Is there no similar proof to change of variables for double integrals as there is for a single integral? The single variable change of variable theorem states that, for any function $\varphi$ with integrable derivative, and a continuous function $f$:
$$\int_{\varphi(a)}^{\varphi(b)}f(x)dx=\int_a^bf(\varphi (x))\varphi '(x)dx$$
The proof is a really simple application of the chain rule and the fundamental theorem of calculus. Is there no similar calculus proof, even if tedious (assuming similar restrictions), for the  double integral case:
$$\int \int_S f(x,y)dxdy =\int \int_R f(g(u,v),h(u,v)) |J|dudv$$
Considering also that a similar theorem to fundamental theorem of calculus exists for the double integrals. See for example: Proof for the "Fundamental Calculus Theorem" for two variables.
 A: I can provide you with this solution:
We first start to integrate with respect to x
$$\int \int f dxdy=\int dy\int f dx.$$
To evaluate it in the other coordinates, we have to express dx using u,v variables, and must be cautious about the meaning of "dx": $dx=(\delta x$ |$\delta y=0$). In terms of Jacobian matrix,
$$
    \begin{pmatrix}
    \delta x \\
    \delta y  \\
    \end{pmatrix}=J \begin{pmatrix}
    \delta u \\
    \delta v  \\
    \end{pmatrix} ,\space  \space \space  J= \begin{pmatrix}
    h'_u & h'_v \\
    g'_u & g'_v \\
    \end{pmatrix},
$$
thus
$$
    \frac{1}{det(J)} \begin{pmatrix}
    g'_v & -h'_v \\
    -g'_u & h'_u \\
    \end{pmatrix}\begin{pmatrix}
    \delta x \\
    \delta y(=0)  \\
    \end{pmatrix} =\begin{pmatrix}
    \delta u \\
    \delta v  \\
    \end{pmatrix},
$$
and
$$dx=\frac{det(J)}{g'_v}du.$$
We can naïvely replace this directly into the integral,
$$\int dy\int f dx= \int dy\int f \frac{det(J)}{g'_v}du \space  \space \space   (wrong),$$
however, this is wrong due to not respecting the g(u,v)=y condition (we are not actually on y=const. line by only sweeping u variable). We must instead force this condition using Dirac Delta function:
$$\int dy\int f dx= \int dy\int \int f \frac{det(J)}{g'_v}\biggl(\delta  (y-g)g'_vdv \biggr) du \space  \space \space   (correct).$$
Note that integration of the expression in the parenthesis amounts to one. One last step is to integrate y variable; finally
$$\int \int f dxdy=\int \int fdet(J) dudv.$$.
