Chapter 3, Problem 59P

(a)

To determine

The solenoidal vector fields among the vector fields $A$, $B$, and $C$.

Explanation of Solution

Given:

The vector field $\left(A\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

The vector field $\left(B\right)$ is $2\rho \cos \varphi a_{\rho }-4\rho \sin \varphi a_{\varphi }+3a_z$.

The vector field $\left(C\right)$ is $\sin \theta a_r+r\sin \theta a_{\varphi }$.

Calculation:

Calculate the divergence of the vector $A$ $\left(\nabla \cdot A\right)$ using the relation.

  $\nabla \cdot A=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}$

  $\begin{array}{c}\nabla \cdot A=\frac{\partial \left(x\right)}{\partial x}+\frac{\partial \left(y\right)}{\partial y}+\frac{\partial \left(z\right)}{\partial z}\\ =1+1+1\\ =3\\ \ne 0\end{array}$

Calculate the divergence of the vector $B$ $\left(\nabla \cdot B\right)$ using the relation.

  $\nabla \cdot B=\frac{1}{\rho }\frac{\partial \left(\rho B_{\rho }\right)}{\partial \rho }+\frac{1}{\rho }\frac{\partial B_{\varphi }}{\partial \varphi }+\frac{\partial B_z}{\partial z}$

  $\begin{array}{c}\nabla \cdot B=\frac{1}{\rho }\frac{\partial \left(\rho 2\rho \cos \varphi \right)}{\partial \rho }+\frac{1}{\rho }\frac{\partial \left(-4\rho \sin \varphi \right)}{\partial \varphi }+\frac{\partial \left(3\right)}{\partial z}\\ =4\cos \varphi -4\cos \varphi +0\\ =0\end{array}$

Calculate the divergence of the vector $C$ $\left(\nabla \cdot C\right)$ using the relation.

  $\nabla \cdot C=\frac{1}{r^2}\frac{\partial \left(r^2{C}_r\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(C_{\theta }\sin \theta \right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial C_{\varphi }}{\partial \varphi }$

  $\begin{array}{c}\nabla \cdot C=\frac{1}{r^2}\frac{\partial \left(r^2\sin \theta \right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(\left(0\right)\sin \theta \right)}{\partial \theta }+\frac{1}{r\sin \theta }\frac{\partial \left(r\sin \theta \right)}{\partial \varphi }\\ =\frac{2r\sin \theta }{r^2}+0+0\\ =\frac{2\sin \theta }{r}\\ \ne 0\end{array}$

Thus, the vector field $\underset{\_}{B}$ is solenoidal.

(b)

To determine

The irrotational vector field among the vector fields $A$, $B$, and $C$.

Explanation of Solution

Given:

The vector field $\left(A\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

The vector field $\left(B\right)$ is $2\rho \cos \varphi a_{\rho }-4\rho \sin \varphi a_{\varphi }+3a_z$.

The vector field $\left(C\right)$ is $\sin \theta a_r+r\sin \theta a_{\varphi }$.

Calculation:

Calculate the curl of the vector $A$ $\left(\nabla \times A\right)$ using the relation.

  $\nabla \times A=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)a_x+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)a_y+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)a_z$

  $\begin{array}{c}\nabla \times A=\left(\frac{\partial \left(z\right)}{\partial y}-\frac{\partial \left(y\right)}{\partial z}\right)a_x+\left(\frac{\partial \left(x\right)}{\partial z}-\frac{\partial \left(z\right)}{\partial x}\right)a_y+\left(\frac{\partial \left(y\right)}{\partial x}-\frac{\partial \left(x\right)}{\partial y}\right)a_z\\ =0+0+0\\ =0\end{array}$

Calculate the curl of the vector $B$ $\left(\nabla \times B\right)$ using the relation.

  $\nabla \times B=\left[\frac{1}{\rho }\frac{\partial B_z}{\partial \varphi }-\frac{\partial B_{\varphi }}{\partial z}\right]a_{\rho }+\left[\frac{\partial B_{\rho }}{\partial z}-\frac{\partial B_z}{\partial \rho }\right]a_{\varphi }+\frac{1}{\rho }\left[\frac{\partial \left(\rho B_{\varphi }\right)}{\partial \rho }-\frac{\partial B_{\rho }}{\partial \varphi }\right]a_z$

  $\begin{array}{c}\nabla \times B=\left[\begin{array}{l}\left[\frac{1}{\rho }\frac{\partial \left(3\right)}{\partial \varphi }-\frac{\partial \left(-4\rho \sin \varphi \right)}{\partial z}\right]a_{\rho }+\left[\frac{\partial \left(\rho 2\rho \cos \varphi \right)}{\partial z}-\frac{\partial \left(3\right)}{\partial \rho }\right]a_{\varphi }+\\ \frac{1}{\rho }\left[\frac{\partial \left(\rho \left(-4\rho \sin \varphi \right)\right)}{\partial \rho }-\frac{\partial \left(\rho 2\rho \cos \varphi \right)}{\partial \varphi }\right]a_z\end{array}\right]\\ =2{\rho }^2\sin \varphi -8\sin \varphi \\ \ne 0\end{array}$

Calculate the curl of the vector $C$ $\left(\nabla \times C\right)$ by using the relation.

  $\nabla \times C=\left[\begin{array}{l}\frac{1}{r\sin \theta }\left[\frac{\partial \left(A_{\varphi }\sin \theta \right)}{\partial \theta }-\frac{\partial \left(A_{\theta }\right)}{\partial \varphi }\right]a_r+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial A_r}{\partial \varphi }-\frac{\partial \left(r{A}_{\varphi }\right)}{\partial r}\right]a_{\theta }+\\ \frac{1}{r}\left[\frac{\partial \left(r{A}_{\theta }\right)}{\partial r}-\frac{\partial A_r}{\partial \theta }\right]a_{\varphi }\end{array}\right]$

  $\begin{array}{c}\nabla \times C=\left[\begin{array}{l}\frac{1}{r\sin \theta }\left[\frac{\partial \left(r\sin \theta \sin \theta \right)}{\partial \theta }-\frac{\partial \left(0\right)}{\partial \varphi }\right]a_r+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial \left(\sin \theta \right)}{\partial \varphi }-\frac{\partial \left(r\left(r\sin \theta \right)\right)}{\partial r}\right]a_{\theta }+\\ \frac{1}{r}\left[\frac{\partial \left(r\left(0\right)\right)}{\partial r}-\frac{\partial \left(\sin \theta \right)}{\partial \theta }\right]a_{\varphi }\end{array}\right]\\ =2\cos \theta a_r-2a_{\theta }+\frac{\cos \theta }{r}{a}_{\varphi }\\ \ne 0\end{array}$

Thus, the vector field $\underset{\_}{A}$ is irrotational.