Suppose I have an integral of the form:
$$\idotsint_D f({\bf{x}})dA = \int_{a_n}^{b_n}\ldots\int_{a_1}^{b_1}f({\bf{x}})dx_1\ldots dx_n$$
Can I perform a single variable calculus style integration by substition? If I fix all other variables in $f({\bf{x}})$ and consider $g(x_1) = f(x_1, c_2 \ldots, c_n)$, then I can make the substitution $u = h(x_1)$ and write $dx = \frac{1}{h'(x_1)}du$ to conclude:
$$\int_{a_1}^{b_1}g(x_1)dx_1 = \int_{h(a_1)}^{h(b_1)}\hat{g}(u)du$$
Assuming the substitution is well chosen s.t. $g(x_1, u)\frac{1}{h'(x_1)} = \hat{g}(u)$ and the problem is more tractable.
But if I try to carry this logic over to an iterated integral then I am transforming several variables, even if they are constant in this integral. Can I naively put these in the limits? I am not sure if I should start drawing pictures in the new system of coordinates to determine the limits I defined and if I should calculate Jacobians, or if I should just apply the formula from the 1 variable case.
If it helps, the integral that confused me was:
$$\int_{0}^{1} \int_{0}^{2}\frac{xy(x^2-y^2)}{(x^2+y^2)^3}dydx$$
With the substition $u = x^2 + y^2$. I'm not sure how I would even calculate a Jacobian for this, without trying to horribly force square roots to write $x$ as a $x(u)$?, for that I need enough relations for a $2\times2$ matrix of partial derivatives, this is just one.
You can make a single variable substitution just like you did in single variable calculus.
you can say
$x = y\tan t$ and $dx = y\sec^2 t \ dt$
And it might help if you think of it as a two variable substitution, e.g.
$x = s\tan t\\ y = s$
and calculate the Jacobian.
$\left|\det\begin {bmatrix} \frac {\partial x}{\partial s} & \frac {\partial x}{\partial t}\\\frac {\partial y}{\partial s}&\frac {\partial y}{\partial t}\end{bmatrix}\right| = \left|\begin {bmatrix} \tan t & s\sec^2 t\\1&0\end{bmatrix}\right| = s \sec^2 t\ dt$
But what you are suggesting:
$u = x^2 + y^2$
looks like you are really reaching for a full change to polar coordinates.
$x = r\cos \theta\\y = r\sin\theta\\r^2 = x^2 + y^2\\ dx\ dy = r\ dr\ d\theta$