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What is the Jacobian?

Illustration of area change
The Jacobian matrix is the matrix of first partial derivatives of a change of variables (x,y)↦(u,v)(x,y)\mapsto(u,v). Its determinant, called the Jacobian determinant, measures how the map stretches area locally. The Jacobian takes its name from the German mathematician Carl Jacobi (1804-1851).
JACOBIAN MATRIX AND DETERMINANT
Let u=u(x,y)u=u(x,y) and v=v(x,y)v=v(x,y) be differentiable functions. We define the Jacobian matrix
(∂u∂x∂u∂y∂v∂x∂v∂y).\begin{equation*}\begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix}.\end{equation*}
Its determinant, called the Jacobian determinant, is
∂(u,v)∂(x,y)=det⁡(∂u∂x∂u∂y∂v∂x∂v∂y).\begin{equation*}\frac{\partial(u,v)}{\partial(x,y)}=\det\begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix}.\end{equation*}
Illustration of how area changes under mapping x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta
Let x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta where rr and θ\theta are polar co-ordinates. Then
∂(x,y)∂(r,θ)=det⁡(∂x∂r∂x∂θ∂y∂r∂y∂θ)=det⁡(cos⁡θ−rsin⁡θsin⁡θrcos⁡θ)=r(cos⁡2θ+sin⁡2θ)=r.\begin{equation*}\frac{\partial(x,y)}{\partial(r,\theta)} = \det\begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix} = \det\begin{pmatrix}\cos\theta & -r\sin\theta \\\sin\theta & r\cos\theta\end{pmatrix} = r(\cos^2\theta + \sin^2\theta) = r.\end{equation*}
A small rectangle [r,r+dr]×[θ,θ+dθ][r,r+dr]\times[\theta,\theta+d\theta] in the (r,θ)(r,\theta)-plane has area dr dθdr\,d\theta. Its image is an annular sector with exact area
ΔA=12((r+dr)2−r2)dθ=(r+dr2)dr dθ≈r dr dθ.\begin{equation*}\Delta A= \frac{1}{2}\left((r+dr)^2-r^2\right)d\theta= \left(r+\frac{dr}{2}\right)dr\,d\theta \approx r\,dr\,d\theta.\end{equation*}
So the local scale factor is rr, i.e.
dx dy=∣∂(x,y)∂(r,θ)∣dr dθ=r dr dθ.\begin{equation*}dx\,dy = \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|dr\,d\theta = r\,dr\,d\theta.\end{equation*}
For the disk D={(x,y):x2+y2≤9}D=\{(x,y):x^2+y^2\le 9\}, this gives the explicit computation
∬D(9−x2−y2) dx dy=∫02π∫03(9−r2) r dr dθ=∫02π[92r2−14r4]03dθ=∫02π814 dθ=81π2.\begin{equation*}\iint_D (9-x^2-y^2)\,dx\,dy= \int_0^{2\pi}\int_0^3 (9-r^2)\,r\,dr\,d\theta= \int_0^{2\pi}\left[\frac{9}{2}r^2-\frac{1}{4}r^4\right]_{0}^{3}d\theta= \int_0^{2\pi}\frac{81}{4}\,d\theta= \frac{81\pi}{2}.\end{equation*}
THEOREM
Let u=u(x,y)u=u(x,y) and v=v(x,y)v=v(x,y) define a bijection from a region RR in the xyxy-plane to a region SS in the uvuv-plane, with differentiable inverse x=x(u,v)x=x(u,v) and y=y(u,v)y=y(u,v) and with
∂(u,v)∂(x,y),∂(x,y)∂(u,v),\begin{equation*}\frac{\partial(u,v)}{\partial(x,y)},\qquad \frac{\partial(x,y)}{\partial(u,v)},\end{equation*}
defined and non-zero everywhere. Further, let ψ(x,y)=Ψ(u,v)\psi(x,y)=\Psi(u,v), where u=u(x,y)u=u(x,y) and v=v(x,y)v=v(x,y). Then
∬(u,v)∈SΨ(u,v) du dv=∬(x,y)∈Rψ(x,y)∣∂(u,v)∂(x,y)∣ dx dy,\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)\,du\,dv = \iint_{(x,y)\in R} \psi(x,y)\left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy,\end{equation*}∬(x,y)∈Rψ(x,y) dx dy=∬(u,v)∈SΨ(u,v)∣∂(x,y)∂(u,v)∣ du dv.\begin{equation*}\iint_{(x,y)\in R} \psi(x,y)\,dx\,dy = \iint_{(u,v)\in S} \Psi(u,v)\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv.\end{equation*}
Sketch of proof. It is sufficient to prove the first integral identity. Divide the region RR into NN square elements QiQ_i of equal area δxδy\delta x\delta y with ψi=ψ(xi+δx/2,yi+δy/2)\psi_i=\psi(x_i+\delta x/2,y_i+\delta y/2) the scalar field at the centre of the ithi^{th} element, i∈{1,…,N}i\in\{1,\ldots,N\}.
