Chapter 12 Jacobian

12.1 Change of variables in the integral of a univariate function

Consider the definite integral of some univariate function f(x) given by I=x2x1f(x)dx. Suppose further that x is a function of t, that is x=g(t), where g is injective, g is continuous and g1 is differentiable. In the Autumn semester we saw how to rewrite I in terms of t rather than x. In particular, there are three steps to rewriting this general integral:

  • Rewrite the function to be integrated in terms of t by replacing every occurrence of x with g(t);

  • Rearranging dxdt=g(t), obtain dx=g(t)dt. Use this identity to write dx in terms of dt in I;

  • Replace the limits x1,x2 of I by t1,t2 respectively where t1=g1(x1) and t2=g1(x2).

As a result I=g1(x2)g1(x1)f(g(t))g(t)dt.

The aim of this chapter is to generalise this theory to a real function of two variables.

12.2 Jacobian

Consider x,y given in terms of u,v by x=x(u,v),y=y(u,v). Further suppose this transformation is one-to-one, that is any one pair (x0,y0) is obtained uniquely from some pair (u0,v0), and similarly each pair (u0,v0) determines exactly one pair (x0,y0). This transform has an inverse: u=u(x,y),v=v(x,y).

Let f(x,y) be a function of x and y, and let R be some region in the (x,y)-plane. Consider the integral I=Rf(x,y)dxdy. How is I given in terms of u and v? The three steps followed for a univariate change of variables of an integral must be followed again in this multivariate setting.

  1. The function f(x,y) being integrated must be written in terms of u and v;

  2. The region R must be understood in terms of u,v;

  3. The differentials dxdy must be rewritten in terms of dudv.

The first two steps are relatively straightforward. The function f(x,y) can be written in terms of u and v, by replacing every occurance of x or y by x(u,v) or y(u,v) respectively. The region R is replaced by the region S belonging to the (u,v)-plane defined as the inverse image of R under the transformation. Formally S={(u0,v0):(x0,y0)=(x(u0,v0),y(u0,v0))R}.

On the other hand, the third step is more involved and requires us to study the Jacobian of the transformation.

When defining the double integral I=Rf(x,y)dxdy, the region R was partitioned into small rectangles of length Δx and width Δy. The integral I was then approximated by Irectanglesf(x,y)ΔxΔy. However there is no intrinsic reason to use rectangles. Indeed any subdivision of R into small consistently-shaped sub-regions will do.

Consider a subdivision of R into small areas defined by families of the level curves of u(x,y) and v(x,y). Specifically fix two small positive values Δu and Δv. Further fix values u0,v0, and consider the area A bounded by the lines u(x,y)=u0, u(x,y)=u0+Δu, v(x,y)=v0 and v(x,y)=v0+Δv. Denote the area of A by ΔA.

The quantity ΔA is called the area element.

The region R can be partitioned with copies of A.


It follows that the integral I can be approximated by Icopies of Af(x,y)ΔA, where each f in the summation has been evaluated at a point within the appropriate copy of A.

Since Δu and Δv are sufficiently small, it follows that A is approximately a parallelogram.


Formally, defining a parametrisation r(u,v)=(x(u,v),y(u,v)), then the region A is approximated by the parallelogram with vertices:

L=r(u0,v0),M=r(u0+Δu,v0),N=r(u0,v0+Δv),K=r(u0+Δu,v0+Δv).

It follows that ΔA|LM×LN|. By the theory of small changes seen in Section 5.1: LM=r(u0+Δu,v0)r(u0,v0)=r(u0,v0)uΔu, and similarly LN=r(u0,v0+Δv)r(u0,v0)=r(u0,v0)vΔv.

This allows us to write ΔA in terms of Δu and Δv:

ΔA|r(u0,v0)uΔu×r(u0,v0)vΔv|=|r(u0,v0)u×r(u0,v0)v|ΔuΔv=|det

This motivates the following definition.

The Jacobian of a one-to-one transformation x=x(u,v), y=y(u,v), denoted \frac{\partial (x,y)}{\partial (u,v)}, is given by \frac{\partial (x,y)}{\partial (u,v)} = \left\lvert \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \right\rvert = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}.

Therefore \Delta A = \frac{\partial (x,y)}{\partial (u,v)} \,\Delta u \,\Delta v.

