The divergence of a continuously differentiable vector point function \(\vec{F}\) is denoted by \(\text{div}\,\vec{F}\) or \(\nabla \cdot \vec{F}\) and is defined as:
\[
\text{div}\,\vec{F} = \nabla \cdot \vec{F} = \hat{i}\cdot \frac{\partial \vec{F}}{\partial x} + \hat{j}\cdot \frac{\partial \vec{F}}{\partial y} + \hat{k}\cdot \frac{\partial \vec{F}}{\partial z} = \sum{\hat{i}\cdot \frac{\partial \vec{F}}{\partial x}}
\]
If \(\vec{F} = F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}\) then:
\[
\text{div}\,\vec{F} = \nabla \cdot \vec{F} = \left( \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right) \cdot \left( F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k} \right) = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}
\]
Note:
- The divergence of a vector point function is a scalar point function
- If \(\text{div}\,\vec{F} = 0\) then \(\vec{F}\) is called a solenoidal vector point function
Curl of a Vector Point Function
The curl of a continuously differentiable vector point function \(\vec{F}\) is denoted by \(\text{curl}\,\vec{F}\) or \(\nabla \times \vec{F}\) and is defined as:
\[
\text{curl}\,\vec{F} = \nabla \times \vec{F} = \hat{i}\times \frac{\partial \vec{F}}{\partial x} + \hat{j}\times \frac{\partial \vec{F}}{\partial y} + \hat{k}\times \frac{\partial \vec{F}}{\partial z} = \sum{\hat{i}\times \frac{\partial \vec{F}}{\partial x}}
\]
If \(\vec{F} = F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}\) then:
\[
\text{curl}\,\vec{F} = \nabla \times \vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_{1} & F_{2} & F_{3}
\end{vmatrix}
= \sum{\left( \frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z} \right)\hat{i}}
\]
Note:
- The curl of a vector point function is also a vector point function
- If \(\text{curl}\,\vec{F} = \vec{0}\), the vector point function \(\vec{F}\) is said to be irrotational
Example 1: Find the divergence and curl of \(\vec{v}=(xyz)\hat{i}+(3x^{2}y)\hat{j}+(xz^{2}-y^{2}z)\hat{k}\) at \((2,-1,1)\).
Given: \(\vec{v}=(xyz)\hat{i}+(3x^{2}y)\hat{j}+(xz^{2}-y^{2}z)\hat{k}\)
Here:
\(v_{1}=xyz,\)
\(v_{2}=3x^{2}y,\)
\(v_{3}=xz^{2}-y^{2}z\)
Divergence Calculation:
\[
\text{div}\,\vec{v} = \frac{\partial v_{1}}{\partial x} + \frac{\partial v_{2}}{\partial y} + \frac{\partial v_{3}}{\partial z}
\]
\[
= \frac{\partial}{\partial x}(xyz) + \frac{\partial}{\partial y}(3x^{2}y) + \frac{\partial}{\partial z}(xz^{2}-y^{2}z)
\]
\[
= yz + 3x^{2} + 2xz - y^{2}
\]
\[
\therefore\ \text{div}\,\vec{v}_{(2,-1,1)} = (-1)(1) + 3(2)^{2} + 2(2)(1) - (-1)^{2} = -1 + 12 + 4 - 1 = 14
\]
Curl Calculation:
\[
\text{curl}\,\vec{v} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
xyz & 3x^{2}y & xz^{2}-y^{2}z
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(xz^{2}-y^{2}z) - \frac{\partial}{\partial z}(3x^{2}y)\right)\hat{i}
\]
\[
- \left(\frac{\partial}{\partial x}(xz^{2}-y^{2}z) - \frac{\partial}{\partial z}(xyz)\right)\hat{j}
\]
\[
+ \left(\frac{\partial}{\partial x}(3x^{2}y) - \frac{\partial}{\partial y}(xyz)\right)\hat{k}
\]
\[
= -2yz\,\hat{i} - (z^{2}-xy)\,\hat{j} + (6xy-xz)\,\hat{k}
\]
\[
\therefore\ \text{curl}\,\vec{v}_{(2,-1,1)} = 2\hat{i} - 3\hat{j} - 14\hat{k}
\]
Example 2: Find \(\text{div}\,\vec{F}\) and \(\text{curl}\,\vec{F}\) at \((1,1,-1)\), where \(\vec{F}=x^{2}z\,\hat{i}-2y^{3}z^{3}\,\hat{j}+xyz^{2}\,\hat{k}\).
