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Section 1.6 Substitution: general

In this section we provide a generalization of sorts of the substitution technique from single-variable calculus. Recall the set up in that context: given a continuous function \(f\) and differentiable function \(g\colon [a,b]\colon\rightarrow \R\) we have

\begin{align*} \int_{g(a)}^{g(b)} f(u)\, du \amp =\int_{a}^{b}f(g(x))g'(x)\, dx \amp \left(\begin{array}{c}u=g(x)\\ du=g'(x)\, dx\end{array}\right)\text{.} \end{align*}

Theorem 1.6.6 is our generalization of this result to double and triple integrals.

Definition 1.6.1. Transformations.

Let \(D, D'\subseteq \R^n\text{,}\) where \(n=2\) or \(n=3\text{.}\) A function

\begin{align*} G\colon D \amp\rightarrow D' \\ (x_1,x_2,\dots, x_n) \amp\mapsto (g_1(x_1,x_2,\dots, x_n),\dots, g_n(x_1,x_2,\dots, x_n)) \end{align*}

is a transformation if it satisfies the following properties.

  1. \(G\) is one-to-one (injective) on the interior of \(D\text{.}\)

  2. \(G(D)=D'\text{:}\) i.e., \(G\colon D\rightarrow D'\) is onto (surjective).

The function \(G\) is continuously differentiable at a point \((x_1,x_2,\dots, x_n)\in D\) if for all \(1\leq i,j\leq n\) the partial derivative \(\partial g_i/\partial x_j\) is continuous at \((x_1,x_2,\dots, x_n)\text{.}\)

Definition 1.6.2. Determinant and Jacobian.

Recall that the determinant map on \(n\times n\) matrices, denoted \(\det\text{,}\) is defined in the \(n=2,3\) cases as

\begin{align*} \det\begin{pmatrix} a\amp b\\ c\amp d \end{pmatrix}\amp=ad-bc \\ \det \begin{pmatrix}a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{pmatrix} \amp =a(ei-gh)-b(di-fg)+c(dh-eg) \text{.} \end{align*}

Let \(D, D'\subseteq \R^n\text{,}\) where \(n=2\) or \(n=3\text{.}\) Given a function

\begin{align*} G\colon D \amp\rightarrow D' \\ (x_1,x_2,\dots, x_n) \amp\mapsto (g_1(x_1,x_2,\dots, x_n),\dots, g_n(x_1,x_2,\dots, x_n)) \end{align*}

its Jacobian \(J(G)=\frac{\partial(g_1,g_2,\dots, g_n)}{\partial(x_1,x_2,\dots, x_n)}\) is the function defined as follows:

\begin{align*} J(G)=\frac{\partial(g_1,g_2)}{\partial(x_1,x_2)} \amp=\det \begin{pmatrix} \frac{\partial g_1}{\partial x}\amp \frac{\partial g_1}{\partial y}\\ \frac{\partial g_2}{\partial x}\amp \frac{\partial g_2}{\partial y} \end{pmatrix} \amp (n=2)\\ J(G)=\frac{\partial(g_1,g_2,g_3)}{\partial(x_1,x_2,x_3)}\amp=\det \begin{pmatrix} \frac{\partial g_1}{\partial x}\amp \frac{\partial g_1}{\partial y}\amp \frac{\partial g_1}{\partial z}\\ \frac{\partial g_2}{\partial x}\amp \frac{\partial g_2}{\partial y}\amp \frac{\partial g_2}{\partial z}\\ \frac{\partial g_3}{\partial x}\amp \frac{\partial g_3}{\partial y}\amp \frac{\partial g_3}{\partial z} \end{pmatrix} \amp (n=3) \end{align*}

Remark 1.6.3. Jacobian notation.

To ease notation we will often give names to the outputs of the component functions \(g_1, g_2, \dots, g_n\text{.}\) For example, given a function of the form \(G(x,y,z)=(g_1(x,y,z),g_2(x,y,z),g_3(x,y,z))\) we might write

\begin{align*} u \amp=g_1(x,y,z) \\ v\amp=g_2(x,y,z) \\ w \amp=g_3(x,y,z) \end{align*}

for short, in which case the Jacobian of \(G\) is denoted \(J(G)=\frac{\partial(u,v,w)}{\partial(x,y,z)} \text{.}\)

Definition 1.6.4. Invertible linear transformations.

Functions of the form

\begin{align*} G(x,y) \amp = (ax+by, cx+dy)=(u,v) \amp (n=2)\\ H(x,y,z)\amp= (ax+by+cz,dx+ey+fz, gx+hy+iz)=(u,v,w) \amp (n=3) \end{align*}

where \(a,b,c,\dots, g,h,i\) are fixed constants, are called linear. We can easily compute their Jacobians:

\begin{align*} \frac{\partial(u,v)}{\partial(x,y)} \amp=\det\begin{pmatrix} a\amp b\\ c\amp d \end{pmatrix}=ad-bc\\ \frac{\partial(u,v,w)}{\partial(x,y,z)}\amp= \det \begin{pmatrix}a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{pmatrix} =a(ei-gh)-b(di-fg)+c(dh-eg) \text{.} \end{align*}

In linear algebra you learn that the \(G\) and \(H\) are one-to-one everywhere if and only if the Jacobians \(J(G)\) and \(J(H)\) are nonzero. This gives us a convenient family of continuously differentiable transformations, called invertible linear transformations.

Remark 1.6.5.

The textbook's determinant notation looks a lot like absolute value notation: e.g., it writes \(\det \begin{pmatrix} a\amp b\\ c\amp d \end{pmatrix}\) as \(\abs{\begin{array}{cc} a\amp b\\ c\amp d\end{array}}\text{.}\) This can lead to confusion in the statement of Theorem 1.6.6, which involves taking the absolute value of the Jacobian of a function. Accordingly, we will stick with our \(\det\) notation.

