Change of variables

Many instructors use as an introduction to the general setting of this section and other skip this generalization. We do not address how to find these advantageous changes of coordinates which will transform regions from non-rectangular to rectangular (in the new coordinate space), but rather we develop the Jacobian as the key element in these transitions between coordinates. Also of note, the more time you spent on visualizing regions and setting boundary conditions will yield great benefits in the time needed for students to practice the ideas in this section.

In single variable calculus, we encountered the idea of a change of variable in a definite integral through the method of substitution. For example, given the definite integral \[\begin{aligned}\end{aligned}\] we naturally consider the change of variable \(u = x^2+1\). From this substitution, it follows that \(du = 2x \, dx\), and since \(x = 0\) implies \(u = 1\) and \(x = 2\) implies \(u = 5\), we have transformed the original integral in \(x\) into a new integral in \(u\). In particular, \[\begin{aligned}\end{aligned}\]

The latter integral, of course, is far easier to evaluate.

Through our work with polar, cylindrical, and spherical coordinates, we have already seen some of the issues and advantages that arise from using a change of variables with two or three variables. In the following Preview Activity, we will look geometrically at a couple of examples of changes of coordinates involving polar coordinates. In particular, we will explore exactly what properties an advantageous change of coordinates will have. In what follows, we seek to understand the general ideas behind any change of variables in a multiple integral.

Condensed: the full section is in Boelkins et al., Active Calculus Multivariable.

The general idea behind a change of variables is suggested by Preview Activity. There, we saw that in a change of variables from rectangular coordinates to polar coordinates, a polar rectangle \([r_1, r_2] \times [\theta_1, \theta_2]\) gets mapped to a Cartesian rectangle under the transformation \[\begin{aligned}\end{aligned}\]

Furthermore, it follows generally that for an original polar rectangle \(P = [r_1, r_2] \times [\theta_1, \theta_2]\), the area of the transformed rectangle \(P'\) is given by \(\frac{r_2+r_1}{2} \Delta r \Delta \theta\). Therefore, as \(\Delta r\) and \(\Delta \theta\) go to 0, this area becomes the familiar area element \(dA = r \, dr \, d\theta\) in polar coordinates. When we work with other transformations for different changes in coordinates, we will have to understand how the transformation affects area so that we may use the correct area element in the new system of variables.

We first focus on double integrals. As with single integrals, we may be able to simplify a double integral of the form \[\begin{aligned}\end{aligned}\] by making a change of variables (that is, a substitution) of the form \[\begin{aligned}\end{aligned}\] where \(x\) and \(y\) are functions of new variables \(s\) and \(t\). This transformation introduces a correspondence between a problem in the \(xy\)-plane and one in the the \(st\)-plane. The equations \(x=x(s,t)\) and \(y=y(s,t)\) convert \(s\) and \(t\) to \(x\) and \(y\); we call these formulas the change of variable formulas.

To complete the change of our integral to the new \(s,t\) variables, we need to

1. \(dA\)\(s\)\(t\)
2. transform the integrand (with the change of variable formulas)
3. \(st\)

In Activity, we used a clever relationship to convert our corner points from \(xy\)-coordinates to \(st\)-coordinates, but in general you would need to solve a system of equations to find the corresponding \((s,t)\) points. For example, if we have change of coordinate equations given by \(x=3s-2t\) and \(y =s+4t\), then the transformation of the point \((x,y)=(1,2)\), would come from solving the system of equations \(1= 3s-2t\) and \(2 =s+4t\).

If we shift our attention to the area element associated with a change of variables, we will need to look at how the area is stretched by the transformation between \(xy\)-coordinates to \(st\)-coordinates. Conceptually, derivatives will allow us to measure how quickly coordinates are changing in the context of a change of variables. Let's look back at the single variable example stated in the introduction to this section.

We have seen how in a two dimensional setting, we can convert the region of integration (Activity) and the integrand (the function being integrated) using our change of variable equations. As was suggested by the previous example, the partial derivatives of \(x\) and \(y\) with respect to \(s\) or \(t\) will measure how each variable (\(x\) or \(y\)) is stretched with respect to small a small change in the inputs (\(s\) or \(t\)). We will need to use vectors to quantify how these individual coordinate changes will amount to changes in the way we measure area.

The following result summarizes how to find the area element corresponding to a change of variables in general. We have included the details of the vector argument as a proof to this result.

Condensed: the full section is in Boelkins et al., Active Calculus Multivariable.

The details of an argument for the change of variable formula for triple integrals are more complicated the two dimension case we have just discussed and we omit those here. The general process, though, is the same as the two-dimensional case; Namely, we will use partial derivatives and the determinant of a matrix of the partial derivatives to give the appropriate volume element for the new coordinate.

Given a solid \(S'\) in the \(xyz\)-coordinate system in \(\R^3\), a change of variables transformation \(x=x(s,t,u)\), \(y=y(s,t,u)\), and \(z = z(s,t,u)\) transforms \(S'\) into a region \(S\) in \(stu\)-coordinates. Any function \(f = f(x,y,z)\) defined on \(S'\) can be considered as a function \(f = f(x(s,t,u), y(s,t,u), z(s,t,u))\) in \(stu\)-coordinates defined on \(S\). The volume element \(dV\) in \(xyz\)-coordinates cooresponds to a scaled volume element in \(stu\)-coordinates, where the scale factor is given by the absolute value of the Jacobian, \(\frac{\partial(x,y,z)}{\partial(s,t,u)}\), which is the determinant of the \(3 \times 3\) matrix \[\begin{aligned}\end{aligned}\]. (Recall that this determinant was introduced in Section.) That is, \(\frac{\partial(x,y,z)}{\partial(s,t,u)}\) is given by \[\begin{aligned}\end{aligned}\]

Suppose a change of variables \(x = x(s,t,u)\), \(y = y(s,t,u)\), and \(z = z(s,t,u)\) transforms a closed and bounded region \(S\) in \(stu\)-coordinates into a closed and bounded region \(S'\) in \(xyz\)-coordinates. Under modest conditions (that are studied in advanced calculus), the triple integral \(\iiint_{S'} f(x,y,z) \, dV\) is equal to \[\begin{aligned}\end{aligned}\]

In mathematics, a change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.

Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution).

A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial:

\(x^6 - 9 x^3 + 8 = 0.\)

Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (see Abel-Ruffini theorem). This particular equation, however, may be written

\((x^3)^2-9(x^3)+8=0\)

(this is a simple case of a polynomial decomposition). Thus the equation may be simplified by defining a new variable \(u = x^3\). Substituting x by \(\sqrt[3]{u}\) into the polynomial gives

\(u^2 - 9 u + 8 = 0 ,\)

which is just a quadratic equation with the two solutions:

\(u = 1 \quad \text{and} \quad u = 8.\)

The solutions in terms of the original variable are obtained by substituting x back in for u, which gives

\(x^3 = 1 \quad \text{and} \quad x^3 = 8.\)

\(x = (1)^{1/3} = 1 \quad \text{and} \quad x = (8)^{1/3} = 2.\)

Condensed: the full section is in Wikipedia.

Consider the system of equations

\(xy+x+y=71\)

\(x^2y+xy^2=880\)

where \(x\) and \(y\) are positive integers with \(x>y\). (Source: 1991 AIME)

Solving this normally is not very difficult, but it may get a little tedious. However, we can rewrite the second equation as \(xy(x+y)=880\). Making the substitutions \(s=x+y\) and \(t=xy\) reduces the system to \(s+t=71, st=880\). Solving this gives \((s,t)=(16,55)\) and \((s,t)=(55,16)\). Back-substituting the first ordered pair gives us \(x+y=16, xy=55, x>y\), which gives the solution \((x,y)=(11,5).\) Back-substituting the second ordered pair gives us \(x+y=55, xy=16, x>y\), which gives no solutions. Hence the solution that solves the system is \((x,y)=(11,5)\).

Let \(A\), \(B\) be smooth manifolds and let \(\Phi: A \rightarrow B\) be a \(C^r\)-diffeomorphism between them, that is: \(\Phi\) is a \(r\) times continuously differentiable, bijective map from \(A\) to \(B\) with \(r\) times continuously differentiable inverse from \(B\) to \(A\). Here \(r\) may be any natural number (or zero), \(\infty\) (smooth) or \(\omega\) (analytic).

The map \(\Phi\) is called a regular coordinate transformation or regular variable substitution, where regular refers to the \(C^r\)-ness of \(\Phi\). Usually one will write \(x = \Phi(y)\) to indicate the replacement of the variable \(x\) by the variable \(y\) by substituting the value of \(\Phi\) in \(y\) for every occurrence of \(x\).

Some systems can be more easily solved when switching to polar coordinates. Consider for example the equation

\(U(x, y) := (x^2 + y^2) \sqrt{ 1 - \frac{x^2}{x^2 + y^2} } = 0.\)

This may be a potential energy function for some physical problem. If one does not immediately see a solution, one might try the substitution

\(\displaystyle (x, y) = \Phi(r, \theta)\) given by \(\displaystyle \Phi(r,\theta) = (r \cos(\theta), r \sin(\theta)).\)

Note that if \(\theta\) runs outside a \(2\pi\)-length interval, for example, \([0, 2\pi]\), the map \(\Phi\) is no longer bijective. Therefore, \(\Phi\) should be limited to, for example \((0, \infty] \times [0, 2\pi)\). Notice how \(r = 0\) is excluded, for \(\Phi\) is not bijective in the origin (\(\theta\) can take any value, the point will be mapped to (0, 0)). Then, replacing all occurrences of the original variables by the new expressions prescribed by \(\Phi\) and using the identity \(\sin^2 x + \cos^2 x = 1\), we get

\(V(r, \theta) = r^2 \sqrt{ 1 - \frac{r^2 \cos^2 \theta}{r^2} } = r^2 \sqrt{1 - \cos^2 \theta} = r^2\left|\sin\theta\right|.\)

Now the solutions can be readily found: \(\sin(\theta) = 0\), so \(\theta = 0\) or \(\theta = \pi\). Applying the inverse of \(\Phi\) shows that this is equivalent to \(y = 0\) while \(x \not= 0\). Indeed, we see that for \(y = 0\) the function vanishes, except for the origin.

Note that, had we allowed \(r = 0\), the origin would also have been a solution, though it is not a solution to the original problem. Here the bijectivity of \(\Phi\) is crucial. The function is always positive (for \(x,y\in\reals\)), hence the absolute values.

The chain rule is used to simplify complicated differentiation. For example, consider the problem of calculating the derivative

\(\frac{d}{dx}\sin(x^2).\)

Let \(y = \sin u\) with \(u = x^2.\) Then:

\(\begin{aligned} \frac{d}{dx}\sin(x^2) &= \frac{dy}{dx} \\[6pt] &= \frac{dy}{du} \frac{du}{dx} && \text{This part is the chain rule.} \\[6pt] &= \left( \frac d {du} \sin u \right) \left( \frac{d}{dx} x^2 \right) \\[6pt] &= (\cos u) (2x) \\ &= \left (\cos(x^2) \right) (2x) \\ &= 2x\cos(x^2) \end{aligned}\)

Difficult integrals may often be evaluated by changing variables; this is enabled by the substitution rule and is analogous to the use of the chain rule above. Difficult integrals may also be solved by simplifying the integral using a change of variables given by the corresponding Jacobian matrix and determinant. Using the Jacobian determinant and the corresponding change of variable that it gives is the basis of coordinate systems such as polar, cylindrical, and spherical coordinate systems.

Variable changes for differentiation and integration are taught in elementary calculus and the steps are rarely carried out in full.

The very broad use of variable changes is apparent when considering differential equations, where the independent variables may be changed using the chain rule or the dependent variables are changed resulting in some differentiation to be carried out. Exotic changes, such as the mingling of dependent and independent variables in point and contact transformations, can be very complicated but allow much freedom.

Very often, a general form for a change is substituted into a problem and parameters picked along the way to best simplify the problem.

Probably the simplest change is the scaling and shifting of variables, that is replacing them with new variables that are "stretched" and "moved" by constant amounts. This is very common in practical applications to get physical parameters out of problems. For an n order derivative, the change simply results in

\(\frac{d^n y}{d x^n} = \frac{y_\text{scale}}{x_\text{scale}^n} \frac{d^n \hat y}{d \hat x^n}\)

where

\(x = \hat x x_\text{scale} + x_\text{shift}\)

\(y = \hat y y_\text{scale} + y_\text{shift}.\)

This may be shown readily through the chain rule and linearity of differentiation. This change is very common in practical applications to get physical parameters out of problems, for example, the boundary value problem

\(\mu \frac{d^2 u}{d y^2} = \frac{d p}{d x} \quad ; \quad u(0) = u(L) = 0\)

describes parallel fluid flow between flat solid walls separated by a distance δ; μ is the viscosity and \(d p/d x\) the pressure gradient, both constants. By scaling the variables the problem becomes

\(\frac{d^2 \hat u}{d \hat y^2} = 1 \quad ; \quad \hat u(0) = \hat u(1) = 0\)

where

\(y = \hat y L \qquad \text{and} \qquad u = \hat u \frac{L^2}{\mu} \frac{d p}{d x}.\)

Scaling is useful for many reasons. It simplifies analysis both by reducing the number of parameters and by simply making the problem neater. Proper scaling may normalize variables, that is make them have a sensible unitless range such as 0 to 1. Finally, if a problem mandates numeric solution, the fewer the parameters the smaller the number of computations.

Consider a system of equations

\(\begin{aligned} m \dot v & = - \frac{ \partial H }{ \partial x } \\[5pt] m \dot x & = \frac{ \partial H }{ \partial v } \end{aligned}\)

for a given function \(H(x, v)\). The mass can be eliminated by the (trivial) substitution \(\Phi(p) = 1/m \cdot p\). Clearly this is a bijective map from \(\mathbb{R}\) to \(\mathbb{R}\). Under the substitution \(v = \Phi(p)\) the system becomes

\(\begin{aligned} \dot p & = - \frac{ \partial H }{ \partial x } \\[5pt] \dot x & = \frac{ \partial H }{ \partial p } \end{aligned}\)

Given a force field \(\varphi(t, x, v)\), Newton's equations of motion are

\(m \ddot x = \varphi(t, x, v).\)

Lagrange examined how these equations of motion change under an arbitrary substitution of variables \(x = \Psi(t, y)\), \(v = \frac{\partial \Psi(t, y)}{\partial t} + \frac{\partial\Psi(t, y)}{\partial y} \cdot w.\)

He found that the equations

\(\frac{ \partial{L} }{ \partial y} = \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial{L}}{\partial{w}}\)

are equivalent to Newton's equations for the function \(L = T - V\), where T is the kinetic, and V the potential energy.

In fact, when the substitution is chosen well (exploiting for example symmetries and constraints of the system) these equations are much easier to solve than Newton's equations in Cartesian coordinates.