Given: \(\vec{F}=x^{2}z\,\hat{i}-2y^{3}z^{3}\,\hat{j}+xyz^{2}\,\hat{k}\)
Here:
\(F_{1}=x^{2}z,\)
\(F_{2}=-2y^{3}z^{3},\)
\(F_{3}=xyz^{2}\)
Divergence Calculation:
\[
\text{div}\,\vec{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}
\]
\[
= \frac{\partial}{\partial x}(x^{2}z) + \frac{\partial}{\partial y}(-2y^{3}z^{3}) + \frac{\partial}{\partial z}(xyz^{2})
\]
\[
= 2xz - 6y^{2}z^{3} + 2xyz
\]
\[
\text{At } (1,1,-1):\quad 2(1)(-1) - 6(1)^{2}(-1)^{3} + 2(1)(1)(-1) = -2 + 6 - 2 = 2
\]
\[
\therefore\ \text{div}\,\vec{F}_{(1,1,-1)} = 2
\]
Curl Calculation:
\[
\text{curl}\,\vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^{2}z & -2y^{3}z^{3} & xyz^{2}
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(xyz^{2}) - \frac{\partial}{\partial z}(-2y^{3}z^{3})\right)\hat{i}
\]
\[
- \left(\frac{\partial}{\partial x}(xyz^{2}) - \frac{\partial}{\partial z}(x^{2}z)\right)\hat{j}
\]
\[
+ \left(\frac{\partial}{\partial x}(-2y^{3}z^{3}) - \frac{\partial}{\partial y}(x^{2}z)\right)\hat{k}
\]
\[
= (xz^{2} + 6y^{3}z^{2})\hat{i} - (yz^{2} - x^{2})\hat{j} + (0 - 0)\hat{k}
\]
\[
= (xz^{2} + 6y^{3}z^{2})\hat{i} + (x^{2} - yz^{2})\hat{j}
\]
\[
\text{At } (1,1,-1):\quad (1(-1)^{2} + 6(1)^{3}(-1)^{2})\hat{i} + (1^{2} - 1(-1)^{2})\hat{j}
\]
\[
= (1 + 6)\hat{i} + (1 - 1)\hat{j} = 7\hat{i} + 0\hat{j}
\]
\[
\therefore\ \text{curl}\,\vec{F}_{(1,1,-1)} = 7\hat{i}
\]
Model problems on Divergence and Curl of Vector Functions
- Evaluate \(\text{div}\,\vec{F}\) and \(\text{curl}\,\vec{F}\) at \((1,2,3)\) where \(\vec{F}=x^{2}yz\,\hat{i}+xy^{2}z\,\hat{j}+xyz^{2}\,\hat{k}\).
Answer: \(6xyz\), \(5\hat{i}-16\hat{j}+9\hat{k}\)
-
Find \(\text{div}\,\vec{F}\) and \(\text{curl}\,\vec{F}\) at \((1,2,3)\) where \(\vec{F}=\text{grad}(x^{3}y+y^{3}z+z^{3}x-x^{2}y^{2}z^{2})\).
Answer: \(-32\), \(\vec{0}\)
-
If \(\vec{F}=(x+y+1)\hat{i}-(x+y)\hat{k}\) show that \(\vec{F}\cdot \text{curl}\,\vec{F}=0\).
Answer: Verified (showing the dot product is zero)
-
If \(u=x^{2}yz\), \(v=xy-3z^{2}\) find (i) \(\nabla (\nabla u \cdot \nabla v)\) (ii) \(\nabla \cdot (\nabla u \times \nabla v)\) at \(P(1,2,3)\).
Answer: \(-39\hat{i}+6\hat{j}-3\hat{k}\), \(0\)
-
Find the value of \(a\) for which \(\vec{F}=(2x^{2}y+z^{2})\hat{i}+(xy^{2}-x^{2}z)\hat{j}+(axyz-2x^{2}y^{2})\hat{k}\) is solenoidal.
Answer: \(a=-6\)
-
If \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\) and \(r=|\vec{r}|\) then evaluate (i) \(\nabla (r^{n})\) (ii) \(\nabla r\) (iii) \(\nabla (1/r)\) (iv) \(\nabla (\log r)\).
Answer: (i) \(nr^{n-2}\vec{r}\) (ii) \(\frac{\vec{r}}{r}\) (iii) \(-\frac{\vec{r}}{r^{3}}\) (iv) \(\frac{\vec{r}}{r^{2}}\)
-
(i) Show that \(\nabla (r^{n})=nr^{n-2}\vec{r}\). (ii) Show that \(\text{div}(r^{n}\vec{r})=(n+3)r^{n}\). Hence show that \(\frac{\vec{r}}{r^{3}}\) is solenoidal.