Remark 1.6.7. Substitution as change of variable.

Substitution is often described as a change of variable technique. Let's see why, treating the \(n=2\) case. Suppose we are asked to compute \(\iint\limits_\mathcal{R}f(x,y)\, dA\) over the region \(\mathcal{R}'\text{.}\) If we want to make use of (1.6.1) to transform this integral to a simpler one, we should treat \(\mathcal{R}\) as \(\mathcal{D'}\text{,}\) the target region in our substitution setup.

Next, suppose that in terms of the expressions

\begin{align*} u \amp = h_1(x,y)\\ v \amp = h_2(x,y) \end{align*}

we have \(f(x,y)=q(h_1(x,y),h_2(x,y))=q(u,v)\text{.}\) Substitution allows us to compute the original integral in terms of an integral involving the new function \(q\) as follows:

  1. Let \(H=(h_1(x,y),h_2(x,y))\text{.}\) The function \(H\) maps our original region \(\mathcal{R}\) to a new region \(\mathcal{S}\) that we consider to live in the \(uv\)-plane.

  2. Since \(\mathcal{R}=\mathcal{D}'\) in the Theorem 1.6.6 setup, we identify \(\mathcal{S}=\mathcal{D}\) in this setup.

  3. If \(H\) is a transformation, then it is invertible, and there is an inverse function \(H^{-1}=G=(g_1(u,v),g_2(u,v))\text{,}\) where

    \begin{align*} x \amp = g_1(u,v)\\ y \amp = g_2(u,v) \end{align*}

    are equations expressing \(x\) and \(y\) in terms of \(u\) and \(v\text{.}\) By inverse properties, \(G\) maps \(\mathcal{S}\) (in the \(uv\)-plane) onto \(\mathcal{R}\) (in the \(xy\)-plane).

  4. Using Theorem 1.6.6 we have

    \begin{align*} \iint\limits_{\mathcal{R}}f(x,y)\, dA \amp =\iint\limits_{\mathcal{S}}f(g_1(u,v),g_2(u,v))\vert J(G)\vert\,dA \\ \amp = \iint\limits_{\mathcal{S}}q(u,v)\left\vert\frac{\partial(x,y)}{\partial(u,v)}\right\vert\, dA\text{.} \end{align*}

Let's make the discussion above official with a procedure. As above, it is expressed for \(n=2\) case, but generalizes easily to \(n=3\text{.}\)

Example 1.6.9. Substitution: linear, two variables.

Define \(\mathcal{R}\) be the trapezoid with vertices \((0,1),(0,2),(1,0),(2,0)\text{.}\) Use a substitution to compute

\begin{equation*} \iint\limits_\mathcal{R}e^{(x-y)/(x+y)}\, dA \end{equation*}
Solution.

Use the substitution \(H(x,y)\)

\begin{align*} u \amp =x-y\\ v \amp =x+y \end{align*}

with inverse \(G(u,v)\)

\begin{align*} x\amp=\frac{1}{2}(u+v) \\ y \amp=\frac{1}{2}(-u+v) \end{align*}

and Jacobian \(\partial(x,y)/\partial(u,v)=\frac{1}{2}\text{.}\) The trapezoidal region in the \(xy\)-plane has boundary equations \(x+y=1, x+y=2, x=0, y=0\text{,}\) which are transformed to the boundary equations \(v=1, v=2, u=-v, u=v\text{.}\) This defines the trapezoidal region

\begin{equation*} \mathcal{S}=\{(u,v)\colon 1\leq v\leq 2, -v\leq u\leq v\} \end{equation*}

in the \(uv\)-plane, and by substitution we have

\begin{align*} \iint\limits_{\mathcal{R}}e^{(x-y)/x+y)}\, dA \amp = \iint\limits_\mathcal{S}e^(u/v)\abs{\frac{\partial(x,y)}{\partial(u,v)}}\, dA \\ \amp=\int_1^2\int_{-v}^v\frac{1}{2}e^(u/v)\, du\, dv\\ \amp = \frac{1}{2}\int_1^2v(e-1/e)\, dv\\ \amp =\frac{3}{4}(e-1/e) \end{align*}

Example 1.6.10. Volume of ellipsoid.

Let \(\mathcal{R}\) be the solid ellipsoid defined by the inequality

\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leq 1\text{,} \end{equation*}

where \(a,b,c\in \R\) are fixed constants. Use change of variables to compute

\begin{equation*} \operatorname{vol} \mathcal{R}=\iiint\limits_\mathcal{R}1\, dV\text{.} \end{equation*}
Solution.

The transformation \(G(u,v,w)=(au,bv,cw)\) transforms the the solid ball \(\mathcal{S}\colon u^2+v^2+w^2\leq 1\) in \(uvw\)-space to the given ellispoid \(\mathcal{R}\) in \(xyz\)-space. The Jacobian of \(G\) is

\begin{equation*} J(G)=\det \begin{pmatrix} a\amp 0\amp 0\\ 0\amp b\amp 0\\ 0\amp 0\amp c \end{pmatrix}=abc\text{.} \end{equation*}

Now compute

\begin{align*} \operatorname{vol} \mathcal{R} \amp =\iiint\limits_\mathcal{R}1\, dV \\ \amp =\iiint\limits_\mathcal{S}1 \abs{J(G)}\, dV \amp \text{(subst.)} \\ \amp =abc\iiint\limits_\mathcal{S}1\, dV\\ \amp =abc\operatorname{vol}\mathcal{S}\\ \amp =\frac{4}{3}\pi abc \text{,} \end{align*}

since the volume of the unit ball \(\mathcal{R}\) is \(4\pi/3\text{.}\)