Consider a 2-dimensional domain $\mathcal{D}$ (a subset of $\mathbb{R}^2$) and a function which maps points from that region to some height ($f: \mathcal{D} \to \mathbb{R}$). Now consider the double integral

$$\int\int_{\mathcal{D}} f(x,y) dx dy \tag{1}$$


If I remove an arbitrary 1-dimensional curve $c(t) = (x(t), y(t))$ from the region $\mathcal{D}$ such that $\mathcal{D}'$ includes all the points from $\mathcal{D}$ except those on that curve $c(t)$, then would the double integral

$$\int\int_{\mathcal{D}\;'} f(x,y)dxdy$$ evaluate to the same result as equation $(1)$? I would suspect that the answer is yes. Double integrals calculate volume. The line integral over the curve $c(t)$ would give an area. Removing this area would not affect the volume, just as removing a point from a one-dimensional domain does not affect the area of $\int f dx$. Is this correct? (Motivation: I'm thinking about some introductory physics in more detail. I'm dealing with a 2-dimensional object, but then the object is cut into pieces, or ruptures into fragments for whatever reason. I'm not modeling the object discretely, but as a continuum. I'm worrying about the "points lost" on the boundaries of each shattered fragment in my equations - and what that would mean physically. But since I'm taking the philosophy of continuum mechanics, I need to better understand "continuous mathematics" or maybe topology. On the boundary points, there is mass there. But then it kind of gets lost. Or doesn't affect the math.)


The curve could fill the whole region $D$, in that case $D'$ would be empty, the integral over $D'$ would be zero, while the original could be non-zero.

If you add some conditions (smoothness) to the curve, then the volume of its image is zero, and the result you want follows. Continuity of the curve is not enough.

In general as long as the measure (area) of what you are removing is zero the integral there would be zero, therefore the original integral doesn't change. When the measure is not zero then it depends on the function that you are integrating. In that case removing could potentially change the value of the integral.


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