I'm beyond multi-variable calculus, where this is taught, but I still don't know what the hell a surface integral is. I understand that [itex]d\sigma[/itex] is the surface element, and [itex] | \frac{\partial \vec{r}}{\partial u}du \times \frac{\partial \vec{r}}{\partial v}dv | = d\sigma = |\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} | \, dudv[/itex], and that cross product is equal to the area of the infinitesimally small parallelogram described by the two vectors, so summing across all of that (with double integrals yields a surface area). But when you have a function in there like this [itex]\iint_\Sigma f d\sigma[/itex] I don't know what the heck this is supposed to represent. My intuition tells me it's a volume, because it's a function (height) times an area, but I don't think that's right.(adsbygoogle = window.adsbygoogle || []).push({});

I know (or think I know) a line integral of the form [itex] \int_C f ds [/itex] is equivalent to the area under the function traced by the curve C. So the natural extension of this should be a surface integral equaling a volume.

Basically I know how to calculate surface integrals, but I don't know what in the world I'm calculating.

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# What *is* a surface integral?

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