If $\bar{F}=(xyz)^b (x^ai + y^aj + z^ak)$ be irrotational, then show that $b=0$ or $a=-1$. Find the scalar potential $\phi$.
I can solve the 1st part by solving $curl\,\bar{F}=\bar{0}$. By comparison of compnents of $i,j,k$, and some simplification, I get the conclusion: either $b=0$ or $a=-1$.
My trouble is for the 2nd part. I started with $\bar{F}=grad\,\phi$. But I am unable to get $\phi$. Please help me for $\phi$.
If $b=0$, then the vector field is
$$\vec F = x^a\,\vec\imath + y^a\,\vec\jmath + z^a\,\vec k \quad.$$
Now find a scalar function $f(x,y,z)$ such that $\nabla f = \vec F$, i.e. satisfying $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=\left(x^a,y^a,z^a\right)$.
Integrating both sides of the first component wrt $x$ gives
$$\frac{\partial f}{\partial x} = x^a \implies f(x,y,z) = \frac{x^{a+1}}{a+1} + g(y,z) \quad.$$
Next, we differentiate wrt $y$ and compare to the $y$ component of $\vec F$,
$$\frac{\partial f}{\partial y} = y^a = \frac{\partial g}{\partial y}$$
and integrate both sides wrt $y$ to recover $g$,
$$g(y,z) = \frac{y^{a+1}}{a+1} + C \implies f(x,y,z) = \frac{x^{a+1}+y^{a+1}}{a+1} + h(z)\quad.$$
Now solve for $h$ and hence $f$.
Otherwise, if $a=-1$, the field is
$$\vec F = (xyz)^b \left(\frac{\vec\imath}x + \frac{\vec\jmath}y + \frac{\vec k}z\right)\quad.$$
From the $x$ component of $\nabla f=\vec F$ we have
$$\frac{\partial f}{\partial x} = x^{b-1} y^b z^b \implies f(x,y,z) = \frac{(xyz)^b}b + g(y,z)$$
and so on, repeating the same procedure as in the previous case.
Lastly, if both $a=-1$ and $b=0$, then
$$\vec F=\dfrac{\vec\imath}x+\dfrac{\vec\jmath}y+\dfrac{\vec k}z$$
$$\implies \frac{\partial f}{\partial x} = \frac1x \implies f(x,y,z)=\log|x| + g(y,z)$$
etc.