Calculate the curls of the vector functions in Prob. 1.15.

The curls of the vector functions$v=x^{2}i+3x{z}^{2}j-3xzk,v=xyi+2yzj-3zxk,$$andv=y^{2}i(2xy+z^2)j-2yzk$ has to be evaluated.

The curl of a vector is defined as $\mathbf{\nabla }\mathbf{\times }\mathit{v}$ , expressed as

$\mathbf{\nabla }\mathbf{\times }\mathit{v}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{(}{\mathit{v}}_{\mathbf{x}}\mathit{i}\mathbf{+}{\mathit{v}}_{\mathbf{y}}\mathbf{}\mathit{j}\mathbf{+}{\mathit{v}}_{\mathbf{2}}\mathit{k}\mathbf{)}$

localid="1656138740796" $\mathbf{=}\mathbf{\left|}\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\\ {\mathbf{v}}_{\mathbf{x}}& {\mathbf{v}}_{\mathbf{y}}& {\mathbf{v}}_{\mathbf{z}}\end{array}\mathbf{\right|}$

Compute curl of vector V .

$\nabla \times \mathrm{v}=\nabla \times ({\mathrm{v}}_{\mathrm{x}}\mathrm{i}+{\mathrm{v}}_{\mathrm{y}}\mathrm{j}+{\mathrm{v}}_{\mathrm{z}}\mathrm{k})$

Solve using matrix form as,

$$\begin{gathered} \nabla \times v=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ v_x& v_y& v_z\end{array}\right| \\ =i(\frac{\partial v_x}{\partial y}-\frac{\partial v_y}{\partial z})-j(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z})+k(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}) \end{gathered}$$

(a)

The vector is defined as. $v=x^{2}i+3x{z}^{2}j-3xzk$. The components of vector v are written as

role="math" localid="1656140041010" $$\begin{gathered} v_x=x^2 \\ {v}_y=3x{z}^2 \\ {v}_z=3xz \end{gathered}$$

Substitute $x^2$for $v_x$ , $3x{z}^2$for $v_y$, and 3xz for $v_z$into .

role="math" localid="1656140148512" $$\begin{gathered} \nabla \times v=i(\frac{\partial v_x}{\partial y}-\frac{\partial v_y}{\partial z})-j(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z})+k(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}) \\ \nabla \times v=i(\frac{\partial }{\partial y}\left(-2xz\right)-\frac{\partial }{\partial z}\left(3x{z}^2\right))-j(\frac{\partial }{\partial x}\left(-2xz\right)-\frac{\partial }{\partial z}\left(x^2\right))+k(\frac{\partial }{\partial x}\left(3x{z}^2\right)-\frac{\partial }{\partial y}\left(x^2\right)) \\ =(0-6xz)i+(0+2z)j(3z^2-0)k \\ =-6xzi+2zj+3z^{2}k \end{gathered}$$

(b)

The vector v is defined as. $v=xyi+2yzj-3zxk$. The components of vector v are written as

role="math" localid="1656140562578" $$\begin{gathered} v_x=xy \\ {v}_y=2yz \\ {v}_z=3zx \end{gathered}$$

Substitute xy for $v_x$, $2yz$ for$v_y$ , and 3zx for $v_z$ into .

role="math" localid="1656140718462" $$\begin{gathered} \nabla \times v=i(\frac{\partial v_x}{\partial y}-\frac{\partial v_y}{\partial z})-j(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z})+k(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}) \\ \nabla \times v=i(\frac{\partial }{\partial y}\left(3zx\right)-\frac{\partial }{\partial z}\left(2yz\right))-j(\frac{\partial }{\partial x}\left(3zx\right)-\frac{\partial }{\partial z}\left(xy\right))+k(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial y}\left(xy\right)) \\ =(0-2y)i+(0-3z)j+(0-x)k \\ =-2yi+3zj-xk \end{gathered}$$

Thus, the curl of vector v is obtained as $\nabla \times v=-2yi+3zj-xk$.

(c)

The vector v is defined as. $v=y^{2}i(2xy+z^2)j-2yzk$. The components of vector are written as

$$\begin{gathered} v_x=y^2 \\ {v}_y=2xy+z^2 \\ {v}_z=2yz \end{gathered}$$

Substitute $y^2$for $v_x$,$(2xy+z^2)$for $v_y$ , and 2yz for $v_z$ into

$$\begin{gathered} \nabla \times v=i(\frac{\partial v_x}{\partial y}-\frac{\partial v_y}{\partial z})-j(\frac{\partial v_z}{\partial x}-\frac{\partial v_x}{\partial z})+k(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}) \\ \nabla \times v=i(\frac{\partial }{\partial y}\left(2yz\right)-\frac{\partial }{\partial z}\left(2xy+z^2\right))-j(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial z}\left(xy\right))+k(\frac{\partial }{\partial x}\left(2xy+z^2\right)-\frac{\partial }{\partial y}\left(y^2\right)) \\ =(2z-2z)i+(0-0)j+(2y-2y)k \\ =0 \end{gathered}$$

Thus, the curl of vector v is obtained as$\nabla \times v=0$