Is it possible/how does one find the volume under a surface and over a region bounded by a vector-valued function?

Question

In class we learned how to find the volume under a function f(x,y) and over a rectangular region using double integrals. We then extended this to volume over triangular and circular regions. Is there a way to find the volume under a surface given by f(x,y) and over a general "blob-like" (non-rectangular) region, as long as the region is simple and closed? More specifically, is there a way to find the volume under the surface f(x,y) and over the region in the x-y plane bounded by r(t) = $<f(t),g(t)>$ using integrals, if the region bounded by r(t) is closed and simple for a ≤ t ≤ b? And if it isn't simple, can it still be done?

Answer 1

Certainly, as long as the functions involved are continuous enough (let's leave these details aside though as a full discussion would probably be out of scope).

In the simplest case, we assume that the blob can be defined as the region between two curves

$y = g(x), a <= x <= b$, and $y = h(x), a <= x <= b$

i.e. the blob is the set of points

$B = \{(x,y)| a<= x <= b, g(x) <= y <= h(x) \}$

Then we would have

V = $\int_a^b \int_{g(x)}^{h(x)} f(x,y) dy dx$

If that is not the case you can try to break the blob up into several pieces of this form.

If the blob is not easy to break up in a rectangular fashion but is in a polar fashion (e.g. if the set is star-shaped but not convex), you can do it in polar coordinates:

V = $\int_0^{2\pi} \int_0^{r(t)} f(rcos(t),rsint(t)) r dr dt$

A more general approach would be to define planar integrals in terms of Riemann sums on rectangles enclosed inside the blob, or to fill the domain up with small cubes as completely and take the upper limit of all possible such fillings (which would be the 3d integral approach). This would take longer to explain though.