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I have taught myself Line Integration and when doing some practice questions I came across the following:

Evalute the line integral $$\int_{C}(y^2\:dx+xy\:dy+zx\:dz)$$ Where $C$ is the broken line from $A(0,0,0)$ to $B(1,1,1)$ connecting $(0,0,0)$, $(0,0,1)$, $(0,1,1)$ and $(1,1,1)$.

The first step I took was to split $C$ into 3 smaller curves: $C=C_{1}+C_{2}+C_{3}$, such that:

  • $C_{1}$ can be parametrized by $0\leq t\leq1$ with $x=y=0, z=t \implies dx=dy=0, \:dz=dt$
  • $C_{2}$ can be parametrized by $0\leq t \leq 1$ with $x=0, y=t, z=1 \implies dx=dz=0,\: dy=dt$
  • $C_{3}$ can be parametrized by $0 \leq t \leq 1$ with $x=t, y=z=1 \implies dx=dt,\:dy=dz=0$

I then split my integral up as follows:

$$\begin{align*}\int_{C}(y^2\:dx+xy\:dy+zx\:dz)&=\int_{C_{1}}(y^2\:dx+xy\:dy+zx\:dz) \\ &+\int_{C_{2}}(y^2\:dx+xy\:dy+zx\:dz)\\&+\int_{C_{3}}(y^2\:dx+xy\:dy+zx\:dz)\end{align*}$$

Evaluating these we get:

$$\int_{C}(y^2\:dx+xy\:dy+zx\:dz)=\int_{0}^{1}0\:dt+\int_{0}^{1}0\:dt+\int_{0}^{1}\:dt=1$$

Is this the correct way to go about evaluating such an integral?

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  • $\begingroup$ Yes, your method is correct. $\endgroup$ Oct 24, 2013 at 20:59

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I agree with Mhenni. Strangely enough, I first thought that the answer could not be that simple :-), but reading along, I realized that the function you are integrating along this curve, and the curve itself, seem to have been designed to cancel all interesting contributions and make computation easy.

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