In my calculus course I learned how to calculate the volumes of "simple 2-dimensional regions" that are built up from the even simpler type 1 and type 2 regions, which, for functions $f_1(x)$,$f_2(x)$, $g_1(y)$, $g_2(y)$ and reals $a$ and $b$ ($a<b$, $f_1(x) < f_2(x)$ and $g_1(y) < g_2(y)$), are defined as follows:
Type 1: $R \in \left\{ (x,y)\ |\ a < x < b,\ f_1(x) < y < f_2(x)\right\}$
Type 2: $R \in \left\{ (x,y)\ |\ g_1(y) < x < g_2(y),\ a < y < b\right\}$
(For some reason I don't have a precise specification of what type of functions $g_1$ and $g_2$ have to be, for the rest of this question assume continuous functions, or if it makes things easier, differentiable functions.)
So then the volume of the region is (for example for a type 1 region R)
$\displaystyle{\int_R{dA}=\int_a^b{\int_{f_1(x)}^{f_2(x)}}dy\ dx}$
And similarly, for a three-dimensional solid bounded in the $(x,y)$ plane by some (arbitrarily as) type 1 region R and the two functions $h_1(x,y)$ and $h_2(x,y)$, $h_1(x,y) < h_2(x,y)$ for $(x,y) \in \mathrm{R}$, the volume would be
$\displaystyle{\int_R{\int_{h_1(x,y)}^{h_2(x,y)}}dz\ dA}$
My question is just if I'm thinking correctly by saying that using this method it's possible to compute the area / volume of any region / solid. For example to find the area of a two dimensional region $R_1$ defined as follows (all functions continuous, or if it's needed differentiable):
$R \in \left\{(x,y)\ |\ g_1(y)<x<g_2(y),\ f_1(x)<y<f_2(x) \right\}$
I could take the triple integral over the simple region R in the, say, $xz$ plane and have z be bounded by reals a and b, $a<b$. So the volume of the region would be
$\displaystyle{\int_R{\int_{f_1(x)}^{f_2(x)}dy\ dA} = \int_a^b{\int_{g_1(y)}^{g_2(y)}{\int_{f_1(x)}^{f_2(x)}}}dy\ dx\ dz}$
and the area of the region in the xy plane ("flattening out" z) would be
$\displaystyle{\lim_{b \rightarrow a^{+}}{\int_a^b{\int_{g_1(y)}^{g_2(y)}{\int_{f_1(x)}^{f_2(x)}}}dy\ dx\ dz}} = \displaystyle{\int_{g_1(y)}^{g_2(y)}{\int_{f_1(x)}^{f_2(x)}}dy\ dx}$
Is this the right approach? Does this even make sense, since x and y are dependent on each other? If the last step of getting rid of the limit was faulty, could somebody explain to me why? And, finally, for my original question, would this method also work to calculate volumes of n-dimensional solids bounded only by non-trivial functions by taking $n+1$ integrals and then, like above, taking the limit so that the dimension with the constant-bound parameters would just fall away?
Here is a picture of the region $R$ you have in mind:
In order to compute its area you'll have to cut it up into "simple" regions, as indicated by the dotted lines.
The basic formula for computing multiple integrals "recursively" comes from Fubini's theorem. Given a maybe complicated, but reasonable domain $R\subset{\mathbb R}^2$ and a reasonable function $f:\>R\to{\mathbb R}$ one has $$\int\nolimits_R f(x,y)\ {\rm d}(x,y)=\int\nolimits_{R'}\int\nolimits_{R_x} f(x,y)\ {\rm d}y\ {\rm d}x\ ,$$ where $$R'=\{x\ |\ \exists y:\ (x,y)\in R\}$$ is the projection of $B$ onto the $x$-axis, and $$R_x:=\{y\ |\ (x,y)\in R\}\qquad(x\in R')$$ is essentially the intersection of $R$ with a vertical erected at $x$. Only if the sets $R'$ and $R_x$ are intervals we can use the fundamental theorem of calculus to compute the respective integrals. That's meant by requiring the regions $R$ to be "simple".