕
𝜕𝑧
(
𝜕𝑽1
𝜕𝑧
−
𝜕𝑽3
𝜕𝑥
)]𝒊 + [
𝜕
𝜕𝑧
𝜕𝑽3
𝜕𝑦
−
𝜕𝑽2
𝜕𝑧
−
𝜕
𝜕𝑥
(
𝜕𝑽2
𝜕x
−
𝜕𝑽1
𝜕y
)]𝐣 + [
𝜕
𝜕𝑥
𝜕𝑽1
𝜕𝑧
−
𝜕𝑽3
𝜕𝑥
−
𝜕
𝜕𝑦
(
𝜕𝑽3
𝜕𝑦
−
𝜕𝒗2
𝜕𝑧
)]𝐤
= (−
𝜕2 𝑽1
𝜕𝑦2 −
𝜕2 𝑽1
𝜕𝑧2 )i+ −
𝜕2 𝑽2
𝜕𝑧2 −
𝜕2 𝑽2
𝜕𝑥2 𝐣 + −
𝜕2 𝑽3
𝜕𝑥2 −
𝜕2 𝑽3
𝜕𝑦2 𝒌 + (
𝜕2 𝑽2
𝜕𝑦𝜕𝑥
+
𝜕2 𝑽3
𝜕𝑧𝜕𝑥
)i+
𝜕2 𝑽3
𝜕𝑧𝜕𝑦
+
𝜕2 𝑽1
𝜕𝑦𝜕𝑥
𝐣 +
𝜕2 𝑽1
𝜕𝑧𝜕𝑥
+

12. Laplacian (Laplace operator)
The Laplace operator is defined as
∆= 𝛁. 𝛁 where 𝛁 is the usual del operator.
The laplacian is also written as 𝛁2
.
A function F is said to be satisfy laplacian if ∆F= 𝛁. 𝛁(𝐅)=0
For example ∆ & 𝛁 looks very different in spherical coordinates .In
particular:
𝛁 = (
𝝏
𝝏𝒓
,
𝟏
𝒓
𝜕
𝜕𝜃
,
𝟏
𝒓𝒔𝒊𝒏𝜃
𝜕
𝜕𝜑
)
and
∆= 𝛁. 𝛁= (
𝝏
𝝏𝒓 𝟐
𝝏
𝝏𝒓
𝒓 𝟐 𝝏
𝝏𝒓
+
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜃
𝜕
𝜕𝜃
( sin𝜃
𝜕
𝜕𝜃
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏 𝟐 𝜃
(
𝜕
𝜕𝜑
𝜕
𝜕𝜑
) )

13. Prove that 𝛁 × 𝐕 × 𝐁 = 𝐁. 𝛁 𝐕 − 𝐁 𝛁. 𝑽 − 𝐕. 𝛁 𝐁 +
𝑽 𝛁. 𝐁
Proof: Let two vectors V=𝐕𝟏 𝐢+𝐕𝟐 𝐣+𝐕𝟑 𝐤 & B=𝐁 𝟏 𝐢+𝐁 𝟐 𝐣+𝐁 𝟑 𝒌 then
𝑽 × 𝑩 =
𝑖 𝑗 𝑘
V1 V2 V3
B1 B2 B3
= (V2 𝐵3 − V2 𝐵3)i + (V1 𝐵3 − V3 𝐵1)J + (V1 𝐵2 − V2 𝐵1)k
𝛁 × 𝐕 × 𝐁 =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
V2 𝐵3 − V2 𝐵3 V1 𝐵3 − V3 𝐵1 V1 𝐵2 − V2 𝐵1
𝛁 × 𝐕 × 𝐁 = i ((
𝜕
𝜕𝑦
(V1 𝐵2 − V2 𝐵1) -
𝜕
𝜕𝑧
(V1 𝐵3 − V3 𝐵1))
- j(
𝜕
𝜕𝑥
(V1 𝐵2 − V2 𝐵1) -
𝜕
𝜕𝑧
(V2 𝐵3 − V3 𝐵2)) + k(
𝜕
𝜕𝑦
(V2 𝐵3 − V3 𝐵2) -
𝜕
𝜕𝑥
(V3 𝐵1 − V1 𝐵3))

