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Section 3.1 Change of Variables

Among the single variable integration techniques, substitution was the most useful and most important of them. In this section, I want to extend the notation of substitution to multiple integration. Now, since calculation of multiple integrals is by interated integrals, I could do substitutions individual in each variable, since I do the single-variable integrations one-by-one. That's not what I want to do here: I want to introduce substitution of all the variables at once.

Let me review single variable substitution, formalizing the language and showing the way forward for higher dimensions. Consider a single variable integral.

\begin{equation*} \int_a^b f(x) dx \end{equation*}

Typically, I substitute \(u = g(x)\) using some invertible function \(g\text{.}\) Then I calculate \(du = g^\prime(x) dx\text{.}\) With these two pieces, I try to change all the \(x\) variables to \(u\) variables in the integral. I also change the bounds, starting from \(u = g(a)\) to \(u = g(b)\text{.}\) There is an invertible function, \(u = g(x)\text{,}\) which defined the substitition; in the process, the old variable is the independent variable of the transformation and the new variable is the dependent variable.

However, I could have setup the relationship differently, by having \(x = h(u)\) and \(dx = h^\prime(u) du\text{,}\) making old variable the dependent variable and the new variable the independent variable. There is an example of this in the integration techinques from Calculus II: trigonometric substitution. Consider a trig substitution: \(x = \tan \theta\text{.}\) the function acts on the new variable and the old variable is the dependent variable.

The standard single-variable process, where the old variable is the dependent variable, substitution was mostly about doing the chain rule in reverse. I would try to look for a form that resembled the chain rule, with the derivative of an inside function in the integrand. This isn't a fruitful way to extend the idea to multiple integration, since the chain rule doesn't extend in a way that applies to these integrals. In the process for trigonometric substitution, the substitution was more about changing the situation, turing other functions into trigonometric functions to use trig identities to solve integrals. This is a much more productive direction: substitution is a thing that changes the situation or functions involved to make a more reasonable integral.

Therefore, let me clarify the setps in the single variable case for this second setup. I introduce a new variable \(u\) by \(x = h(u)\) for some invertible function \(h\text{.}\) Then \(dx = h^\prime(u) du\) and the bounds become \(h^{-1}(a)\) and \(h^{-1}(b)\text{.}\) We get a new integral in the variable \(u\text{.}\)

\begin{equation*} \int_{u=h^{-1}(a)}^{u=h^{-1}(b)} f(h(u)) h^\prime(u) du \end{equation*}

The key relationship here is \(dx = h^\prime(u) du\text{:}\) the derivative tells us the relationships between the differential terms. In Calculus III, I did all that work on extending the notion of the derivative. I came to the conclusion that the derivative measures (in local coordinates) the best linear approximation to the function. In one variable, that was just multiplication by a number. In several variables, though, that linear approximation was a matrix, called the Jacobian matrix. Let me recall the definition. (I'll only state it for an equal number of input and output variables, since that is the case that I will use for extending substitution).

Definition 3.1.1.

Let \(F: \RR^n \rightarrow \RR^n\) be a function on \(n\) variables, \(u_1, \ldots, u_n\text{.}\) We can write \(F\) as its component functions \(F_1, \ldots, F_n\text{.}\) Then the Jacobian matrix of \(F\) is the matrix of all partial derivatives.

\begin{equation*} J(F) = \begin{pmatrix} \dfrac{\del F_1}{\del u_1} \amp \dfrac{\del F_1}{\del u_2} \amp \ldots \amp \dfrac{\del F_1}{\del u_n} \\ \dfrac{\del F_2}{\del u_1} \amp \dfrac{\del F_2}{\del u_2} \amp \ldots \amp \dfrac{\del F_2}{\del u_n} \\ \vdots \amp \vdots \amp \vdots \amp \vdots \\ \dfrac{\del F_n}{\del u_1} \amp \dfrac{\del F_n}{\del u_2} \amp \ldots \amp \dfrac{\del F_n}{\del u_n} \end{pmatrix} \end{equation*}

The determinant \(\det J(F)\) is called the Jacobian determinaint. Often, this determinant is just called the Jacobian of the function.