It follows that I \approx \sum\limits_{\text{rectangles}} f \big( x(u,v),y(u,v) \big) \frac{\partial (x,y)}{\partial (u,v)} \,\Delta u \,\Delta v, where the sum is over rectangles of width \Delta u and \Delta v filling S in the (u,v)-plane, and each f in the summation has been evaluated at a point within the appropriate rectangle.

Taking the limit as \Delta u, \Delta v \rightarrow 0, one obtains I = \int \int_{S} f \big( x(u,v),y(u,v) \big) \frac{\partial (x,y)}{\partial (u,v)} \,du \,dv.

In summary, all of the above is an informal proof of the following statement:

Suppose the region R in the (x,y)-plane is mapped to the region S in the (u,v)-plane by a one-to-one transformation x=x(u,v), y=y(u,v) where the first order partial derivatives of x and y are continous and the Jacobian \frac{\partial(x,y)}{\partial (u,v)} is only zero at isolated points of S. If f(x,y) is a function with continuous partial derivatives on R, then \int \int_{R} f(x,y) \,dx \,dy = \int \int_{S} f \left( x(u,v), y(u,v) \right) \left\lvert \frac{\partial(x,y)}{\partial (u,v)} \right\rvert \,du \,dv.

Use the one-to-one transformation x(u,v)=\frac{1}{3} \left(v-u \right), y(u,v)= \frac{1}{3}\left( v+2u \right) to evaluate \int \int_{R} 2x+y \,dx \,dy for the finite region R bounded by the lines y=-2x+4, y =-2x + 6, y=x-2 and y=x+1.


Note f(x,y) = 2x+y, so that f \big( x(u,v), y(u,v) \big) = 2 \cdot \frac{1}{3} \left(v-u \right) + \frac{1}{3}\left( v+2u \right)= v. Now looking at the bounding lines of R, and rewriting them in terms of u and v by substitution, one obtains \begin{align*} y=-2x+4 \qquad &\text{becomes} \qquad v=4, \\[3pt] y=-2x+6 \qquad &\text{becomes} \qquad v=6, \\[3pt] y=x-2 \qquad &\text{becomes} \qquad u=-2, \\[3pt] y=x+1 \qquad &\text{becomes} \qquad u=1. \end{align*} It follows the region S is given as below:


Calculating the Jacobian, one obtains
\frac{\partial(x,y)}{\partial (u,v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix} = \begin{vmatrix} -\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{vmatrix} = - \frac{1}{3}.
Therefore
\begin{align*} \int \int_{R} 2x+y \,dx \,dy &= \int \int_{S} v \left\lvert -\frac{1}{3} \right\rvert \,du \,dv \\[3pt] &=\int_{4}^{6} \int_{-2}^{1} \frac{v}{3} \,du \,dv \\[3pt] &= \int_{4}^{6} \left[ \frac{uv}{3} \right]_{-2}^{1} \,dv \\[3pt] &= \int_{4}^{6} \frac{v}{3} - \left( - \frac{2v}{3} \right) \,dv \\[3pt] &= \int_{4}^{6} v \,dv \\[3pt] &= \left[ \frac{v^2}{2} \right]_{4}^{6} \\[3pt] &= 18 - 8 \\[3pt] &= 10. \end{align*}

12.3 Geometrical Interpretation of the Jacobian

Using the notation from the previous section, the small area \Delta A in R was the result of small changes \Delta u and \Delta v to u and v, and therefore corresponds to a small area \Delta u \Delta v in S. In particular:

\begin{align*} &\Delta A \approx \left\lvert \frac{\partial (x,y)}{\partial (u,v)} \right\rvert \Delta u \Delta v, \\[5pt] \implies \qquad &\left\lvert \frac{\partial (x,y)}{\partial (u,v)} \right\rvert \approx \frac{\Delta A}{\Delta u \Delta v}. \end{align*}

That is the Jacobian is the ratio of the two small corresponding areas. In other words at any point (u,v) \in S, the value \left\lvert \frac{\partial (x,y)}{\partial (u,v)} \right\rvert is the factor by which the area local to (u,v) increases when mapped to R.