14. .Now solving for i component:
= (
𝜕
𝜕𝑦
(V1 𝐵2 − V2 𝐵1) -
𝜕
𝜕𝑧
(V1 𝐵3 − V3 𝐵1)) …..(A)
= 𝑽1
𝜕𝐵2
𝜕𝑦
+ 𝑩 𝟐
𝜕V1
𝜕𝑦
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑩 𝟏
𝜕V2
𝜕𝑦
− 𝑩 𝟑
𝜕V3
𝜕𝑧
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
+ 𝑩 𝟐
𝜕V1
𝜕𝑧
= 𝑽1
𝜕𝐵2
𝜕𝑦
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
….. (1)
+𝑩 𝟐
𝜕v1
𝜕𝑦
− 𝑩 𝟏
𝜕v2
𝜕𝑦
− 𝑩 𝟑
𝜕𝑣3
𝜕𝑧
+ 𝑩 𝟐
𝜕v1
𝜕𝑧
….. (2)
first we Solve equation (1):
= 𝑽1
𝜕𝐵2
𝜕𝑦
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
= 𝑽1
𝜕𝐵2
𝜕𝑦
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
By Adding and Subtracting the term 𝒗1
𝜕𝑩 𝟏
𝜕𝑥
:
= 𝑽1
𝜕𝐵2
𝜕𝑦
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
+ 𝑽1
𝜕𝑩 𝟏
𝜕𝑥
− 𝑽1
𝜕𝑩 𝟏
𝜕𝑥
=[ 𝑽1
𝜕𝑩 𝟏
𝜕𝑥
+ 𝑽1
𝜕𝐵2
𝜕𝑦
+ 𝑽1
𝜕𝑩 𝟑
𝜕𝑧
] − [𝑽1
𝜕𝑩 𝟏
𝜕𝑥
−𝑽2
𝜕𝑩 𝟏
𝜕𝑦
− 𝑽3
𝜕𝑩 𝟑
𝜕𝑧
]
= 𝑽 𝟏 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟏

15. .
By Solving equation ( 2 ):
= 𝐁 𝟐
𝜕v1
𝜕y
− 𝐁 𝟏
𝜕v2
𝜕y
− 𝐁 𝟏
𝜕v3
𝜕z
+ 𝐁 𝟑
𝜕V1
𝜕z
= −𝐁 𝟏
𝜕V2
𝜕y
− 𝐁 𝟏
𝜕V3
𝜕z
+ 𝐁 𝟐
𝜕V1
𝜕y
+ 𝐁 𝟑
𝜕V1
𝜕z
by adding and subtracting the term 𝐁1
𝜕𝑽 𝟏
𝜕x
:
= −𝐁1(
𝜕𝑽 𝟏
𝜕x
+
𝜕V2
𝜕y
+
𝜕V3
𝜕z
) + (𝐁1
𝜕
𝜕x
+ 𝐁 𝟐
𝜕
𝜕y
+ 𝐁 𝟑
𝜕
𝜕z
)𝑉1
= −𝐁 𝟏 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟏
Thus, equation (A) becomes from the result of equation (1) & (2):
= 𝑽 𝟏 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟏 − 𝐁 𝟏 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟏 … (B)
Similarly for j component:
= 𝑽 𝟐 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟐 − 𝐁 𝟐 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟐 … (C)
& similarly for k component:
= 𝑽 𝟑 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟑 − 𝐁 𝟑 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟑 … (D)
Now by adding equations (B),(C) &(D) :

