**Definition**: (Jacobian) If $ T $ is the transformation from the $ uv $-plane to the $ xy $-plane defined by the equations $ x = x(u, v) $ and $ y = y(u, v) $, then the Jacobian of $ T $ is denoted by $ J(u, v) $ or by $ \partial(x, y)/\partial(u, v) $ and is defined by

$ J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \cdot \frac{\partial x}{\partial v} $,

i.e. the determinant of the matrix of the partial derivatives (also known as the Jacobian matrix). Naturally, this can be generalized to more variables.

--------------------

**Theorem**: If the transformation $ x = x(u, v) $, $ y = y(u, v) $ maps the region $ S $ in the $ uv $-plane into the region $ R $ in the $ xy $-plane, and if the Jacobian $ \partial(x, y)/\partial(u, v) $ is nonzero and does not change sign on $ S $, then (with appropriate restrictions on the transformation and the regions) it follows that

$ \displaystyle \int_R \int f(x, y) dA_{xy} = \int_S \int f(x(u, v), y(u, v)) \left|\frac{\partial(x, y)}{\partial(u, v)} \right| dA_{uv} $.

--------------------

**Problem**: Evaluate $ \displaystyle \int_R \int e^{(y-x)/(y+x)} dA $, where $ R $ is the region in the first quadrant enclosed by the trapezoid with vertices $ (0, 1); (1, 0); (0, 4); (4, 0) $.

**Solution**: The bounding lines can be written as $ x = 0 $, $ y = 0 $, $ y = -x+1 $, and $ y = -x+4 $. Now consider the transformation $ u = y+x $ and $ v = y-x $. In the $ uv $-plane, the bounding lines of the new region $ S $ can now be written as $ u = 1 $, $ u = 4 $, $ v = u $, and $ v = -u $.

We can write $ x $ and $ y $ as functions of $ u $ and $ v $: simply $ x = \frac{u-v}{2} $ and $ y = \frac{u+v}{2} $. So the Jacobian $ \displaystyle \frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \cdot \frac{\partial x}{\partial v} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \left(-\frac{1}{2} \right) = \frac{1}{2} $.

Then our original integral becomes $ \displaystyle \int_R \int e^{(y-x)/(y+x)} dA = \frac{1}{2} \int_S \int e^{v/u} dA $. And this is equivalent to

$ \displaystyle \frac{1}{2} \int_S \int e^{v/u} dA = \frac{1}{2} \int_1^4 \int_{-u}^u e^{v/u} dv du = \frac{1}{2} \int_1^4 \big[ u e^{v/u} \big]_{v=-u}^u du = \frac{1}{2} \int_1^4 u\left(e-\frac{1}{e}\right) du = \frac{15}{4}\left(e-\frac{1}{e}\right) $.

QED.

--------------------

Comment: Note that the above theorem is probably very important in multivariable calculus, as it is the equivalent to $ u $-substitution in one variable, which we all know is the ultimate integration technique. It functions in the same way, giving you a lot more flexibility on the function you are integrating and the region you are integrating on.

--------------------

Practice Problem: Evaluate $ \displaystyle \int_R \int (x^2-y^2) dA $, where $ R $ is the rectangular region enclosed by the lines $ y = -x $, $ y = 1-x $, $ y = x $, $ y = x+2 $.