# $dA$ vs $dS$ - what is what?

I'm a bit confused as to the difference between $$dA$$ and $$dS$$. I understand the semantic difference, and the relation between the two formula-wise. But what is the most basic difference between them?

Is $$\int dS$$ equal to the area of the shape in question? If that's the case, then what does $$\int dA$$ give us?

• I think it's just the matter of the notation. – Botond Apr 23 at 17:17
• What is the definition of $S$ and $A$? – callculus Apr 23 at 17:29
• You have not told us what $S$ and $A$ are; as such we can't help you clarify the relationship between their differentials. – Allawonder Apr 23 at 17:48

Let me know if I've misunderstood what you're trying to ask. Loosely, $$dS$$ refers to surface area element of objects which are not necessarily flat, while $$dA$$ typically refers to flat regions. Let me expand on that a bit. Of course they are related: say that $$S$$ is a surface, $$f(x,y,z)$$ is a real-valued function, and you want to compute the integral of $$f$$ over the surface $$S$$, $$\iint_S f(x,y,z) \, dS.$$ Then, if you have a parameterization for the surface, say $$\textbf{r}(u,v)$$ where $$(u,v)$$ is in some domain $$D$$, you can compute the surface integral by computing the magnitude of the normal vector to $$S$$ and integrating $$\iint_S f(x,y,z) \, dS = \iint_D f(\textbf{r}(u,v)) \, |\textbf{r}_u \times \textbf{r}_v | \, dA$$
For instance, if $$S$$ is the upper hemisphere of the unit sphere $$x^2+y^2+z^2=1$$, then $$S$$ can be parameterized by $$\textbf{r}(\theta, \varphi) = (\cos\theta\sin\varphi, \,\,\sin\theta\sin\varphi, \,\,\cos\varphi)$$ for $$(\theta,\varphi)$$ in the region $$D$$ given by $$D = \{(\theta,\varphi) \, : \, 0\leq\theta\leq 2\pi, \, 0\leq \varphi \leq \pi/2\}$$. Now you can compute the surface integral of some function $$f$$ via
$$\iint_S f(x,y,z) \, dS = \iint_D f(\cos\theta\sin\varphi, \,\,\sin\theta\sin\varphi, \,\,\cos\varphi) |\textbf{r}_\theta \times \textbf{r}_\varphi | \, dA$$ and the integral on the right hand side can be computed as a standard iterated integral in whichever order is convenient; i.e $$\iint_D dA = \int_0^{2\pi} \int_0^{\pi/2} d\varphi d\theta = \int_0^{\pi/2}\int_0^{2\pi} d\theta d\varphi$$
I leave the details of the computations to you. Note that you are correct in noting that $$\iint_S 1 \, dS$$ returns the surface area of $$S$$, and to translate it to an integral involving $$dA$$ you could use the above procedure. Very roughly speaking, $$dA$$ is to be used when you have a flat 2-dimensional region of integration as opposed to a surface which lives in 3-dimensions.