So what from what I understand
- Two integrals: The volume underneath a surface
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Now in this course they define it like so
- Let $z = f(x, y)$ be a surface and let R be a region in the $xy$ plane, where $R$ is the projection of surface z onto the $xy$ plane
- Area of a region in the $xy$ plane corresponds to the case where $f(x ,y) = 1$ and hence, $R$ = $\int\int dR$
$$$$ How can this be the area of something? Two integrals indicate a volume, in this description above, it indicates a flat shape with a thickness of 1, hence, 3D
They then say mass of a region in the $xy$ plane with mass per unit area $\rho(x ,y)$ is $$R = \int \int \rho dR$$
which would only make sense if it was 3D, but how do you have a mass of a region, where a region is a 2D planar shape?
$$$$ So if two integrals represent the volume beneath an area, 3 integrals represents the hypervolume in the 4th dimension underneath a 3D object?
$$$$ But then they state: "The volume of a solid region V in $R^3$ corresponds to the case where $f(x, y, z)$ = 1, so volume of $$V = \int \int \int dx dy dz$$ From what I understand, the outcome of this integral should be 4D
$$$$ I'm sure I've gotten something very wrong, but this is incredibly confusing so if anyone can explain this it would be extremely helpful.