maths.free › Multivariable Calculus › 4. Multiple Integrals › Change of Variables
Change of Variables
Many instructors use as an introduction to the general setting of this section and other skip this generalization.
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.
Introduction
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.
Change of Variables in Polar Coordinates
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 \[\begin{aligned}\end{aligned}\]
Example
The vertices of the polar rectangle \(P\) given by \([r_1, r_2] \times [\theta_1, \theta_2]\) are transformed into the vertices of a closed and bounded region \(P'\) in rectangular coordinates. If we view the standard coordinate system as having the horizontal axis represent \(r\) and the vertical axis represent \(\theta\), then the polar rectangle \(P\) appears to us at left in Figure. The image \(P'\) of the polar rectangle \(P\) under the transformation given by is shown at right in Figure. We thus see that there is a correspondence between a simple region (a traditional, right-angled rectangle) and a more complicated region (a fraction of an annulus) under the function \(T\) given by \(T(r, \theta) = (r\cos(\theta), r\sin(\theta))\).
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.
General Change of Coordinates
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
- \(dA\)\(s\)\(t\)
- transform the integrand (with the change of variable formulas)
- \(st\)
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.
Change of Variables in a Triple Integral
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 \[\begin{aligned}\end{aligned}\]
Activity
Consider the solid \(S'\) defined by the inequalities \(0 \leq x \leq 2\), \(\frac{x}{2} \leq y \leq \frac{x}{2}+1\), and \(0 \leq z \leq 6\). Consider the transformation defined by \(s = \frac{x}{2}\), \(t = \frac{x-2y}{2}\), and \(u = \frac{z}{3}\). Let \(f(x,y,x) = x-2y+z\).
The transformation turns the solid \(S'\) in \(xyz\)-coordinates into a box \(S\) in \(stu\)-coordinates. Apply the transformation to the boundries of the solid \(S'\) to find \(stu\)-coordinate descriptions of the box \(S\).
Compute and simplify the Jacobian \(\frac{\partial(x,y,z)}{\partial(s,t,u)}\).
Use the transformation to perform a change of variables and evaluate \(\iiint_{S'} f(x,y,z) \, dV\) by evaluating \[\begin{aligned}\end{aligned}\]
Practice (1)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Consider the change of variables \[\begin{aligned}\end{aligned}\] which is the transformation from spherical coordinates to rectangular coordinates. Determine the Jacobian of the transformation. How is the result connected to our earlier work with iterated integrals in spherical coordinates?
Жауап беріңіз
Here we have \[\begin{aligned}\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}\right| \amp = \left|\frac{\partial x}{\partial \rho}\left[\frac{\partial y}{\partial \theta}\frac{\partial z}{\partial \phi} - \frac{\partial y}{\partial \phi}\frac{\partial z}{\partial \theta}\right] - \frac{\partial x}{\partial \theta}\left[\frac{\partial y}{\partial \rho}\frac{\partial z}{\partial \phi} - \frac{\partial y}{\partial \phi}\frac{\partial z}{\partial \rho}\right] + \frac{\partial x}{\partial \phi}\left[\frac{\partial y}{\partial \rho}\frac{\partial z}{\partial \theta} - \frac{\partial y}{\partial \theta}\frac{\partial z}{\partial \rho}\right]\right| \\ \amp = \left|\sin(\phi)\cos(\theta)[(\rho\sin(\phi)\cos(\theta))(-\rho \sin(\phi)) - (\rho \cos(\phi) \sin(\theta))(0)] \right. \\ \amp \qquad + \rho \sin(\phi) \sin(\theta)[(\sin(\phi)\sin(\theta))(-\rho \sin(\phi)) - (\rho \cos(\phi) \sin(\theta))(\cos(\phi))] \\ \amp \qquad + \left. \rho \cos(\phi) \cos(\theta)[(\sin(\phi)\sin(\theta))(0) - (\rho \sin(\phi) \cos(\theta))(\cos(\phi))]\right| \\ \amp = \left|-\rho^2 \sin^3(\phi) \cos^2(\theta) - \rho^2 \sin^3(\phi) \sin^2(\theta) \right. \\ \amp \qquad - \left. \rho^2 \sin(\phi)\cos^2(\phi) \sin^2(\theta) - \rho^2 \sin(\phi) \cos^2(\phi) \cos^2(\theta) \right| \\ \amp = \left| -\rho^2 [ \sin^3(\phi) + \sin(\phi) \cos^2(\phi)] \right| \\ \amp = \left|-\rho^2 [ \sin^3(\phi) + \sin(\phi) (1-\sin^2(\phi))] \right| \\ \amp = \rho^2 \sin(\phi)\end{aligned}\].
This is exactly the volume element in spherical coordinates that we derived earlier.
Symbols used here
Derivative with respect to x, holding the other variables fixed.
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
Inequalities that allow equality; < and > exclude it.
Vector of partial derivatives; points uphill.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Change of Variables
- What is a change of variables in a multivariable coordinate system?
- How can we measure how area and volume are distorted for a change of variables in two or three dimensions?
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Өзіңіздіңіңізді сынап көріңіз
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.
Келесіде Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems