## Section4.7Use What You Know

Suppose you want to find the work done by the force $$\FF=y^2\,\xhat+y\,\yhat$$ when moving along a given curve $$C\text{.}$$ Curves can be specified in several different ways; let us consider some examples, all of which refer to the same curve, starting at $$(1,0)$$ and ending at $$(0,1)\text{.}$$

1. Consider first the parametric curve $$\rr=(1-u^2)\,\xhat+u\,\yhat\text{.}$$ It is straightforward to compute

\begin{equation*} d\rr = (-2u\,\xhat + \yhat) \, du . \end{equation*}

Since a curve is described by a single parameter, in this case $$u\text{,}$$ we write everything in terms of that parameter. Since $$\rr=x\,\xhat+y\,\yhat\text{,}$$ we have $$x=1-u^2\text{,}$$ $$y=u\text{,}$$ and therefore

\begin{align*} \Lint \FF\cdot d\rr \amp= \int_0^1 \left( u^2\,\xhat + u\,\yhat \right) \cdot (-2u\,\xhat + \yhat) \> du\\ \amp= \int_0^1 (-2u^3 + u) \,du = 0 . \end{align*}
2. In physical applications, one is rarely given an explicit parameterization of the curve, but rather some other description. For instance, the curve just discussed might have been defined by the equation $$x=1-y^2\text{.}$$ Finding the differential of both sides of this expression yields $$dx=-2y\,dy\text{,}$$ and substituting into (2.2.2) leads to

\begin{equation*} d\rr = (-2y\,\xhat + \yhat) \,dy . \end{equation*}

The computation is exactly the same as before, using $$y$$ instead of $$u\text{.}$$

3. Alternatively, one can solve for $$y\text{,}$$ obtaining $$y=\sqrt{1-x}\text{,}$$ then compute $$dy$$ in terms of $$dx\text{,}$$ then substitute into (2.2.2), obtaining

\begin{equation*} d\rr = dx\,\xhat - {{dx}\over{2\sqrt{1-x}}}\,\yhat \end{equation*}

so that  1

\begin{equation*} \Lint\FF\cdot d\rr = \int_1^0 \left({1\over2}-x\right)\,dx = 0 . \end{equation*}

It is important to realize that all of these methods work; using what you know will always yield correct answers — eventually.

In a “use what you know” strategy, you may not be sure when to stop! A common error is to substitute for $$dy$$ in terms of $$dx$$ in $$d\rr\text{,}$$ but to forget to substitute for $$y$$ in terms of $$x$$ in $$\FF\text{.}$$ The rule of thumb is that you shouldn't start integrating until you have the integral in terms of a single parameter — including correctly determining the limits in terms of that parameter. Curves are one-dimensional!

Here is another example. Suppose you want to integrate $$\Lint\FF\cdot d\rr\text{,}$$ where $$\FF=y\,\xhat\text{,}$$ and $$C$$ is the line segment from $$(1,0)$$ to $$(0,-1)\text{.}$$ Start with expression (2.2.2) for $$d\rr\text{.}$$ What do you know? Well, the slope of the line segment is clearly $$+1\text{,}$$ and its $$y$$-intercept is $$-1\text{,}$$ so the equation of the line is $$y=x-1\text{.}$$ Taking the differential of both sides, $$dy=dx\text{,}$$ so that $$d\rr=dx\,\xhat+dx\,\yhat\text{.}$$ Finally, note that $$x$$ runs from 1 to 0! Thus,

\begin{align*} \Lint\FF\cdot d\rr \amp= \int_1^0 (x-1)\,\xhat \cdot (\xhat+\yhat)\> dx\\ \amp= \int_1^0 (x-1) \,dx = \left( {~x^2\over2} - x \right) \Bigg|_1^0 = {1\over2} . \end{align*}

Yes, of course, this particular curve is easy to parameterize, but this is not always the case. Note also how easy it was to get the limits right, and thus get the correct sign, simply by always starting with expression (2.2.2) for $$d\rr\text{,}$$ and integrating from your starting point to your final point.

This is important: Vector line integrals of the form $$\Lint\FF\cdot d\rr$$ are directed integrals; the sign of the answer depends on which way you traverse the curve. You will obtain the correct sign automatically if you integrate from the beginning point to the final point, without putting in any artificial signs. As in the last example, this may result in an integral which goes from a larger value of the integration variable to a smaller one.

This sign dependence is not the case for line integrals with respect to arclength. Since $$ds=|d\rr|\text{,}$$ such integrals do not depend on which way the curve is traversed. Standard examples are arclength and mass, which must be positive! Other times you may have to think about it to get the sign right; the total charge on a wire could be positive or negative. One way to keep this straight is to remember that $$ds=|d\rr|\text{,}$$ which has an absolute value in it, which requires care with signs.

We summarize this discussion by writing

$$\Int_{-C} f \, ds = + \Lint f \, ds \quad\hbox{but}\quad \Int_{-C} \FF\cdot d\rr = - \Lint \FF\cdot d\rr\tag{4.7.1}$$

where $$-C$$ denotes the reversed curve. You will always get the sign right for vector line integrals if you are careful to put the limits in correctly (from starting point to ending point), whereas you may have to think about signs for the scalar line integrals.

Note that $$y$$ decreases along the curve.