From the single variable case \(dx = h^\prime(u) du\text{,}\) the Jacobian is part of the new differential \(du\text{.}\) If I start with variables \(x_1, x_2, \ldots, x_n\) and we have a function \(F: \RR^n \rightarrow \RR^n\) that has \(x_i\) as its output (the original variables are the range of the transformation), I can write \(x_1 = F_1(u_1, \ldots, u_n)\text{,}\) \(x_2 = F_2(u_1, \ldots, u_n)\) up to \(x_n = F_n(u_1, \ldots, u_n)\text{.}\) The Jacobian determines the relationship between the differential in these variables.

\begin{equation*} dx_1 dx_2 \ldots dx_n = (\det J(F)) du_1 du_2 \ldots du_n \end{equation*}

I can interpret the Jacobian as the change in area/volume/hyper-volume due to the change in variables. Its appearance in the integral makes sense with this interpretation: the integral is measureing area/volume/hyper-volume, so when I change variables, I need a term that keeps track of the relative change in area/volume/hypervolume.

Let \((x,y) = F(u,v) = (3u, 4v)\text{.}\) Here are the partial derivatives, the Jacobian matrix, and the Jacobian determinant.

\begin{align*} \frac{\del F_1}{\del u} \amp = 3\\ \frac{\del F_1}{\del v} \amp = 0\\ \frac{\del F_2}{\del u} \amp = 0\\ \frac{\del F_2}{\del v} \amp = 4\\ J(F) \amp = \begin{pmatrix} 3 \amp 0 \\ 0 \amp 4 \end{pmatrix}\\ \det J(F) \amp = (3)(4) - (0)(0) = 12\\ dx dy \amp = 12 du dv \end{align*}

The function is a dialation by \(3\) in \(u\) and by \(4\) in \(v\text{,}\) so the effect on area is multiplication by \(12\text{,}\) which makes sense. The function is linear already, so the linear approximation is constant: it is exactly the function itself.

Let \((x,y) = F(u,v) = (u^2, v)\text{.}\) Here are the partial derivatives, the Jacoabian matrix, and the Jacobian determinant.

\begin{align*} \frac{\del F_1}{\del u} \amp = 2u\\ \frac{\del F_1}{\del v} \amp = 0\\ \frac{\del F_2}{\del u} \amp = 0\\ \frac{\del F_2}{\del v} \amp = 1\\ J(F) \amp = \begin{pmatrix} 2u \amp 0 \\ 0 \amp 1 \end{pmatrix} \\ \det J(F) \amp = 2u\\ dx dy \amp = 2u du dv \end{align*}

This Jacobian isn't constant. This is a stretch in \(u\text{,}\) the but effect is exagerated away from the origin due to the square term. Therefore, as \(u\) get larger, the stretch effect is greater. The Jacobian reflects that.

Let \((x,y) = F(r,\theta) = (r \cos \theta, r \sin \theta)\text{.}\) (There are the polar cooredinates, which I will discuss at length in Section 3.2.) Here are the partial derivatives, the Jacobian matrix and the Jacobian determinant.

\begin{align*} \frac{\del F_1}{\del r} \amp = \cos \theta\\ \frac{\del F_1}{\del \theta} \amp = -r\sin \theta\\ \frac{\del F_2}{\del r} \amp = \sin \theta\\ \frac{\del F_2}{\del \theta} \amp = r \cos \theta\\ J(F) \amp = \begin{pmatrix} \cos \theta \amp - r \sin \theta \\ \sin \theta \amp r \cos \theta \end{pmatrix} \\ \det J(F) \amp = r\\ dx dy \amp = r dr d\theta \end{align*}

This Jacobian shows that the radius term gives the effect on area. This make sense: for larger circles, the differential area is part of a larger arc. Figure 3.1.5 illustrates this. There are two regions; each region has the same increase in radii between the inside and outside of the region, and each region has the same inscribed angle. However, the two regions clearly have a different area: a region with the same ‘width’ (different in radius) and the same ‘height’ (difference in angle) is a larger region the further out it is on the circle. Ths figure shows this at a marcoscopic level; the Jacobian is essentially measuring the same reality at an infinitesimal level. The Jacobian multiplies by the radius term \(r\) to indicate this scaling of area as the radius increases.

