Chapter 3, Problem 41P

(a)

To determine

The curl of vector field $A$ and divergence of the curl.

Explanation of Solution

Given:

The vector field $\left(A\right)$ is $x^{2}y{a}_x+y^{2}z{a}_y-2xz{a}_z$.

Calculation:

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

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

  $\begin{array}{c}\nabla \times A=\left(\frac{\partial \left(-2xz\right)}{\partial y}-\frac{\partial \left(y^{2}z\right)}{\partial z}\right)a_x+\left(\frac{\partial \left(x^{2}y\right)}{\partial z}-\frac{\partial \left(-2xz\right)}{\partial x}\right)a_y+\left(\frac{\partial \left(y^{2}z\right)}{\partial x}-\frac{\partial \left(x^{2}y\right)}{\partial y}\right)a_z\\ =\left(0-y^2\right)a_x+\left(0+2z\right)a_y+\left(0-x^2\right)a_z\\ =-y^2{a}_x+2z{a}_y-x^2{a}_z\end{array}$

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

  $\nabla \cdot \left(\nabla \times A\right)=\frac{\partial {\left(\nabla \times A\right)}_x}{\partial x}+\frac{\partial {\left(\nabla \times A\right)}_y}{\partial y}+\frac{\partial {\left(\nabla \times A\right)}_z}{\partial z}$

  $\begin{array}{c}\nabla \cdot \left(\nabla \times A\right)=\frac{\partial \left(-y^2\right)}{\partial x}+\frac{\partial \left(2z\right)}{\partial y}+\frac{\partial \left(-x^2\right)}{\partial z}\\ =0\end{array}$

Thus, the curl of vector field $A$ is $\underset{\_}{-y^2{a}_x+2z{a}_y-x^2{a}_z}$ and divergence of the curl is $\underset{\_}{0}$.

(b)

To determine

The curl of vector field $A$ and divergence of the curl.

Explanation of Solution

Given:

The vector field $\left(A\right)$ is ${\rho }^{2}z{a}_{\rho }+{\rho }^3{a}_{\varphi }+3\rho z^2\varphi a_z$.

Calculation:

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

  $\nabla \times A=\left(\frac{1}{\rho }\frac{\partial A_z}{\partial \varphi }-\frac{\partial A_{\varphi }}{\partial z}\right)a_{\rho }+\left(\frac{\partial A_{\rho }}{\partial z}-\frac{\partial A_z}{\partial \rho }\right)a_{\varphi }+\frac{1}{\rho }\left(\frac{\partial A_{\varphi }}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)a_z$

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

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

  $\nabla \cdot \left(\nabla \times A\right)=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho {\left(\nabla \times A\right)}_{\rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{\left(\nabla \times A\right)}_{\varphi }+\frac{\partial }{\partial z}{\left(\nabla \times A\right)}_z$

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

Thus, the curl of vector field $A$ is $\underset{\_}{\left({\rho }^2-3z^2\right)a_{\varphi }+4{\rho }^2{a}_z}$ and divergence of the curl is $\underset{\_}{0}$.

(c)

To determine

The curl of vector field $A$ and divergence of the curl.

Explanation of Solution

Given:

The vector field $\left(A\right)$ is $\frac{\sin \varphi }{r^2}{a}_r-\frac{\cos \varphi }{r^2}{a}_{\theta }$.

Calculation:

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

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

  $\begin{array}{c}\nabla \times A=\left[\begin{array}{l}\frac{1}{r\sin \theta }\left(\frac{\partial \left(0\right)}{\partial \theta }-\frac{\partial \left(-\frac{\cos \varphi }{r^2}\right)}{\partial \varphi }\right)a_r+\frac{1}{r}\left(\frac{1}{\sin \theta }\frac{\partial \left(\frac{\sin \varphi }{r^2}\right)}{\partial \varphi }-\frac{\partial \left(0\right)}{\partial r}\right)a_{\theta }+\\ \frac{1}{r}\left(\frac{\partial \left(r\left(-\frac{\cos \varphi }{r^2}\right)\right)}{\partial r}-\frac{\partial \left(\frac{\sin \varphi }{r^2}\right)}{\partial \theta }\right)a_{\varphi }\end{array}\right]\\ =\left[\frac{1}{r\sin \theta }\left(0-\frac{\sin \varphi }{r^2}\right)a_r+\frac{1}{r}\left(\frac{1}{\sin \theta }\frac{\cos \varphi }{r^2}\right)a_{\theta }+\frac{1}{r}\left(\frac{\cos \varphi }{r^2}-0\right)a_{\varphi }\right]\\ =-\frac{\sin \varphi }{r^3\sin \theta }{a}_r+\frac{\cos \varphi }{r^3\sin \theta }{a}_{\theta }+\frac{\cos \varphi }{r^3}{a}_{\varphi }\end{array}$

  $\nabla \cdot \left(\nabla \times A\right)=0$

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

  $\nabla \cdot \left(\nabla \times A\right)=\left[\begin{array}{l}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2{\left(\nabla \times A\right)}_r\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left({\left(\nabla \times A\right)}_{\theta }\sin \theta \right)+\\ \frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{\left(\nabla \times A\right)}_{\varphi }\end{array}\right]$

  $\begin{array}{c}\nabla \cdot \left(\nabla \times A\right)=\left[\begin{array}{l}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\left(-\frac{\sin \varphi }{r^3\sin \theta }\right)\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\frac{\cos \varphi }{r^3\sin \theta }\sin \theta \right)+\\ \frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(\frac{\cos \varphi }{r^3}\right)\end{array}\right]\\ =\frac{1}{r^2}\left(\frac{\sin \varphi }{r^2\sin \theta }\right)+\frac{1}{r\sin \theta }\left(0\right)-\frac{1}{r\sin \theta }\left(\frac{\cos \varphi }{r^3}\right)\\ =\frac{\sin \varphi }{r^4\sin \theta }-\frac{\sin \varphi }{r^4\sin \theta }\end{array}$

Thus, the curl of vector field $A$ is $\underset{\_}{\left(-\frac{\sin \varphi }{r^3\sin \theta }{a}_r+\frac{\cos \varphi }{r^3\sin \theta }{a}_{\theta }+\frac{\cos \varphi }{r^3}{a}_{\varphi }\right)}$ and divergence of the curl is $\underset{\_}{0}$.