# Consider the vector field u = 16xi - (2-y^2)j a. Find div u. Is the vector field solenoidal? b....

## Question:

Consider the vector field

{eq}u = 16x\mathbf i - (2-y^2)\mathbf j {/eq}

a. Find {eq}div \,u {/eq}. Is the vector field solenoidal?

b. Find {eq}curl \,u {/eq}. Is the vector field irrotational?

c. If possible, find a scalar potential {eq}\phi(x,y) {/eq} such that {eq}u = \nabla \phi {/eq} .

## Curl and Divergence of a Vector Field

{eq}{/eq}

Consider a vector field given by

$$\displaystyle \vec{u} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \\$$

{eq}A_x \ , \ A_y \ and \ A_z \text{ are functions of x, y and z} \\ {/eq}

The divergence of this vector field is defined as :

$$\displaystyle \bigtriangledown \cdot \vec{u} = (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}) \cdot (A_x \hat{i} + A_y \hat{j} + A_z \hat{k}) \\$$

and the curl of this vector field is defined as :

$$\displaystyle{ \bigtriangledown \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \\ \end{vmatrix} \\ }$$

{eq}{/eq}

A. To find the divergence of vector field u :

{eq}\displaystyle{\bigtriangledown \cdot \vec{u} = (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j}) \cdot (16x\hat{i} - (2-y^2)\hat{j}) \\ \Rightarrow \bigtriangledown \cdot \vec{u} = \frac{\partial}{\partial x} (16x) \hat{i} + \frac{\partial}{\partial y} (2-y^2) \hat{j} \\ \Rightarrow \bigtriangledown \cdot \vec{u} = 16 \hat{i} -2y \hat{j} \\ } {/eq}

Since the value of divergence is not zero for all x and y, the vector field is not solenoidal.

{eq}{/eq}

B. To find the curl of vector field u :

{eq}\displaystyle{ \bigtriangledown \times \vec{u} = (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j}) \times (16x\hat{i} - (2-y^2)\hat{j}) \\ \Rightarrow \bigtriangledown \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 16x & 2-y^2 & 0 \\ \end{vmatrix} \\ \Rightarrow \bigtriangledown \times \vec{u} = (\frac{\partial }{\partial y} (0) - \frac{\partial }{\partial z} (2-y^2)) \hat{i} - (\frac{\partial }{\partial x} (0) - \frac{\partial }{\partial z} (16x)) \hat{j} + (\frac{\partial }{\partial x} (2-y^2) - \frac{\partial }{\partial y} (16x)) \hat{k} \\ \Rightarrow \bigtriangledown \times \vec{u} = (0-0) \hat{i} + (0-0) \hat{j} + (0-0) \hat{k} \\ \Rightarrow \bigtriangledown \times \vec{u} = \vec{0} \\ } {/eq}

Since the curl of vector field u is 0, it is irrotational in nature.

{eq}{/eq}

C. To find the scalar potential of vector field u :

{eq}\displaystyle{\bigtriangledown \phi = \vec{u} \\ \Rightarrow (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j}) \phi = 16x \hat{i} + (2-y^2)\hat{j} \\ \Rightarrow \frac{\partial \phi}{\partial x} \hat{i} + \frac{\partial \phi}{\partial y} \hat{j} = 16x \hat{i} + (2-y^2)\hat{j} \\ \Rightarrow \frac{\partial \phi}{\partial x} = 16x \quad and \quad \frac{\partial \phi}{\partial y} = 2-y^2 \\ } {/eq}

{eq}\displaystyle{Now, \ \frac{\partial \phi}{\partial x} = 16x \\ \Rightarrow \partial \phi = 16x \partial x \\ \Rightarrow \phi = \int 16x \ dx + g(y) \\ \Rightarrow \phi = 8x^2 + g(y) + c \\ } {/eq}

Where g(y) is a function of y and c is the constant of integration.

{eq}\displaystyle{Also, \ \frac{\partial \phi}{\partial y} = 2-y^2 \\ \Rightarrow \frac{\partial }{\partial y} (8x^2 + g(y) + c ) = 2-y^2 \\ \Rightarrow 0+g'(y) + 0 = 2-y^2 \\ \Rightarrow g(y) = \int (2-y^2) dy \\ \Rightarrow g(y) = 2y - \frac{y^3}{3} + c' \\ } {/eq}

Therefore, the scalar potential can be written as

$$\displaystyle \phi (x, \ y) = 8x^2 +2y - \frac{y^3}{3} + C\\$$

{eq}\displaystyle C = c+c' \\ {/eq}