2- Show that if $C$ is a vertical line segment $c \leq y \leq d,$ and if $F$ is a function of 2 variables defined on $C$, then $$\int_{C} F(x,y)dx = 0$$.

I understand that the integration is calculating the area under the curve and because we are integrating on a vertical line so the area is zero. is this the meaning of the question ?

Also I am stucked in the proof as all the theorems and definitions in this section include parametrizations, or shall I use this question Proving a Line-integration along a parametrized curve identitiy. in proving it?if so how?. Any help in the proof will be appreciated.

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    $\begingroup$ You should write $\int_C F(x, y)\, dxdy$; otherwise, the integral need not vanish. $\endgroup$ – Giuseppe Negro Apr 18 at 11:00
  • $\begingroup$ @GiuseppeNegro why the integral would not vanish if $dy$ is not written? $\endgroup$ – Smart Apr 18 at 14:26
  • $\begingroup$ Because I read "the area is zero", so I assumed we were talking about a double integral. Actually, the integral of the differential form $F(x, y)\, dx$ also vanishes; this I had overlooked. $\endgroup$ – Giuseppe Negro Apr 18 at 17:13

Yes the area under this curve will be zero. You can do this a bit more rigorously by consider the following limiting process. Consider a line defined by $$y(x) = c + \frac{x}{\epsilon}(d-c), \quad x\in [0,\epsilon]$$ so that $y(0) = c$ and $y(\epsilon) = d$ and consider $$|\int_0^{\epsilon} F(x,y(x))dx| \leq \max_{(x,y)\in [0,\epsilon]\times[c,d]} |F(x,y)|\int_0^{\epsilon} dx \leq C\epsilon \rightarrow 0$$ This is just taking the line connecting $(0,c)$ and $(\epsilon,d)$, so in the limit that $\epsilon \rightarrow 0$, this approaches the appropriate segment at $x=0$. This holds for any $x$ so the integral must be zero just as the area under the curve is zero.

  • $\begingroup$ Also can you look at this math.stackexchange.com/questions/3191845/… if you have time please? $\endgroup$ – Intuition Apr 18 at 10:22
  • $\begingroup$ Can it be done by the question mentioned by the OP? $\endgroup$ – Smart Apr 18 at 14:23
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    $\begingroup$ That is precisely what I used as $$\int_C F(x,y)dx = \int_a^bF(x,y(x))dx$$ $\endgroup$ – Dayton Apr 18 at 14:27
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    $\begingroup$ It stated "a vertical line segment", so you must describe where that segment is. I chose $x=0$, but any value will give the same result. ie $$C = \{0\}\times [c,d]$$ $\endgroup$ – Dayton Apr 18 at 14:31
  • $\begingroup$ what do you mean by approaching the appropriate segment at $x = 0$? $\endgroup$ – Smart Apr 18 at 14:48

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