How to get the Jacobian for Double Integrals So in my textbook, it has the following equation for when you are changing the variables in a double integral where $x=g(u, v)$ and $y=h(u, v)$.
\begin{align}
\int \int_R f(x,y) dA = \int \int_S f(g(u, v), h(u, v)) \left|\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}\right| du dv
\end{align}
I understand that the following is the Jacobian, $\displaystyle \left|\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}\right|.$
The textbook also says that we get the above Jacobian from this determinant:
$\begin{array}{|ccc|} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}.$
Why is the Jacobian this determinant? I don't understand where this determinant is coming from.
 A: Even though this question is very old I will try to give an answer:
The confusion seems to be caused by the ambiguous use of the word Jacobian.
The term Jacobian is used for two things: The Jacobian matrix or the Jacobian determinant (that is the determinant of the Jacobian matrix).
The Jacobian matrix is always associated with some (vector valued) function. Its entries are the partial derivatives of this function. Depending on the dimension of the domain and co-domain of the function that the Jacobian matrix is associated with, it does not have to be a square matrix!
Now, if the Jacobian matrix is a square matrix, we can calculate its determinant. In the case of a change of variables in a multiple integral, we need the absolute value (!) of the Jacobian determinant.
Thus when you are saying:

I understand that the following is the Jacobian: $|\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}|$

You need to understand that it is the absolute value of the determinant of the Jacobian matrix of the form: $$\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}$$
(Here, $x$ and $y$ are the scalar functions $g(u,v)$ and $h(u,v)$ as you noted yourself.)
To get some handwavy intuition on why this absolute value of the Jacobian determinant is needed when performing a change of variables you can try to think of it this way: In our original double integral, the term $dxdy$ described the small area element that we get when multiplying $dx$ and $dy$. This is untuitively obvious as $dx$ and $dy$ are orthogonal. However, if we change our coordinate system to polar coordinates for example the expression $drd\theta$ no longer describes our area element in question. To counteract this, we need to multiply $drd\theta$ with just the absolute value of the Jabobian determinant, which is often denoted as $\lvert J\rvert$.
Thus in this particular case we have: $$\lvert J\rvert = |\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}| = \lvert det\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} \rvert = \lvert \begin{array}{|ccc|} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\rvert$$
Here you can see that $\lvert \cdot \rvert$ is used for both, the absolute value and the determinant which is an unlucky coincidence, try to keep this in mind to avoid further confusion.
