Answer to Question #237968 in Classical Mechanics for Emma

Question

Q8. Calculate the divergence and Curl, of the following functions

a. V= x2 i+3xy^2 j -2xyz k

b. V=xyi+2yz j+3zx k

c. V= y2i +(2xy+z2)j+ 2yz k , where i , j , k are unit vectors along x, y and z axis

Answer 1

$${\rm div}{\bf V}=\frac{\partial V_x}{\partial x}+\frac{\partial V_y }{\partial y}+\frac{\partial V_z}{\partial z}$$

$$\rm curl\:{\bf V}=\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x&V_y&V_z \end{vmatrix}$$

(a)

$${\rm div}{\bf V}=\frac{\partial (x^2)}{\partial x}+\frac{\partial (3xy^2) }{\partial y}+\frac{\partial (-2xyz)}{\partial z}\\=2x+6xy-2xy=2x+4xy$$

$${\rm curl\:{\bf V}}=\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2&3xy^2&-2xyz \end{vmatrix}$$ $$=-2xy\hat i+2yz\hat j+3x\hat k$$

(b)

$${\rm div}{\bf V}=\frac{\partial (xy)}{\partial x}+\frac{\partial (2yz) }{\partial y}+\frac{\partial (3zx)}{\partial z}\\=y+2z+3x$$

$${\rm curl\:{\bf V}}=\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy&2yz&3zx \end{vmatrix}$$

$$=-2y\hat i-3z\hat j-x\hat k$$

(c)

$${\rm div}{\bf V}=\frac{\partial (y^2)}{\partial x}+\frac{\partial (2xy+z^2) }{\partial y}+\frac{\partial (2yz)}{\partial z}\\=2x+2y$$

$${\rm curl\:{\bf V}}=\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2&2xy+z^2&2yz \end{vmatrix}$$

$$=0\hat i+0\hat j+0\hat k=\bf 0$$