Consider the one-to-one transformation from Theorem 2.1.1 to convert polar coordinates to Cartesian coordinates:

\begin{align*} x &= x (\rho,\varphi) = \rho \cos (\varphi), \\ y &= y (\rho,\varphi) = \rho \sin (\varphi). \end{align*}

Calculate the Jacobian of this transformation as

\frac{\partial (x,y)}{\partial (\rho,\varphi)} = \begin{vmatrix} x_{\rho} & x_{\varphi} \\ y_{\rho} & y_{\varphi} \\ \end{vmatrix} = \begin{vmatrix} \cos (\varphi) & -\rho \sin(\varphi) \\ \sin(\varphi) & \rho \cos(\varphi) \end{vmatrix} = \rho \cos^{2}(\varphi) + \rho \sin^{2}(\varphi) = \rho.

Consider the regions A_{1},A_{2} in the (\rho,\varphi)-plane, and the regions A_{3},A_{4} in the (x,y)-plane, as depicted:



Via standard area calculations, obtain

\begin{align*} \text{Area}(A_{1}) &= \frac{\pi}{4} \\ \text{Area}(A_{2}) &= \frac{\pi}{4} \\ \text{Area}(A_{3}) &= \frac{3 \pi}{8} \\ \text{Area}(A_{4}) &= \frac{7 \pi}{8} \end{align*}

Note A_{3} is the image of A_{1} under the polar coordinate transformation. Consider the ratio of A_{3} to A_{1}: \frac{A_{3}}{A_{1}} = \frac{\frac{3 \pi}{8}}{\frac{\pi}{4}} = \frac{3}{2}. The value \frac{3}{2} is the average value taken by the Jacobian \frac{\partial(x,y)}{\partial(\rho, \varphi)} =\rho over the region A_{1}.

Also note A_{4} is the image of A_{2} under the polar coordinate transformation. Consider the ratio of A_{4} to A_{2}: \frac{A_{4}}{A_{2}} = \frac{\frac{7 \pi}{8}}{\frac{\pi}{4}} = \frac{7}{2}. The value \frac{7}{2} is the average value taken by the Jacobian \frac{\partial(x,y)}{\partial(\rho, \varphi)} =\rho over the region A_{2}.

12.4 Jacobian of the Inverse

It is known that given a univariate function z=z(t), that \frac{dt}{dz} = \frac{1}{\left( \frac{dz}{dt} \right)} provided \frac{dz}{dt} \neq 0. However for the multivariate one-to-one transformation x=x(u,v), y=y(u,v), it is not necessarily true that \frac{\partial u}{\partial x} = \frac{1}{\left( \frac{\partial x}{\partial u} \right)}.

The two-dimensional analogue of the identity in the univariate case is a relationship between the Jacobian of an inverse transformation in terms of the Jacobian of the original transformation.

Let x=x(u,v), y=y(u,v) be a one-to-one transformation such that \frac{\partial (x,y)}{\partial(u,v)} \neq 0. Then the Jacobian of the inverse transformation u=u(x,y), v=v(x,y) is given by \frac{\partial (u,v)}{\partial(x,y)} = \frac{1}{\left( \frac{\partial (x,y)}{\partial(u,v)} \right)}.

Often an application of Lemma 12.4.1 can present an easier method to calculate the Jacobian of a transformation than direct computation does.

Consider the transformation

\begin{align*} \rho &= \rho (x,y) = \sqrt{x^2 + y^2}, \\ \varphi &= \varphi(x,y) = \arctan \left( \frac{y}{x} \right). \end{align*}

Calculate the Jacobian of the transformation.

The transformation is recognised as the transformation to convert from Cartesian coordinates to polar coordinates. In particular it the inverse of the transformation to convert polar coordinates to Cartesian coordinates given by

\begin{align*} x &= x (\rho,\varphi) = \rho \cos (\varphi), \\ y &= y (\rho,\varphi) = \rho \sin (\varphi). \end{align*}

We saw in Example 14.3.1 that \frac{\partial (x,y)}{\partial (\rho,\varphi)} = \rho. Hence

\frac{\partial (\rho,\varphi)}{\partial (x,y)} = \frac{1}{\left( \frac{\partial (x,y)}{\partial (\rho,\varphi)}\right)} = \frac{1}{r}.