How does an Iterated Integral Work? I am simply confused because of planes now. 
Consider:
$$J = \int_{R}\int_{R} e^{-(x^2 + y^2)} dxdy$$
What is the geometrical aspect of this integral? 
This represents the volume under $h(x,y) = e^{-(x^2 + y^2)}$ I believe.
Then how is:
$$I = \int_{R} e^{-x^2} dx = \int_{R} e^{-y^2} dy$$?
How does that statement hold true? Which are independent/dependent variables? I am utterly confused.
 A: Consider $$I = \int_{-\infty}^\infty e^{-x^2} dx$$
This is an important integral in probability theory, so it pays to understand what is happening here.  You can't get the indefinite integral in terms of elementary functions, but you can find the value of the integral with limits $-\infty$ to $+\infty$.
First, the statement that you are confused about is, as Harald says, merely a matter of switching what you call the integration variable; that does not change the value of the integral.  Try it:  What is 
$$
\int_0^1 x\, dx$$ and what is
$$\int_0^1 v\, dv$$
You can see the integration variable disappears in the end.
Next, why in the world are we introducing that equivalent but differently "named" integral 
$\int_{-\infty}^\infty e^{-y^2} dy$? We are doing it just so that we can multiply the two integrals together, getting 
$$
I^2 = \int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy
$$
and then since when doing the $x$ integral the entire $y$ integral is just a constant (it does not depend on $x$, after all), we can re-write this as a double integral
$$I^2 = \int_{x=-\infty}^\infty \int_{y=-\infty}^\infty e^{-(x^2+y^2)} dy\,dx $$
Why is this progress?  The trick is (and I like to think of this as having been discovered by Gauss, but probably others did this before him) that this double integral is over the whole $x,y$ plane, and we can therefore easily transform to polar coordinates, because the limits will be $r$ going from $0$ to $\infty$ and $\theta$ from 
$0$ to $2\pi$.
The $(x^2+y^2)$ is conveniently $r^2$, and -- here is the zinger -- 
$$
dx\,dy -\longrightarrow r\,d\theta\,dr $$
That extra factor of $r$ is golden because now we can do the integral:
$$
I^2 = \int_{\theta = 0}^{2\pi} d\theta \int_{r=0}^{\infty} r e^{-r^2} dr
= 2\pi \left[ -\frac{1}{2} e^{-r^2} \right]_{r=0}^{r=\infty} = \pi(1-0) = \pi
$$
SO $$
I = \sqrt{\pi}$$
