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### Section 16.6 : Conservative Vector Fields

7. Find the potential function for the following vector field.

$\vec F = \frac{{2xy}}{{{z^3}}}\vec i + \left( {2y - {z^2} + \frac{{{x^2}}}{{{z^3}}}} \right)\vec j - \left( {4{z^3} + 2yz + \frac{{3{x^2}y}}{{{z^4}}}} \right)\vec k$

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Start Solution

Now, by assumption from how the problem was asked, we can assume that the vector field is conservative and because we don’t know how to verify this for a 3D vector field we will just need to trust that it is.

Let’s start off the problem by labeling each of the components to make the problem easier to deal with as follows.

\begin{align*}P & = \frac{{2xy}}{{{z^3}}}\\ Q & = 2y - {z^2} + \frac{{{x^2}}}{{{z^3}}}\\ R & = - \left( {4{z^3} + 2yz + \frac{{3{x^2}y}}{{{z^4}}}} \right) = - 4{z^3} - 2yz - \frac{{3{x^2}y}}{{{z^4}}}\end{align*}

Be careful with these problems and watch the signs on the vector components. One of the biggest mistakes that students make with these problems is to miss the minus sign that is in front of the third component of the vector field. There won’t always be a minus sign of course, but on occasion there will be one and if we miss it the rest of the problem will be very difficult to do. In fact, if we miss it we won’t be able to find a potential function for the vector field!

Show Step 2

To find the potential function for this vector field we know that we need to first either integrate $$P$$ with respect to $$x$$, integrate $$Q$$ with respect to $$y$$ or $$R$$ with respect $$z$$. It doesn’t matter which one we use chose to use and, in this case, it looks like none of them will be any harder than the other.

In this case the $$R$$ has quite a few terms in it so let’s integrate that one first simply because it might mean less work when dealing with $$P$$ and $$Q$$ in later steps.

\begin{align*}f\left( {x,y,z} \right) & = \int{{R\,dz}}\\ & = \int{{ - 4{z^3} - 2yz - 3{x^2}y\,{z^{ - 4}}\,dz}}\\ & = - {z^4} - y{z^2} + {x^2}y\,{z^{ - 3}} + h\left( {x,y} \right)\end{align*}

Don’t forget that, in this case, because we were integrating with respect to $$z$$ the “constant of integration” will be a function of $$x$$ and/or $$y$$!

Show Step 3

Next, we can differentiate the function from the previous step with respect to $$x$$ and set equal to $$P$$ or differentiate the function with respect to $$y$$ and set equal to $$Q$$.

Let’s differentiate with respect to $$y$$ in this case.

${f_y} = - {z^2} + {x^2}\,{z^{ - 3}} + {h_y}\left( {x,y} \right) = 2y - {z^2} + {x^2}{z^{ - 3}} = Q\hspace{0.25in} \Rightarrow \hspace{0.25in}{h_y}\left( {x,y} \right) = 2y$

Now, recall that because we integrated with respect to $$z$$ in Step 2 $$h\left( {x,y} \right)$$, and hence $${h_y}\left( {x,y} \right)$$ , should only be a function of $$x$$’s and $$y$$’s (as it is in this case). If there had been any $$z$$’s in $${h_y}\left( {x,y} \right)$$ we’d know there was something wrong at this point. Either we’d made a mistake somewhere or the vector field was not conservative.

Also note that there is no reason to expect $${h_y}\left( {x,y} \right)$$ to have both $$x$$’s and $$y$$’s in it. It is completely possible for one (or both) of the variables to not be present!

Show Step 4

We can now integrate both sides of the formula for $${h_y}\left( {x,y} \right)$$ with respect to $$y$$ to get,

$\,h\left( {x,y} \right) = {y^2} + g\left( x \right)$

Now, because $$h\left( {x,y} \right)$$ was a function of both $$x$$ and $$y$$ and we integrated with respect to $$y$$ here the “constant of integration” in this case would need to be a function of $$x$$, $$g\left( x \right)$$ in this case.

The potential function is now,

$f\left( {x,y,z} \right) = - {z^4} - y{z^2} + {x^2}y\,{z^{ - 3}} + {y^2} + g\left( x \right)$ Show Step 5

Next, we’ll need to differentiate the potential function from Step 4 with respect to $$x$$ and set equal to $$P$$. Doing this gives,

${f_x} = 2xy\,{z^{ - 3}} + g'\left( x \right) = 2xy{z^{ - 3}} = P\hspace{0.25in} \Rightarrow \hspace{0.25in}g'\left( x \right) = 0$

Remember, that as in Step 3, we have to recall what variable we are differentiating with respect to here. In this case we are differentiating with respect to $$x$$ and $$g\left( x \right)$$ should only be a function of $$x$$. Had $$g'\left( x \right)$$ contained either $$y$$’s or $$z$$’s we’d know that either we’d made a mistake or the vector field was not conservative.

Also, as shown in this problem it is completely possible for there to be no $$x$$’s at all in $$g'\left( x \right)$$.

Show Step 6

Integrating both sides of the formula for $$g'\left( x \right)$$ from Step 5 and we can see that we must have $$g\left( x \right) = c$$.

Show Step 7

Finally, putting everything together we get the following potential function for the vector field.

$\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( {x,y,z} \right) = - {z^4} - y{z^2} + {x^2}y\,{z^{ - 3}} + {y^2} + c}}$