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This paper presents an educational system called CADACKS for learning CAD for design and drawing for construction.Q: How to integrate this function over a regular curve Let $X$ be a two dimensional manifold and $f:X\to\mathbb{R}$ a smooth function. Show that $\int_C f\,d\sigma=0$ for any regular curve $C\subset X$ if $X$ is oriented. I’m completely lost here. I’ve been watching some videos of Geometry of Riemannian Manifolds and I saw something like this, but I still have not got a clear picture of the problem. Thank you for any suggestion. A: Let $C:[a,b] \to X$ be a regular curve. Then there exists an orientation preserving diffeomorphism $h:[a,b] \to [a’,b’]$ such that $h(a)=a’$ and $h(b)=b’$. By choosing $h$ to be smooth and applying chain rule we have $$ \int_Cf d\sigma = \int_C f \circ h (h^\sharp ds) =\int_C f \circ h \cdot (d\text{area} \circ h) $$ where $h^\sharp ds$ is the differential of area on $[a’,b’]$(this can be seen as the induced vector field of $h$) and $\text{area} \circ h$ is the area of $[a’,b’]$ Now we have: $$ \text{area} \circ h = \int_a^{b’} ds = b’-a’ $$ and since $h$ is orientation preserving we have $h^\sharp ds \equiv \pm 1$, so $$ \int_C f d\sigma = \int_C f \circ h \cdot (d\text{area} \circ h) = \pm \int_a^{b’} f \circ h (d\text{area} \circ h) $$ Now by the fundamental theorem of calculus, we know that: $$ \pm \int_a^{b’} f \circ h (d\text{area} \circ h)