16. .
𝛁 × 𝐕 × 𝐁 = 𝑽 𝟏 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟏 − 𝐁 𝟏 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟏
+ 𝑽 𝟐 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟐 − 𝐁 𝟐 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟐
+ 𝑽 𝟑 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟑 − 𝐁 𝟑 𝛁. 𝑽 − 𝐁. 𝛁 𝑽 𝟑
By rearranging the terms we have
= (𝑽 𝟏 i + 𝑽 𝟐j +𝑽 𝟑 k) 𝛁. 𝐁 - 𝐕. 𝛁 𝐁 𝟏i + 𝐁 𝟐j + 𝐁 𝟑k − 𝐁 𝟏i + 𝐁 𝟐j + 𝐁 𝟑k 𝛁. 𝑽 −
𝐁. 𝛁 (𝑽 𝟏i + 𝑽 𝟐j +𝑽 𝟑 k)
Hence proved that 𝛁 × 𝐕 × 𝐁 = 𝐁. 𝛁 𝐕 − 𝐁 𝛁. 𝑽 − 𝐕. 𝛁 𝐁 + 𝑽 𝛁. 𝐁

17. Example:
L.H.S
Let A=x𝑧2
𝐢 +2yj -3xzk & B=3xzi +2yzj -𝑧2
k
A × B =
𝑖 𝑗 𝑘
𝑥𝑧2
2𝑦 −3𝑥𝑧
3xz 2yz −𝑧2
A × B =(-2y𝑧2
+6xy𝑧2
) i + (-9𝑥2
𝑧2
+x𝑧4
) j + (2xy𝑧3
-2xyz) k
𝛁 × 𝐀 × 𝐁 =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
(−2𝑦𝑧2
+ 6xy𝑧2
) (−9𝑥2
𝑧2
+x𝑧4
) (2xy𝑧3
−2xyz)
𝛁 × 𝐀 × 𝐁 =(-2𝑧3
x-2𝑥𝑧+18𝑥2
z) i + (12xyz+6yz-2y𝑧3
) j +
(𝑧4
+2𝑧2
-24x𝑧3
) k …..(1)

18. R.H.S
𝐁. 𝛁 𝑨=x𝑧3I + 4yj - 6𝑧2k
𝐀. 𝛁 𝐁=(3x𝑧3-9𝑥2z) i +4yj +6x𝑧2 k
A 𝛁. 𝐁 =3x𝑧3
I +6yz -9x𝑧2
k
𝐁 𝛁. 𝑨 =(3x𝑧3+6x𝑧2 − 9𝑥2z) i + (4y𝑧2 −6xyz+2y𝑧3) j + (−𝑧4 −2𝑧2+3x𝑧2)k
Now 𝐁. 𝛁 𝐀 − 𝐁 𝛁. 𝑨 − 𝐀. 𝛁 𝐁 + 𝑨 𝛁. 𝐁 = {x𝑧3
I + 4yj - 6𝑧2
k -
((3x𝑧3
+6x𝑧2
− 9𝑥2
z) i + (4y𝑧2
−6xyz+2y𝑧3
) j + (−𝑧4
−2𝑧2
+3x𝑧2
) k) – ((3x𝑧3
-
9𝑥2
z) i +4yj +6x𝑧2
k) + 3x𝑧3
I +6yz -9x𝑧2
k}
𝐁. 𝛁 𝐀 − 𝐁 𝛁. 𝑨 − 𝐀. 𝛁 𝐁 + 𝑨 𝛁. 𝐁 ={(−2𝑧3
x−2𝑥𝑧+18𝑥2
z) i +
(12xyz+6yz−2y𝑧3
) j + (𝑧4
+2𝑧2
−24x𝑧3
) k} …..(2)
Hence proved that 𝛁 × 𝐕 × 𝐁 = 𝐁. 𝛁 𝐕 − 𝐁 𝛁. 𝑽 − 𝐕. 𝛁 𝐁𝑽 𝛁. 𝐁

19. MCQs
1.Curl of Gradient …
(a) Scalar (b) Vector (c) Real number
2. Del operator is a
(a) Vector field (b) differential operator (c) an object