Answer: Verified proofs
-
Show that \(r^{n}\vec{r}\) is irrotational for any value of \(n\).
Answer: Verified (curl is zero)
-
Show that \(\text{div}(\text{grad}\,r^{n})=\nabla^{2}(r^{n})=n(n+1)r^{n-2}\), where \(r=|\vec{r}|\).
Answer: Verified proof
-
Find the value of \(a\) if the vector \(\vec{F}=(ax^{2}y+yz)\hat{i}+(xy^{2}-xz^{2})\hat{j}+(2xyz-2x^{2}y^{2})\hat{k}\) has zero divergence. Find the curl of this vector.
Answer: \(a=-1\), \(\text{curl}\,\vec{F}=(2xy-4x^{2}y)\hat{i}+y\hat{j}+(y^{2}-2axy)\hat{k}\)
-
Show that each of the following vectors are solenoidal:
i) \((-x^{2}+yz)\hat{i}+(4y-xz^{2})\hat{j}+(2xz-4z)\hat{k}\)
ii) \(3y^{4}z^{2}\hat{i}+4x^{3}z^{2}\hat{j}+3x^{2}y^{2}\hat{k}\)
Answer: Both verified (divergence is zero)
-
Calculate \(\text{curl}(\text{grad}\,f)\), given \(f(x,y,z)=x^{2}+y^{2}-z\).
Answer: \(\vec{0}\)
-
Calculate \(\text{curl}(\text{curl}\,A)\), given \(A(x,y,z)=x^{2}y\hat{i}+y^{2}z\hat{j}+z^{2}y\hat{k}\).
Answer: \((2z-2y)\hat{i}+0\hat{j}+(-2x)\hat{k}\)
-
15. Find the constants \(a,b,c\) such that \(\vec{A}=(x+2y+az)\hat{i}+(bx-3y-z)\hat{j}+(4x+cy+2z)\hat{k}\) is irrotational.
Answer: \(a=4\), \(b=2\), \(c=-1\)
The curl of a continuously differentiable vector point function \(\vec{F}\) is denoted by \(\text{curl}\,\vec{F}\) or \(\nabla \times \vec{F}\) and is defined as:
\[
\text{curl}\,\vec{F} = \nabla \times \vec{F} = \hat{i}\times \frac{\partial \vec{F}}{\partial x} + \hat{j}\times \frac{\partial \vec{F}}{\partial y} + \hat{k}\times \frac{\partial \vec{F}}{\partial z} = \sum{\hat{i}\times \frac{\partial \vec{F}}{\partial x}}
\]
If \(\vec{F} = F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}\) then:
\[
\text{curl}\,\vec{F} = \nabla \times \vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_{1} & F_{2} & F_{3}
\end{vmatrix}
= \sum{\left( \frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z} \right)\hat{i}}
\]
Note:
- The curl of a vector point function is also a vector point function
- If \(\text{curl}\,\vec{F} = \vec{0}\), the vector point function \(\vec{F}\) is said to be irrotational
Example 1: Find the divergence and curl of \(\vec{v}=(xyz)\hat{i}+(3x^{2}y)\hat{j}+(xz^{2}-y^{2}z)\hat{k}\) at \((2,-1,1)\).
Given: \(\vec{v}=(xyz)\hat{i}+(3x^{2}y)\hat{j}+(xz^{2}-y^{2}z)\hat{k}\)
Here:
\(v_{1}=xyz,\)
\(v_{2}=3x^{2}y,\)
\(v_{3}=xz^{2}-y^{2}z\)
Divergence Calculation:
\[
\text{div}\,\vec{v} = \frac{\partial v_{1}}{\partial x} + \frac{\partial v_{2}}{\partial y} + \frac{\partial v_{3}}{\partial z}
\]
\[
= \frac{\partial}{\partial x}(xyz) + \frac{\partial}{\partial y}(3x^{2}y) + \frac{\partial}{\partial z}(xz^{2}-y^{2}z)
\]
\[
= yz + 3x^{2} + 2xz - y^{2}
\]
\[
\therefore\ \text{div}\,\vec{v}_{(2,-1,1)} = (-1)(1) + 3(2)^{2} + 2(2)(1) - (-1)^{2} = -1 + 12 + 4 - 1 = 14
\]
Curl Calculation:
\[
\text{curl}\,\vec{v} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
xyz & 3x^{2}y & xz^{2}-y^{2}z
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(xz^{2}-y^{2}z) - \frac{\partial}{\partial z}(3x^{2}y)\right)\hat{i}
\]
\[
- \left(\frac{\partial}{\partial x}(xz^{2}-y^{2}z) - \frac{\partial}{\partial z}(xyz)\right)\hat{j}
\]
\[
+ \left(\frac{\partial}{\partial x}(3x^{2}y) - \frac{\partial}{\partial y}(xyz)\right)\hat{k}
\]
\[
= -2yz\,\hat{i} - (z^{2}-xy)\,\hat{j} + (6xy-xz)\,\hat{k}
\]
\[
\therefore\ \text{curl}\,\vec{v}_{(2,-1,1)} = 2\hat{i} - 3\hat{j} - 14\hat{k}
\]
Example 2: Find \(\text{div}\,\vec{F}\) and \(\text{curl}\,\vec{F}\) at \((1,1,-1)\), where \(\vec{F}=x^{2}z\,\hat{i}-2y^{3}z^{3}\,\hat{j}+xyz^{2}\,\hat{k}\).
