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Section 5-3 : Line Integrals - Part II

In the previous section we looked at line integrals with respect to arc length. In this section we want to look at line integrals with respect to \(x\) and/or \(y\).

As with the last section we will start with a two-dimensional curve \(C\) with parameterization,

\[x = x\left( t \right)\hspace{0.25in}y = y\left( t \right)\hspace{0.25in}a \le t \le b\]

The line integral of \(f\) with respect to \(x\) is,

\[\int\limits_{C}{{f\left( {x,y} \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right)} \right)x'\left( t \right)\,dt}}\]

The line integral of \(f\) with respect to \(y\) is,

\[\int\limits_{C}{{f\left( {x,y} \right)\,dy}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right)} \right)y'\left( t \right)\,dt}}\]

Note that the only notational difference between these two and the line integral with respect to arc length (from the previous section) is the differential. These have a \(dx\) or \(dy\) while the line integral with respect to arc length has a \(ds\). So, when evaluating line integrals be careful to first note which differential you’ve got so you don’t work the wrong kind of line integral.

These two integral often appear together and so we have the following shorthand notation for these cases.

\[\int\limits_{C}{{Pdx\, + Q\,dy}} = \int\limits_{C}{{P\left( {x,y} \right)dx}} + \int\limits_{C}{{Q\left( {x,y} \right)\,dy}}\]

Let’s take a quick look at an example of this kind of line integral.

Example 1 Evaluate \( \displaystyle \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}}\) where \(C\) is the line segment from \(\left( {0,2} \right)\) to \(\left( {1,4} \right)\).
Show Solution

Here is the parameterization of the curve.

\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle {1,4} \right\rangle = \left\langle {t,2 + 2t} \right\rangle \,\hspace{0.25in}0 \le t \le 1\]

The line integral is,

\[\begin{align*}\int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}} & = \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\,}} + \int\limits_{C}{{y{x^2}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\sin \left( {\pi \left( {2 + 2t} \right)} \right)\left( 2 \right)\,dt}} + \int_{{\,0}}^{{\,1}}{{\left( {2 + 2t} \right){{\left( t \right)}^2}\left( 1 \right)\,dt}}\\ & = - \left. {\frac{1}{\pi }\cos \left( {2\pi + 2\pi \,t} \right)} \right|_0^1 + \left. {\left( {\frac{2}{3}{t^3} + \frac{1}{2}{t^4}} \right)} \right|_0^1\\ & = \frac{7}{6}\end{align*}\]

In the previous section we saw that changing the direction of the curve for a line integral with respect to arc length doesn’t change the value of the integral. Let’s see what happens with line integrals with respect to \(x\) and/or \(y\).

Example 2 Evaluate \( \displaystyle \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}}\) where \(C\) is the line segment from \(\left( {1,4} \right)\) to \(\left( {0,2} \right)\).
Show Solution

So, we simply changed the direction of the curve. Here is the new parameterization.

\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1,4} \right\rangle + t\left\langle {0,2} \right\rangle = \left\langle {1 - t,4 - 2t} \right\rangle \,\hspace{0.25in}0 \le t \le 1\]

The line integral in this case is,

\[\begin{align*}\int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}} & = \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\,}} + \int\limits_{C}{{y{x^2}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\sin \left( {\pi \left( {4 - 2t} \right)} \right)\left( { - 2} \right)\,dt}} + \int_{{\,0}}^{{\,1}}{{\left( {4 - 2t} \right){{\left( {1 - t} \right)}^2}\left( { - 1} \right)\,dt}}\\ & = - \left. {\frac{1}{\pi }\cos \left( {4\pi - 2\pi t} \right)} \right|_0^1 - \left. {\left( { - \frac{1}{2}{t^4} + \frac{8}{3}{t^3} - 5{t^2} + 4t} \right)} \right|_0^1\\ & = - \frac{7}{6}\end{align*}\]

So, switching the direction of the curve got us a different value or at least the opposite sign of the value from the first example. In fact this will always happen with these kinds of line integrals.

Fact

If \(C\) is any curve then,

\[\int\limits_{{ - C}}{{f\left( {x,y} \right)\,dx}} = - \int\limits_{C}{{f\left( {x,y} \right)\,dx}}\hspace{0.25in}\hspace{0.25in}{\rm{and}}\hspace{0.25in}\,\,\,\,\,\,\int\limits_{{ - C}}{{f\left( {x,y} \right)\,dy}} = - \int\limits_{C}{{f\left( {x,y} \right)\,dy}}\]

With the combined form of these two integrals we get,

\[\int\limits_{{ - C}}{{Pdx\, + Q\,dy}} = - \int\limits_{C}{{Pdx\, + Q\,dy}}\]

We can also do these integrals over three-dimensional curves as well. In this case we will pick up a third integral (with respect to \(z\)) and the three integrals will be.

\[\begin{align*}\int\limits_{C}{{f\left( {x,y,z} \right)\,dx}} & = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)x'\left( t \right)\,dt}}\\ \int\limits_{C}{{f\left( {x,y,z} \right)\,dy}} & = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)y'\left( t \right)\,dt}}\\ \int\limits_{C}{{f\left( {x,y,z} \right)\,dz}} & = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)z'\left( t \right)\,dt}}\end{align*}\]

where the curve \(C\) is parameterized by

\[x = x\left( t \right)\hspace{0.25in}y = y\left( t \right)\hspace{0.25in}z = z\left( t \right)\hspace{0.25in}a \le t \le b\]

As with the two-dimensional version these three will often occur together so the shorthand we’ll be using here is,

\[\int\limits_{C}{{Pdx\, + Q\,dy + R\,dz}} = \int\limits_{C}{{P\left( {x,y,z} \right)dx}} + \int\limits_{C}{{Q\left( {x,y,z} \right)\,dy}} + \int\limits_{C}{{R\left( {x,y,z} \right)dz}}\]

Let’s work an example.

Example 3 Evaluate \( \displaystyle \int\limits_{C}{{y\,dx + x\,dy + z\,dz}}\) where \(C\) is given by \(x = \cos t\), \(y = \sin t\), \(z = {t^2}\), \(0 \le t \le 2\pi \).
Show Solution

So, we already have the curve parameterized so there really isn’t much to do other than evaluate the integral.

\[\begin{align*}\int\limits_{C}{{y\,dx + x\,dy + z\,dz}} &= \int\limits_{C}{{y\,dx}} + \int\limits_{C}{{x\,dy}} + \int\limits_{C}{{z\,dz}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\sin t\left( { - \sin t} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{\cos t\left( {\cos t} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{{t^2}\left( {2t} \right)\,dt}}\\ & = - \int_{{\,0}}^{{\,2\pi }}{{{{\sin }^2}t\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{{{\cos }^2}t\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{2{t^3}\,dt}}\\ & = - \frac{1}{2}\int_{{\,0}}^{{\,2\pi }}{{\left( {1 - \cos \left( {2t} \right)} \right)\,dt}} + \frac{1}{2}\int_{{\,0}}^{{\,2\pi }}{{\left( {1 + \cos \left( {2t} \right)} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{2{t^3}\,dt}}\\ & = \left. {\left( { - \frac{1}{2}\left( {t - \frac{1}{2}\sin \left( {2t} \right)} \right) + \frac{1}{2}\left( {t + \frac{1}{2}\sin \left( {2t} \right)} \right) + \frac{1}{2}{t^4}} \right)} \right|_0^{2\pi }\\ & = 8{\pi ^4}\end{align*}\]