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Examples of Line Integrals (Example 2)
Sometimes you may be asked to find a line integral that has multiple line segments.
Let’s find the integral \[\int\limits_C^{} {(y + z)dx + (x + z)dy + (x + y)dz} \], given that C is the line segments joining (0,0,0) to (1,0,1), and (1,0,1) to (0,1,2).
The first thing to do is make a parametric equation for the first line segment, (0,0,0) to (1,0,1) and also find its derivative.
\[r(t) = \left\langle {t,0,t} \right\rangle \]
\[r'(t) = \left\langle {1,0,1} \right\rangle \]
\[0 \le t \le 1\]
And each component in r(t) represents x, y, and z.
One problem we run into is that we don’t have dt in our integral, we instead have dx, dy, and dz.
So just set the components in r(t) equal to x, y, and z, then find the derivatives.
\[x = t\]
\[y = 0\]
\[z = t\]
\[\frac{{dx}}{{dt}} = 1\]
\[\frac{{dy}}{{dt}} = 0\]
\[\frac{{dz}}{{dt}} = 1\]
\[dx = dt\]
\[dz = dt\]
Now we can substitute all this into the integral.
As well, t goes from t = 0 to t = 1 for the first line segment.
Let’s call this first integral I\(_1\).
\[{I_1} = \int_0^1 {(0 + t)dt + (x + z)(0) + (t + 0)dt} \]
\[{I_1} = \int_0^1 {2tdt} \]
\[{I_1} = 1\]
And we find another integral representing the second line segment.
So, you need to make equations that go from (1,0,1) to (0,1,2), and “reset the clock”, so that it goes from t = 0 to t = 1 again.
Basically, make your equations such that when you substitute in t = 0 you get (1,0,1), then when you substitute in t = 1 you get (0,1,2).
\[x = – t + 1\]
\[y = t\]
\[z = 1 + t\]
\[0 \le t \le 1\]
And find their derivatives as before.
\[dx = – dt\]
\[dy = dt\]
\[dz = dt\]
And substitute in all this information to make an integral, we’ll call this second integral I\(_2\).
\[{I_2} = \int_0^1 {(t + (1 + t))( – dt) + (( – t + 1) + (1 + t))dt + (( – t + 1) + t)dt} \]
\[{I_2} = \int_0^1 {( – 2t – 1)dt + 2dt + 1dt} \]
\[{I_2} = 1\]
And you sum each of the individual integrals to get the total integral value.
\[{I_{total}} = {I_1} + {I_2}\]
The total integral value is 1+1 = 2
Try the next example, where we solve a line integral over a parametric curve!