Given: \(\vec{F}=x^{2}z\,\hat{i}-2y^{3}z^{3}\,\hat{j}+xyz^{2}\,\hat{k}\)
Here:
\(F_{1}=x^{2}z,\)
\(F_{2}=-2y^{3}z^{3},\)
\(F_{3}=xyz^{2}\)
Divergence Calculation:
\[
\text{div}\,\vec{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}
\]
\[
= \frac{\partial}{\partial x}(x^{2}z) + \frac{\partial}{\partial y}(-2y^{3}z^{3}) + \frac{\partial}{\partial z}(xyz^{2})
\]
\[
= 2xz - 6y^{2}z^{3} + 2xyz
\]
\[
\text{At } (1,1,-1):\quad 2(1)(-1) - 6(1)^{2}(-1)^{3} + 2(1)(1)(-1) = -2 + 6 - 2 = 2
\]
\[
\therefore\ \text{div}\,\vec{F}_{(1,1,-1)} = 2
\]
Curl Calculation:
\[
\text{curl}\,\vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^{2}z & -2y^{3}z^{3} & xyz^{2}
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(xyz^{2}) - \frac{\partial}{\partial z}(-2y^{3}z^{3})\right)\hat{i}
\]
\[
- \left(\frac{\partial}{\partial x}(xyz^{2}) - \frac{\partial}{\partial z}(x^{2}z)\right)\hat{j}
\]
\[
+ \left(\frac{\partial}{\partial x}(-2y^{3}z^{3}) - \frac{\partial}{\partial y}(x^{2}z)\right)\hat{k}
\]
\[
= (xz^{2} + 6y^{3}z^{2})\hat{i} - (yz^{2} - x^{2})\hat{j} + (0 - 0)\hat{k}
\]
\[
= (xz^{2} + 6y^{3}z^{2})\hat{i} + (x^{2} - yz^{2})\hat{j}
\]
\[
\text{At } (1,1,-1):\quad (1(-1)^{2} + 6(1)^{3}(-1)^{2})\hat{i} + (1^{2} - 1(-1)^{2})\hat{j}
\]
\[
= (1 + 6)\hat{i} + (1 - 1)\hat{j} = 7\hat{i} + 0\hat{j}
\]
\[
\therefore\ \text{curl}\,\vec{F}_{(1,1,-1)} = 7\hat{i}
\]
Answer: \(6xyz\), \(5\hat{i}-16\hat{j}+9\hat{k}\)
Answer: \(-32\), \(\vec{0}\)
Answer: Verified (showing the dot product is zero)
Answer: \(-39\hat{i}+6\hat{j}-3\hat{k}\), \(0\)
Answer: \(a=-6\)
Answer: (i) \(nr^{n-2}\vec{r}\) (ii) \(\frac{\vec{r}}{r}\) (iii) \(-\frac{\vec{r}}{r^{3}}\) (iv) \(\frac{\vec{r}}{r^{2}}\)
Answer: Verified proofs
Answer: Verified (curl is zero)
Answer: Verified proof
Answer: \(a=-1\), \(\text{curl}\,\vec{F}=(2xy-4x^{2}y)\hat{i}+y\hat{j}+(y^{2}-2axy)\hat{k}\)
Answer: Both verified (divergence is zero)
Answer: \(\vec{0}\)
Answer: \((2z-2y)\hat{i}+0\hat{j}+(-2x)\hat{k}\)
Answer: \(a=4\), \(b=2\), \(c=-1\)