Note: There is a question here with same title but the contents are entirely different.
Integration by substitution says:
$$\int^{r(b)}_{r(a)} f(r)dr=\int ^b_a f[r(x)] \dfrac{dr(x)}{dx}dx$$
where $f$ is a function of $r$ and $r$ is a function of $x$.
Now if $f$ is a function of $r$ and $r$ is a multivariable function of $x$ and $y$, are the following true:?
$$\int^{r(b,b')}_{r(a,a')}f(r)dr=\int ^b_a f[r(x,y)] \dfrac{\partial r(x,y)}{\partial x}dx=\int^{b'}_{a'} f[r(x,y)] \dfrac{\partial r(x,y)}{\partial y}dy$$
where $a$ and $b$ are lower and upper limits of $x$; $a'$ and $b'$ are lower and upper limits of $y$
Here's my take on this.
As @saulspatz eluded to in their comment, explicitly solving the definite integral in the way you suggest doesn't make sense as when integrating w.r.t. $x$ you don't know how to evaluate $y$ and vice versa.
However, I believe your relation does hold if you first calculate the indefinite integrals by integrating w.r.t. $x$ or $y$, and then sub back in for $r$ at the end, and evaluate the limits.
Example
For a simple example, say we want to solve
$$ \int_{0}^{1} r^2 \mathrm{d}r = \frac{1}{3}$$ (n.b. $\int r^2 \mathrm{d}r = \frac{r^3}{3} + c$) using the multivariable substitution $r^2 = x^2+y^2$. Then $$ \frac{\partial r}{\partial x} = \frac{x}{(x^2+y^2)^{\frac{1}{2}}}, \quad \frac{\partial r}{\partial y} = \frac{y}{(x^2+y^2)^{\frac{1}{2}}}. $$ So according to the indefinite version of your stated formula we get $$ \int r^2 \mathrm{d}r = \int (x^2+y^2)^{\frac{1}{2}} x \mathrm{d}x = \int (x^2+y^2)^{\frac{1}{2}} y \mathrm{d}y, $$ The latter two integrals evaluate to $\frac{(x^2+y^2)^{\frac{3}{2}}}{3} + c$, which then gives $\frac{r^3}{3} + c$ after substituting for $r$. The definite value of the integral can then be calculated by evaluating the integration limits with $r=0$ and $r=1$ to get the answer $\frac{1}{3}$.