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I have been learning MV calculus and have become really confused by what these things actually mean, for example:

$$\iint f(x,y)~dA$$ should be calculating the volume under a surface in 3 dimensions so then what does

$$\iint f(x,y,z)$$ mean and how does this relate to $$\iiint f(x,y,z)$$ I don't see why you can do a double integral with $3$ variables I thought the whole point of the double integral was to do functions of $2$ variables.

Also with line integrals I was under the understanding that $$\int_c f(x,y)~ ds$$ was the area of a sheet (like a fence or a curtain) along a path under the surface defined by $f(x,y)$ so why people do this in three dimensions also how does that even have any meaning? You can use it to calculate work done and things like that but I don't get how? Is the work done equal to the area of the "curtain"?

Also for Green's theorem I get the idea and how to use it but what is it physically saying? The area of the curtain along $c$ under the surface is equal to the double integral of the partials minus each other with respect to the area. So would this be the volume or just the area of the curve in the $xy$ plane. Also how does this change if you move into some function of $3$ variables because aren't you already having $3$ dimensions anyway so what does it actually mean for $$\oint_c A \,dx + B \,dy +C \,dz=\iint \frac{\partial A}{\partial x}-\frac{\partial B}{\partial y}-\frac{\partial C}{\partial z}\,dx\,dy\,dz$$

Any help is greatly appreciated thank you very much.

Dean
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    $$\iint f(x,y,z)$$ is completely meaningless. Unless you tell what you are integrating over, the expression is not mathematically valid. – 5xum Jun 01 '15 at 10:14
  • So say it's $dA$ how would this change from $dV$ or $ds$ say? – Dean Jun 01 '15 at 10:18
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    You still need to tell what you are integrating over. – 5xum Jun 01 '15 at 10:19
  • These are three different things: $$ \iint f(x,y,z),d(x,y), \qquad \iint f(x,y,z) , d(x,z), \qquad \iint f(x,y,z) , d(y,z) $$ – Michael Hardy Jun 05 '18 at 18:15

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