Q50P

Question

(a) Let $F_{1} = x^{2} i$ and $F_{2} = x i + y j + z k$ Calculate the divergence and curl of $F_{1}$ and $F_{2}$ which one can be written as the gradient of a scalar? Find a scalar potential that does the job. Which one can be written as the curl of a vector? Find a suitable vector potential.

(b) Show that $F_{3} = y z i + z x j + x y k$ can be written both as the gradient of a scalar and as the curl of a vector. Find scalar and vector potentials for this function.

Step-by-Step Solution

Verified

Answer

(a) $F_{2}$ can be expressed as gradient of a scalar. $F_{1}$ can be expressed as curl

of some vector. The vector potential can be written as .

(b) The vector potential is $G = G_{x} i + G_{y} j + G_{z} k$ can be written as $G = \frac{1}{4} \{x (z^{2} - y^{2}) i + y (x^{2} - z^{2}) j + z (y^{2} - x^{2}) k\} .$ The scalar potential for $F_{3}$ is $ϕ = x y z + c$

1Step 1: Describe the given information

Write the given vectors.

$F_{1} = x^{2} k F_{2} = x i + y j + z k F_{3} = y z i + z x j + x y k$

2Step 2: Define the line integral

The gradient of a scalar function $F$ is defined as $\nabla F$ and curl of a vector function is defined as $\nabla \times V$ .

3Step3: Find the divergence and curl of vector R for part (a)

(a)

The dot product of vector R is obtained as follows:

$\nabla . R = [\frac{\partial r_{1}}{\partial x} + \frac{\partial r_{2}}{\partial y} + \frac{\partial r_{3}}{\partial z}]$

The curl of vector R is obtained as follows:

$\nabla \times R = |\begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ r_{1} & r_{2} & r_{3} \end{matrix}| = i (\frac{\partial r_{1}}{\partial y} - \frac{\partial r_{3}}{\partial z}) - j (\frac{\partial r_{3}}{\partial x} - \frac{\partial r_{1}}{\partial z}) + k (\frac{\partial r_{2}}{\partial x} - \frac{\partial r_{1}}{\partial y})$

The Divergence of vector $F_{1}$ is obtained as follows:

$\nabla . F_{1} = [\frac{\partial (0)}{\partial x} + \frac{\partial (0)}{\partial y} + \frac{\partial (x^{2})}{\partial z}] = 0$

The Divergence of vector $F_{2}$ is obtained as follows:

$\nabla . F_{2} = [\frac{\partial (x)}{\partial x} + \frac{\partial (y)}{\partial y} + \frac{\partial (z)}{\partial z}] = 1 + 1 + 1 = 3$

Thus, divergence of vectors $F_{1}$ and $F_{2}$ is obtained as $\nabla . F_{1} = 0$ and $\nabla . F_{2} = 3$ .

The Curl of vector is obtained as follows:

$\nabla . F_{1} = | \begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & x^{2} \end{matrix} | = i (\frac{\partial (x^{2})}{\partial y} - \frac{\partial (0)}{\partial z}) - j (\frac{\partial (x^{2})}{\partial x} - \frac{\partial (0)}{\partial z}) + k (0) = - 2 x j$

The curl of vector $F_{2}$ is obtained as follows:

$\nabla \times F_{1} = |\begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & x^{2} \end{matrix}| = i (\frac{\partial (z)}{\partial y} - \frac{\partial (y)}{\partial z}) - j (\frac{\partial (z)}{\partial x} - \frac{\partial (x)}{\partial z}) + k (\frac{\partial (y)}{\partial x} - \frac{\partial (x)}{\partial z}) = 0$

Thus, curl of vectors $F_{1}$ and $F_{2}$ is obtained as $\nabla \times F_{1} = (- 2 x) j$ and $\nabla \times F_{2} = 0$ .

The divergence of vector $F_{1}$ is 0. So, $F_{1}$ can be expressed as curl of some vector. The divergence of vector $F_{2}$ is 0. So, $F_{2}$ it can be expressed as gradient of a scalar.

Out of the all the given functions, the curl of vector $F_{2}$ is 0. So, it can be expressed as gradient of scalar function.

Let us assume $F_{2} = \nabla V$ . Expand $F_{2} = \nabla V$ as,

$x i + y j + z k = \frac{\partial V_{x}}{\partial X} i + \frac{\partial V_{y}}{\partial y} j + \frac{\partial V_{z}}{\partial Z} k$

On comparing the right and left side of the above equation, we obtain,

$\frac{\partial V_{X}}{\partial X} = X \frac{\partial V_{y}}{\partial y} = y \frac{\partial V_{z}}{\partial Z} = Z$

Integrate the above functions as,

$\int \partial V_{x} = \int x \partial x v_{x} = \frac{x^{2}}{2} + C_{1} \int \partial V_{y} = \int y \partial y v_{y} = \frac{y^{2}}{2} + C_{2} \int \partial V_{Z} = \int Z \partial Z v_{z} = \frac{z^{2}}{2} + C_{3}$

Thus, the scalar function v can be written as $v = \frac{1}{2} (x^{2} + y^{2} + z^{2}) + c$ . Thus the scalar function is obtained as $v = \frac{1}{2} (x^{2} + y^{2} + z^{2}) + c$ .