Figure 3.1.5. Similar Regions with Different Areas

Let \((x,y) = F(u,v) = (u, uv)\text{.}\) This is the change in variables we used in Example 2.2.11, just now treated as a two-variable changes of variables instead of a one variable substitution. Here are the partial derivatives, the Jacobian matrix, and the Jacobian determinant.

\begin{align*} \frac{\del F_1}{\del u} \amp = 1\\ \frac{\del F_1}{\del v} \amp = 0\\ \frac{\del F_2}{\del u} \amp = v\\ \frac{\del F_2}{\del v} \amp = u\\ J(F) \amp = \begin{pmatrix} 1 \amp 0 \\ v \amp u \end{pmatrix} \\ \det J(F) \amp = u\\ dx dy \amp = u du dv \end{align*}

For single variable integrals, substitution was used to change the integrand to make it more reasonable. For multiple integration, there is difficult both in the integrand and in the region of integration. I can use a change of variables to help us with either problem. Sometime, as in the following examples, the same transformation helps both. When I are choosing a change of variables to simplify the region of integration, I am looking for a change of variables where some of the non-constant bounding curves of the region become \(u=c\) or \(v=c\text{.}\) This can let me setup reasonable bounds of integration for an integral over a complicated region.

Figure 3.1.8. The region in the first quadrant betwee \(x+y=1\) and \(x+y=2\text{.}\)

Let \(D\) be the region in the first quadrant between the lines \(x+y = 1\) and the line \(x+y =2\text{,}\) shown in Figure 3.1.8. In the following integral, I want to simply the integrand and integrate over a simpler region if possible.

\begin{equation*} \int_D \cos \left( \frac{y-x}{y+x} \right) dA \end{equation*}

To simplify the integrand, I can look for transformations where \(u = x+y\) and \(v = x-y\text{.}\) Inverting these (either by linear algebra, since the transformation is linear, or just by solving with conventional algebra), I get \(x = \frac{u+v}{2}\) and \(y = \frac{u-v}{2}\text{.}\)

Then I look at the bounds. If I take \(x+y=1\) and replace \(x\) and \(y\text{,}\) I get \(u=1\text{.}\) Likewise, \(x+y=2\) is \(u=2\text{.}\) This is excellent, since I need at least one variable with constant bounds; I'll take \(u\) to be the variable of the outside integral. The other bounds are the axes, which I can express by the equations \(y=0\) and \(x=0\text{.}\) These turn into \(u=v\) and \(u=-v\text{,}\) respective. Therefore, the bounds of \(v\) will be these two lines: \(v \in [u, -u]\text{.}\)

Finally, we calculate the Jacobian.

\begin{align*} x_u \amp = \frac{1}{2}\\ x_v \amp = \frac{1}{2}\\ y_u \amp = \frac{1}{2}\\ y_v \amp = -\frac{1}{2}\\ \det J \amp = -\frac{1}{2}\\ dx dy \amp = \frac{1}{-2} du dv \end{align*}

We have the new integrand, the new bounds and the Jacobian. We can proceed to the new integral.

\begin{align*} \int_D \cos \left( \frac{y-x}{y+x} \right) dA \amp = \int_1^2 \int_{-u}^u \cos \left( \frac{-v}{u} \right) \frac{-1}{2} dv du\\ \amp = \frac{-1}{2} \int_1^2 \left. -\sin \left( \frac{-v}{u} \right) (-u) \right|_{-u}^{u} du\\ \amp = \frac{-1}{2} \int_1^2 u (\sin (-1) - \sin(1)) du\\ \amp = \sin (1) \left. \frac{u^2}{2} \right|_1^2\\ \amp = \sin (1) (2 - \frac{1}{2}) = \frac{3\sin(1)}{2} \end{align*}
Figure 3.1.10. A region of integration.

Consider \(D\) be the region in the plane bounded by the curves \(y = \frac{1}{x}\text{,}\) \(y = \frac{3}{x}\text{,}\) \(y = 3x\) and \(y = x\text{,}\) as shown in Figure 3.1.10. Consider the following integral.

\begin{equation*} \int_D xy dA \end{equation*}

The integrand is reasonable, so I will look for a substitution to changes the bounds. I would like to produce constant bounds in at least one variable. To do that, I'll use this change of variables.

