Vanishing of the integration along vertical line

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.

• You should write $\int_C F(x, y)\, dxdy$; otherwise, the integral need not vanish. – Giuseppe Negro Apr 18 at 11:00
• @GiuseppeNegro why the integral would not vanish if $dy$ is not written? – Smart Apr 18 at 14:26
• 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. – 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.
• That is precisely what I used as $$\int_C F(x,y)dx = \int_a^bF(x,y(x))dx$$ – Dayton Apr 18 at 14:27
• 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 $$C = \{0\}\times [c,d]$$ – Dayton Apr 18 at 14:31
• what do you mean by approaching the appropriate segment at $x = 0$? – Smart Apr 18 at 14:48