Out of the all the given functions, the dot product of vector $F_{1}$ is 0. So, it can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume $\nabla \times F_{1} = A$ . Expand $\nabla \times F_{1} = A$ as,

$\nabla \times A = F_{1}$

$|\begin{matrix} i & i & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_{X} & A_{Y} & A_{Z} \end{matrix}| = x^{2} (\frac{\partial A_{Z}}{\partial y} - \frac{\partial A y}{\partial y}) i - (\frac{\partial A_{Z}}{\partial x} - \frac{\partial A_{x}}{\partial z}) j + (\frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y}) k = x^{2} k$

On comparing the right and left side of the above equation, we obtain,

$(\frac{\partial A_{Z}}{\partial y} - \frac{\partial A y}{\partial z}) = 0 . . . . . . . . . . . . . . . . (1) (\frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}) = 0 . . . . . . . . . . . . . . . . (2) (\frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y}) = x^{2} . . . . . . . . . . . . . . . . (3)$

On comparing above result, we obtain, $A_{x} = 0$ and $A_{y} = 0$ . Comparing

equation (1),

$(\frac{\partial A_{y}}{\partial x} - \frac{\partial A x}{\partial y}) = x^{2} \frac{\partial A_{y}}{\partial x} = x^{2}$

Integrate the resulting (1),

$\int \frac{\partial A_{y}}{\partial x} \partial x = \int x^{2} \partial x \int \partial A_{y} = \int x^{2} \partial x A_{y} = \frac{x^{3}}{3} + C$

Thus, the vector potential can be written as $A = (\frac{x^{3}}{3} + c) j$ .

4Step 4: Find the divergence and curl of vector F 3 in part (b)

(b)

The curl of vector $F_{3}$ is obtained as follows:

$\nabla \times F_{3} = |\begin{matrix} i & i & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y z & z x & x y \end{matrix}| = i (\frac{\partial (x y)}{\partial y} - \frac{\partial (z x)}{\partial z}) - j (\frac{\partial (x y)}{\partial y} - \frac{\partial (y z)}{\partial z}) + k (\frac{\partial (z x)}{\partial x} - \frac{\partial (y z)}{\partial z}) = 0$

Thus, curl of vector $F_{3}$ is obtained as $\nabla \times F_{3} = 0$ . Thus, can also be written as gradient of scalar function.

The vector $F_{3}$ can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume $F_{3} = \nabla \times G$ . Expand $F_{3} = \nabla \times G$ as,

$F_{3} = \nabla \times G$

$(\frac{\partial G_{x}}{\partial y} - \frac{\partial G_{y}}{\partial z}) i - (\frac{\partial G_{z}}{\partial x} - \frac{\partial G_{x}}{\partial z}) j + (\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y}) k = y z i + z x j + x y k$

On comparing the right and left side of the above equation, we obtain,

$(\frac{\partial G_{z}}{\partial y} - \frac{\partial G_{y}}{\partial z}) = y z . . . . . . . . . . . . . (1) (\frac{\partial G_{x}}{\partial z} - \frac{\partial G_{z}}{\partial x}) = x z . . . . . . . . . . . . . (2) (\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y}) = x y . . . . . . . . . . . . . (3)$

Integrate $(\frac{\partial G_{z}}{\partial y} - \frac{\partial G_{y}}{\partial z}) = y z$

$G_{Z} = \frac{y^{2} z}{4} + f (x, z) . . . . . . . . . . (1) G_{Y} = \frac{y z^{2}}{4} + g (x, y) . . . . . . . . . (2)$

Integrate $(\frac{\partial G_{x}}{\partial z} - \frac{\partial G_{z}}{\partial x}) = x z$ and $(\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y}) = x y$

$G_{x} = \frac{z^{2} x}{4} + h (x, y) . . . . . . . . . . (3) G_{y} = \frac{z^{2} x}{4} + j (x, y) . . . . . . . . . (4)$

Integrate $(\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y})$ =xy

$G_{y} = \frac{x^{2} y}{4} + f (x, y) . . . . . . . . . . (5) G_{x} = \frac{x^{2} y}{4} + I (x, y) . . . . . . . . . (6)$

From the equations, (1), (2), (3), (4), (5), (6), the vector $G = G_{x} i + G_{y} j + G_{z} k$

can be written as $G = \frac{1}{4} \{x (z^{2} - y^{2}) i + y (x^{2} - z^{2}) j + z (y^{2} - x^{2}) k\}$ .

Let the function $F_{3}$ is written in terms of scalar potential as $F_{3} = \nabla ϕ$ . The divergence of scalar potential $ϕ$ is written as,

$\nabla ϕ = \frac{\partial ϕ}{\partial x} \overset{\land}{x} + \frac{\partial ϕ}{\partial y} \overset{\land}{y} + \frac{\partial ϕ}{\partial z} \overset{\land}{z}$

On comparing equation $F_{3} = y z i + z x j + x y k,$ with $\nabla ϕ = \frac{\partial ϕ}{\partial x} \overset{\land}{x} + \frac{\partial ϕ}{\partial y} \overset{\land}{y} + \frac{\partial ϕ}{\partial z} \overset{\land}{z}$ , we