\begin{equation*} (x,y) = F(u,v) = (uv,v) \end{equation*}

The lines \(y=3x\) and \(y=x\) become \(v=3uv\) and \(v=uv\text{.}\) Away from \(v=0\text{,}\) these lines are \(u=\frac{1}{3}\) and \(u=1\text{,}\) so there are constant bounds in \(u\) and I will treat \(u\) as the outside variable. (I am safe with the assumption \(v=0\) since \(v=y\) and the region is disjoint from the \(x\)-axis, where \(y=0\)). The curves \(y = \frac{1}{x}\) and \(y = \frac{3}{x}\) become \(v = \frac{1}{uv}\) and \(v = \frac{3}{uv}\text{,}\) which simplify into \(uv^2 =1\) and \(uv^2=3\text{.}\) If I solve for \(v\) in each, I get \(v = \sqrt{\frac{1}{u}}\) and \(v = \sqrt{\frac{3}{u}}\text{.}\) That means I can take \(u \in [\frac{1}{3}, 3]\) and \(v \in \left[\sqrt{\frac{1}{u}}, \sqrt{\frac{3}{u}} \right]\text{.}\) The integrand \(xy\) becomes \(uv^2\text{.}\) Now I need to calculate the partial derivatives and the Jacobian to complete the change of variables.

\begin{align*} x_u \amp = v\\ x_v \amp = u\\ y_u \amp = 0\\ y_v \amp = 1\\ \det J \amp = (v)(1) - (u)(0) = v\\ dx dy \amp = v du dv \end{align*}

Now that I have all the pieces, I go ahead and do the change of variables: inputing the new bounds for the region, changing the variables in the integrand, and using the Jacobian to change the differential. After the change of variables, I calculate the iterated integral as usual.

\begin{align*} \int_D xy dA \amp = \int_{\frac{1}{3}}^1 \int_{\sqrt{\frac{1}{u}}}^{\sqrt{\frac{3}{u}}} uv^2 v dv du\\ \amp = \int_{\frac{1}{3}}^1 \left. u\frac{v^4}{4} \right|_{\sqrt{\frac{1}{u}}}^{\sqrt{\frac{3}{u}}} du\\ \amp = \int_{\frac{1}{3}}^1 \frac{u}{4} \left( \frac{9}{u^2} - \frac{1}{u^2} \right) du\\ \amp = \int_{\frac{1}{3}}^1 \frac{2}{u} du\\ \amp = 2 \ln |u| \bigg|_{\frac{1}{3}}^1\\ \amp = 2 \ln 1 - 2 \ln \frac{1}{3} = -2 \ln \frac{1}{3} = 2 \ln 3 \end{align*}

For the same integral, I could have use the change of variables \(y=v\) and \(x= \frac{u}{v}\text{.}\) Under this transformation, the bounds become \(u \in [1,3]\) and \(v \in [\sqrt{u}, \sqrt{3u}]\text{.}\) The Jacobian is now \(\frac{1}{v}\) and the integrand is \(u\text{.}\) Here is the calculate with this alternative change of variables.

\begin{align*} \int_D xy dA \amp = \int_1^3 \int_{\sqrt{u}}^{\sqrt{3u}} \frac{1}{v} u dv du\\ \amp = \int_1^3 u \ln |v| \bigg|_{\sqrt{u}}^{\sqrt{3u}} du\\ \amp = \int_1^3 u (\ln |\sqrt{3u}| - \ln |\sqrt{u}|du\\ \amp = \int_1^3 \frac{1}{2} (u \ln 3u - u \ln u) du\\ \amp = \left. \frac{1}{2} \left(\frac{1}{4} u^2 (2 \ln 3u - 1) \frac{1}{3} - \frac{1}{4} u^2 (2 \ln u - 1) \right) \right|_1^3 \end{align*}

If I evaluate this, it will evaluate, eventually, to \(2 \ln 3\text{.}\) However, the expression that results from this change of variable is obviously more complicated than the first change of variables. Different changes of variables can have very different effects on the integral; some will make it easier, some more difficult. Also, the answers from different changes of variables can look quite differet, such as the expression above, while actually being the same value.

As an aside, there is a simplification in the second version of this example that removes the complication. In the second last step, I could write \(\ln 3u\) as \(\ln 3 + \ln u\text{.}\) If I do that, the calculation proceeds as follows.

\begin{align*} \amp = \int_1^3 \frac{1}{2} (u \ln 3u - u \ln u) du\\ \amp = \frac{1}{2} \int_1^3 u (\ln 3 + \ln u - \ln u) du = \frac{1}{2} \int_1^3 u \ln 3 du\\ \amp = \left. \frac{\ln 3}{2} \frac{u^2}{2} \right|_1^3 = \frac{\ln 3}{4} (9-1) = \frac{8 \ln 3}{4} = 2 \ln 3 \end{align*}