Consider the mapping of the ithi^{th} element, which is bounded by the co-ordinate lines x=xix = x_i and x=xi+δxx = x_i + \delta x and y=yiy = y_i and y=yi+δyy = y_i + \delta y. The value of the scalar field at the mapped element centre is
Ψi=Ψ(u(xi+δx/2,yi+δy/2),v(xi+δx/2,yi+δy/2))=ψi.\begin{equation*}\Psi_i = \Psi(u(x_i + \delta x/2, y_i + \delta y/2), v(x_i + \delta x/2, y_i + \delta y/2)) = \psi_i.\end{equation*}
Also, for sufficiently small δx\delta x and δy\delta y, the image of QiQ_i is approximately a parallelogram, which we denote by PiP_i, and its area is given by the absolute value of the determinant of the linearized change in (u,v)(u,v):
Area⁡(Pi)≈∣det⁡(∂u∂x(xi,yi)∂u∂y(xi,yi)∂v∂x(xi,yi)∂v∂y(xi,yi))∣δx δy=∣∂(u,v)∂(x,y)(xi,yi)∣δx δy.\begin{equation*}\operatorname{Area}(P_i)\approx\left|\det\begin{pmatrix}\frac{\partial u}{\partial x}(x_i,y_i) & \frac{\partial u}{\partial y}(x_i,y_i) \\\frac{\partial v}{\partial x}(x_i,y_i) & \frac{\partial v}{\partial y}(x_i,y_i)\end{pmatrix}\right| \delta x\,\delta y= \left|\frac{\partial(u,v)}{\partial(x,y)}(x_i,y_i)\right| \delta x\,\delta y.\end{equation*}
Thus, before taking limits, partitioning SS by the parallelograms PiP_i, we have an approximation for
∬(u,v)∈SΨ(u,v)du dv\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)du\,dv\end{equation*}
which is
∑i=1NΨiArea(Pi)=∑i=1NΨi∣∂(u,v)∂(x,y)(xi,yi)∣δxδy=∑i=1Nψi∣∂(u,v)∂(x,y)(xi,yi)∣δxδy.\begin{equation*}\sum_{i=1}^N \Psi_i\text{Area}(P_i) = \sum_{i=1}^N \Psi_i\left|\frac{\partial(u,v)}{\partial(x,y)}(x_i,y_i)\right|\delta x\delta y = \sum_{i=1}^N \psi_i\left|\frac{\partial(u,v)}{\partial(x,y)}(x_i,y_i)\right|\delta x\delta y.\end{equation*}
Taking limits gives
∬(u,v)∈SΨ(u,v)du dv=∬(x,y)∈Rψ(x,y)∣∂(u,v)∂(x,y)∣dx dy.\begin{equation*}\iint_{(u,v)\in S} \Psi(u,v)du\,dv = \iint_{(x,y)\in R} \psi(x,y)\left|\frac{\partial(u,v)}{\partial(x,y)}\right|dx\,dy.\end{equation*}

References