Using Parameterizations to Calculate Line Integrals

Let \(\vF=\langle xy,y^2\rangle\), let \(C_1\) be the line segment from \((1,1)\) to \((4,1)\), let \(C_2\) be the line segment from \((4,1)\) to \((4,3)\), and let \(C_3\) be the line segment from \((1,1)\) to \((4,3)\). Also let \(C = C_1 + C_2\). This vector field and the curves are shown in .

Every point along \(C_1\) has \(y=1\). Therefore, along \(C_1\), the vector field \(\vF\) can be viewed purely as a function of \(x\). In particular, along \(C_1\), we have \(\vF(x,1) = \langle x,1\rangle\). Since every point along \(C_2\) has the same \(x\)-value, write \(\vF\) as a function of \(y\) only (for the points on \(C_2\)).

Recall that \(d\vr \approx \Delta \vr\), and along \(C_1\), we have that \(\Delta\vr = \Delta x\vi \approx dx\vi\). Thus, \(d\vr = \langle dx,0\rangle\). We know that along \(C_1\), \(\vF = \langle x,1\rangle\).

Write \(\vF\cdot d\vr\) along \(C_1\) without using a dot product.

What interval of \(x\)-values describes \(C_1\)?

Write \(\int_{C_1} \vF\cdot d\vr\) as an integral of the form \(\int_a^b f(x)\, dx\) and evaluate the integral.

Use an analogous approach to write \(\int_{C_2} \vF\cdot d\vr\) as an integral of the form \(\int_c^d g(y)\, dy\) and evaluate the integral.

Use the previous parts and a property of line integrals to calculate \(\int_C\vF\cdot d\vr\) without having to evaluate any additional integrals.

Since \(C_3\) is from \((1,1)\) to \((4,3)\), we can determine that the line segment has slope \(2/3\), so we can write an equation for the line as \(y-1 = (2/3)(x-1)\) or \(y=\frac{2}{3}x+\frac{1}{3}\). Thus, along the curve \(C_3\), we can write \[\begin{aligned}\end{aligned}\]. Thinking of the slope of \(C_3\) as \(\Delta y/\Delta x\), we can write \(\Delta y/\Delta x = 2/3\), which can be rearranged to \(\Delta y = \frac{2}{3}\Delta x\). We may view \(\Delta\vr\) as \(\langle \Delta x,\Delta y\rangle\). Since \(\Delta x\approx dx\) and \(\Delta y\approx dy\), we use the fact that \(\Delta y = \frac{2}{3}\Delta x\) to write \(d\vr = \langle dx,\frac{2}{3}\, dx\rangle = \left\langle 1,\frac{2}{3}\right\rangle\, dx\). Along \(C_3\), we have \(1\leq x\leq 4\), so having rewritten \(\vF\) in terms of \(x\) and \(\vr\) in terms of \(dx\), we can now write \[\begin{aligned}\int_{C_3}\vF\cdot d\vr \amp = \int_1^4 \left\langle \frac{2}{3}x^2+\frac{1}{3}x,(\frac{2}{3}x+\frac{1}{3})^2\right\rangle \cdot \left\langle 1,\frac{2}{3}\right\rangle\, dx \\ \amp = \int_1^4 \frac{2}{3}x^2 + \frac{1}{3}x + \frac{2}{3}\left(\frac{2}{3}x + \frac{1}{3}\right)^2\, dx = \frac{151}{6}\end{aligned}\]. Notice that this result is different than what you obtained for \(\int_{C}\vF\cdot d\vr\) in , even though \(C\) and \(C_3\) both start at \((1,1)\) and end at \((4,3)\).

Is \(\vF=\langle xy,y^2\rangle\) a gradient vector field? Why or why not?

Let \(\vF = x\vi\). We consider two different oriented curves from \((0,1)\) to \((3,3)\). The first oriented curve \(C\) travels horizontally to \((3,1)\) and then proceeds vertically to \((3,3)\). The second oriented curve \(C_3\) is the line segment from \((0,1)\) to \((3,3)\). Notice that, as depicted in , we can break \(C\) up into two oriented curves \(C_1\) (the horizontal portion) and \(C_2\) (the vertical portion) so that \(C = C_1 + C_2\).

We first note that since \(\vF\) is orthogonal to \(C_2\), \(\int_{C_2}\vF\cdot d\vr=0\); therefore \(\int_C\vF\cdot d\vr =\int_{C_1}\vF\cdot d\vr\). We can parametrize \(C_1\) as \(t\vi+\vj\) for \(0\leq t\leq 3\) (\(t\) is treated like the coordinate \(x\)), which leads to \[\begin{aligned}\end{aligned}\].

Thus, \(\int_C\vF\cdot d\vr = 9/2\).

Now we look at \(\int_{C_3}\vF\cdot d\vr\), but we parametrize \(C_3\) in a nonstandard way by letting \(\vr(t) = \langle 3\sin(t),1+2\sin(t)\rangle\) for \(0\leq t\leq \frac{\pi}{2}\). (You should use a graphing utility to plot this parametrization to help convince yourself that it really does give \(C_3\).) This gives \(\vr'(t) = \langle 3\cos(t),2\cos(t)\rangle\), and \[\begin{aligned}\int_{C_3}\vF\cdot d\vr \amp = \int_0^{\pi/2}\langle 3\sin(t),0\rangle\cdot\langle 3\cos(t),2\cos(t)\rangle\, dt \\ \amp = \int_0^{\pi/2} 9\sin(t)\cos(t)\, dt = \frac{9}{2}\end{aligned}\].

In the next activity, you are asked to consider the more typical parametrization of \(C_3\) and verify that using it gives the same value for the line integral.

It's also worth observing here that \(\int_C\vF\cdot d\vr = \int_{C_3}\vF\cdot d\vr\), so at least two (very different) paths from \((0,1)\) to \((3,3)\) give the same value of the line integral here. The next section will further investigate when line integrals over different paths (with the same initial point and final point) will evaluate to the same value.