(Vector-fields question) Are these two essentially the same question?

Question

I've been given these two questions by my lecturer:

The vector field $\space F \space$ is defined by $\space F= {x^3}{y^2}(4yz\underline{i} + 3xz\underline{j} + xy\underline{k}) \space$

(i) Show that the vector field $ \space F \space$ is conservative.

(ii) Find the scalar function of position $\space \phi(x,y,z) \space$ such that: $\space F= \nabla\phi$

For $\space F \space$ to be conservative I basically need to find a function $\space f(x,y,z) \space$ that satisfies $\space F= \nabla f$.

Which is essentially number two right? Or have I gone horribly wrong somewhere?

Answer 1

You can check that the curl vanishes $\;\nabla\times F=0\;$ (zero vector), and since the first partial derivatives of the component functions are continuous $\;F\;$ is conservative.

Now, suppose $\;F=\nabla\phi\;$, then:

$$\phi=\int 4x^3y^3z\,dx=x^4y^3z+K(y,z) \;(=\text{constant wrt}\;\;x)\implies$$

$$\frac{\partial\phi}{\partial y}=3x^4y^2z+K'_y(y,z)\stackrel{\text{must be}}=3x^4y^2z\implies K_y'(y,z)=0\implies K(y,z)=C+C(z)\implies$$

$$\frac{\partial\phi}{\partial z}=x^4y^3+C'(z)\stackrel{\text{must be}}=x^4y^3\implies C'(z)=0\implies C(z)=C=(\text{ constant}\,)$$

and thus $\;\phi(x,y,z)=x^4y^3z+C\;$

Thus, in fact, with (i) you make sure $\;\phi\;$ in (ii) exists, and then you go out to find it.

Answer 2

As DonAnotonio has pointed out in his answer, the the two parts of the problem are related, but not quite the same. In part (i) you show that $\phi$ exists, and in part (ii) you find it. You could solve part (i) by solving part (ii) first, but it’s more usual to solve (i) by showing that the curl of $\mathbf F$ vanishes.

For part (ii), a different way to appraoch it is to define $\phi(x,y,z)=\int_\Gamma\mathbf F(\mathbf r)\cdot d\mathbf r$, where $\Gamma$ is a smooth path joining the origin to the point $(x,y,z)$. By part (i), the value of this integral depends only on the endpoints of $\Gamma$. It’s not very difficult to show that $\nabla\phi=\mathbf F$.

To find a closed form for $\phi$, choose a convenient path $\Gamma$, such as the line segment from the origin to $(x,y,z)$ with the simple parametrization $\gamma:t\mapsto(tx,ty,tz)$ for $t\in[0,1]$. This gives us $$\phi(x,y,z)=\int_0^1\mathbf F(tx,ty,tz)\cdot(x,y,z)\,dt.$$ I’ll leave evaluating this integral with the specific $\mathbf F$ in the problem to you.

I find this approach less tedious than the usual alternating integration/differentiation method once the number of dimensions goes beyond 2 or 3. It also has the virtue of being extendable to a method for finding antiderivatives of $k$-forms.