Question

(a) Which of the vectors in Problem 1.15 can be expressed as the gradient of a scalar? Find a scalar function that does the job.

(b) Which can be expressed as the curl of a vector? Find such a vector.

Short answer

Write the given vectors.

$$\begin{gathered} v_a=x^{2}i+3x{z}^{2}j+(-2xz)k \\ {v}_b=y^{2}i+2yzj+3zxk \\ {v}_c=y^{2}i+(2xy+z^2)j+2yzk \end{gathered}$$

Step-by-step solution

Describe the given information

Write the given vectors.

$$\begin{gathered} v_a=x^{2}i+3x{z}^{2}j+(-2xz)k \\ {v}_b=y^{2}i+2yzj+3zxk \\ {v}_c=y^{2}i+(2xy+z^2)j+2yzk \end{gathered}$$

Define the line integral

The gradient of a scalar function $\mathit{F}$ is defined as$\mathbf{\nabla }\mathit{F}$and curl of a vector function$v$is defined as $\mathbf{\nabla }\mathbf{\times }\mathit{v}$.

Find the scalar function for part (a)

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)+\frac{\partial }{\partial y}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\\ +\frac{\partial }{\partial z}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\end{array}\right] \\ =2x+0-2x \\ =0 \end{gathered}$$

The dot product of vector $v_b$is obtained as follows:

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(y^{2}i+2yzj+3zxk)+\frac{\partial }{\partial y}(y^{2}i+2yzj+3zxk)\\ +\frac{\partial }{\partial z}(y^{2}i+2yzj+3zxk)\end{array}\right] \\ =y+2z+3x \end{gathered}$$

The dot product of vector $v_c$is obtained as follows:

$$\begin{gathered} \nabla .v_a=\left[\begin{array}{c}\frac{\partial }{\partial x}(y^{2}i+(2xy+z^2)j+2yzk)+\frac{\partial }{\partial y}(y^{2}i+(2xy+z^2)j+2yzk)\\ +\frac{\partial }{\partial z}(y^{2}i+(2xy+z^2)j+2yzk)\end{array}\right] \\ =0+(2x+0)+2y \\ =2(x+y) \end{gathered}$$

Step: 4 Find the cross product of the given vectors

The cross product of vector $v_a$is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^2& 3x{z}^2& -2xz\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}(-2zx)-\frac{\partial }{\partial z}\left(3x{z}^2\right)\right)-j\left(\frac{\partial }{\partial x}(-2xz)-\frac{\partial }{\partial z}\left(x^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}\left(3x{z}^2\right)-\frac{\partial }{\partial y}\left(x^2\right)\right)\end{array}\right]/ \\ =-6xzi+2zj+3z^{2}k \end{gathered}$$

The cross product of vector $v_b$is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y^2& 2yz& 3zx\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}\left(3zx\right)-\frac{\partial }{\partial z}\left(2yz\right)\right)-j\left(\frac{\partial }{\partial x}\left(3zx\right)-\frac{\partial }{\partial z}\left(y^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial y}\left(y^2\right)\right)\end{array}\right] \\ =-2yi-3zj-xk \end{gathered}$$

The cross product of vector$v_c$ is obtained as follows:

$$\begin{gathered} \nabla \times v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y^2& 2xy+z^2& x\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{\partial }{\partial y}\left(2yz\right)-\frac{\partial }{\partial z}(2xy+z^2)\right)-j\left(\frac{\partial }{\partial x}\left(2yz\right)-\frac{\partial }{\partial z}\left(y^2\right)\right)\\ +i\left(\frac{\partial }{\partial x}(2xy+z^2)-\frac{\partial }{\partial y}\left(y^2\right)\right)\end{array}\right] \\ =0 \end{gathered}$$

Out of the all the given functions, the curl of vector $v_c$is 0. So, it can be expressed as gradient of scalar function.

