Integration is about chopping things up, and adding the pieces. This “chopping and adding” viewpoint is often helpful in setting up problems involving integration, especially those involving multiple integrals.

For example, consider finding the mass of a straight wire, given the (linear) mass density $\lambda$. Using $x$ to measure distance along the wire, chop the wire into small pieces of length $dx$. 1) What is the mass of each piece? Clearly, $\lambda \,dx$, where in general $\lambda$ depends on $\rr$, and hence can be thought of as a function of $x$, that is, $\lambda=\lambda(x)$. The total mass $M$ is given by adding up the mass of each piece, so that \begin{equation} M = \int \lambda(x) \,dx \end{equation}

The same principles apply in higher dimensions: chop and add. For example, to find the mass of a rectangular plate given its (surface) mass density $\sigma$, first chop the plate into small rectangular pieces of size \begin{equation} dA = dx \,dy \end{equation} What is the mass of each such piece? Clearly, $\sigma \,dA$. Since $\sigma$ is not constant, the total mass is given by \begin{equation} M = \int \sigma(\rr) \,dA \end{equation} Since we are working in rectangular coordinates, we can write $\sigma=\sigma(x,y)$, so that the total mass is now 2) \begin{equation} M = \dint \sigma(x,y) \,dx\,dy \end{equation}

Had the plate (and the density function) been round, we would have been better off using polar pieces, with 3) \begin{equation} dA = r \,dr \,d\phi \end{equation} Expressing the mass density in terms of polar coordinates, $\sigma=\sigma(r,\phi)$, then yields the total mass in the form \begin{equation} M = \dint \sigma(r,\phi) \>r\,dr\,d\phi \end{equation}

You may have learned that integration is antidifferentiation, and that integrals are areas. Yes, the total mass of the wire, thought of as a function of the distance from one end, is indeed an antiderivative of the function $\lambda(x)$. And yes, this integral represents the “area” under the graph of the function $\lambda(x)$, although the dimensions are not those of area. But the “infinitesmal mass” $\lambda \,dx$ better represents the relevant physical process, and should therefore be regarded as fundamental. Note the importance of $dx$ (and its units) to this argument!