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Given a vector field $\mathbf{F}=(A, B, C)$ in Cartesian coordinates, I understand that the line integral along an oriented curve $\mathcal{C}$ is $$ \int_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r}=\int_{\mathcal{C}} A d x+B d y+C d z $$

But why is the surface integral over an oriented surface the following (according to my physics textbook)? I have done some multivariable calculus before but haven't seen this specific formulation being mentioned elsewhere. Some pointers would be helpful, thanks. $$ \int_{\mathcal{S}} \mathbf{F} \cdot \mathbf{d S}=\int_{\mathcal{S}} \int A d y d z+B d z d x+C d x d y $$

Ted Shifrin
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user1143399
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  • In general surface integrals are seen in the courses "Vectorial calculus". This may be helpful https://tutorial.math.lamar.edu/classes/calciii/surfaceintegrals.aspx – Brian Britos Simmari Mar 06 '23 at 18:19
  • @BrianBritosSimmari Thanks - I can solve particular problems of surface integrals (like the ones in the link) but I still don't see how the particular form of dydz / dzdx / dxdy came to be. This formula was shown while introducing differential forms and I don't understand how it's derived. – user1143399 Mar 06 '23 at 18:23
  • A little hint: $\mathbf{F}\cdot{\mathbf {dS}}$ involves the normal vector to the surface and that has a lot to do with a cross product of tangent vectors to the surface which in turn has a lot to do with the wedge products $dy\wedge dz,dz\wedge dx,dx\wedge dy,.$ – Kurt G. Mar 06 '23 at 19:33
  • If you're willing to look at things more generally. – peek-a-boo Mar 06 '23 at 19:43
  • If at some point you decide to learn a bit about differential forms and see this in a coherent manner, some of my YouTube lectures may be helpful. You can get the gist of the computations in lectures 34-35 (MATH 3510) and the discussion of flux (which is what your integral above is calculating). Starting at minute 20 of lecture 41 is a dictionary between the differential form approach and the standard "physics" approach in $\Bbb R^3$. – Ted Shifrin Mar 06 '23 at 19:46

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