Let us assume $\nabla A-v_c$. Expand $\nabla A-v_c$ as,

$\frac{\partial A}{\partial x}i+\frac{\partial A}{\partial y}j+\frac{\partial A}{\partial z}k=y^{2}i+(2xy+z^2)j+2yzk$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \frac{\partial A}{\partial x}=y^2 \\ A=y^{2}x+g(y,z) \end{gathered}$$

Where, $g(y,z)$ is a function of y , z.

Differentiate the above function with respect to y.\

$\frac{\partial A}{\partial y}=2yx+\frac{\partial g(y,z)}{\partial y}$

Where, role="math" localid="1655958836210" $g(y,z)$is a function of y , z.

comparing the right and left side of the above equation, as,

$2xy+z^2=2yx+\frac{\partial g(y,z)}{\partial y}$

We obtain the result, $\frac{\partial g(y,z)}{\partial y}=z^2$, On integrating this equation with respect to y on both sides, we get,

$g(y,z)=z^{2}y+f\left(z\right)$

Thus, the scalar function can be written as $A=y^{2}x+z^{2}y+f\left(z\right)$.

Differentiate the above function with respect to z.

$\frac{\partial A}{\partial z}-2yz+f^{\text{'}}\left(z\right)$

Comparing the right and left side of the above equation, as,

$2yz=2yz+f^{\text{'}}\left(z\right)$

Obtain the result, $f^{\text{'}}\left(z\right)=0$, On integrating this equation with respect to z on both sides, we get,

role="math" localid="1655958018234" $f^{\text{'}}\left(z\right)=C$

Here, C is some constant. Thus the scalar function is obtained as $A=y^{2}x+z^{2}y+C$ .

 Find the vector function for part (b)

Out of the all the given functions, the dot product of vector $v_a$is 0. So, it can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume$\nabla \times F=v_a$. Expand $\nabla \times F=v_a$as,

$$\begin{gathered} \nabla \times F=v_a \\ \left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right|-x^{2}i+3z^{2}j+(-2xz)k \\ \left(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right)i-\left(\frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z}\right)j+\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)k=x^{2}i+3z^{2}j+(-2xz)k \end{gathered}$$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \left(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right)=x^2......\left(1\right) \\ \left(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\right)=3x{z}^2.......\left(2\right) \\ \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)=2xz.........\left(3\right) \end{gathered}$$

To calculate $F_x$,$F_y$ , assume that $F_z=0$ . Substitute 0 for $F_z$int equation (2).

$$\begin{gathered} \left(\frac{\partial F_z}{\partial x}-\frac{\partial \left(0\right)}{\partial z}\right)=-3x{z}^2 \\ \frac{\partial F_z}{\partial x}=-3x{z}^2 \\ {F}_z=-\frac{3}{2}{x}^2{z}^2+f(x,y) \end{gathered}$$

Substitute 0 for $F_x$ int equation (3).

$$\begin{gathered} \left(\frac{\partial F_y}{\partial x}-\frac{\partial \left(0\right)}{\partial }\right)=-2xz \\ \frac{\partial Fy}{\partial x}=-2xz \\ {F}_y=-x^{2}z+g(y,z) \end{gathered}$$

Substitute $-x^{2}z+g(y,z)$ for $F_{y,}-\frac{3}{2}-x^2{z}^2+f(x,y)$for $F_z$ into equation (1).

$$\begin{gathered} \left(\frac{\partial }{\partial y}\left(-\frac{3}{2}-x^2{z}^2+f(x,y)\right)-\frac{\partial }{\partial z}(-x^{2}z+g(y,z\left)\right)\right)=x^2 \\ \frac{\partial }{\partial y}f(y,z)+x^2-\frac{\partial }{\partial z}g(y,z)=x^2 \\ \frac{\partial }{\partial y}f(y,z)+x^2-\frac{\partial }{\partial z}g(y,z)=0 \end{gathered}$$

The resulting expression $\frac{\partial }{\partial y}f(y,z)-\frac{\partial }{\partial z}g(y,z)=0$ is possible only when .

$f(y,z)=g(y,z)=0$

Thus, the vector $F=F_{x}i+F_{y}j+F_{z}k$ can be written as $F=x^{2}zj-\frac{3}{2}{x}^2{z}^{2}k$.