Section 15.9 of Calculus, Early Transcendentals by Stewart.
Introduction.
The main new challenge in multiple integrals is dealing with domains of shapes other that rectangles.
Iterated integrals with the limits of "inner" integrals depending on "outer" variables help in some cases, as do polar, cylindrical and spherical coordinates, but we now develop a more general strategy.
As usual we deal with two-dimensional integrals befoer going on to thre dimensions, but in fact we first revisit one-dimensional integrals with modified notation that fits better with what happens in multiple dimensions.
Subsection5.7.1Changing Variables in 1D Integrals
Inverse Substitution.
To extend changes of variables in multiple integrals beyond those seen for polar, cylindrical and spherical coordinates, let us first review and rephrase the 1D version.
Consider a function \(f(x)\) defined on an interval \([a,b]\text{.}\) The idea is to change from the original variable \(x\) to a new variable \(u\) related by a function \(x=g(u)\) where \(g\) is one-to-one on the interval \([a,b]\text{,}\) so that \(g\) is either increasing or decreasing there.
Note: we express the "old" variable as a function of the "new" one, the inverse substitution approach seen with the use of trigonometric substitutions like \(x = a\sin\theta\) for integrals in involving \(\sqrt{a^2 - x^2}\text{;}\) see Section 3.3 of OpenStax Calculus Volume 2 2
When one changes variables in a definite integral, the domain is changed. In the 1D case, this is just changing the upper and lower limits of integration:
One catch here is that given the original domain \(a \leq b\text{,}\) the endpoints of the new domain must be found by equation solving or with inverses: \(c=g^{-1}(a)\text{,}\)\(d=g^{-1}(b)\text{.}\) Usually this will not be a problem because the change of variables will be chosen precisely for the sake of getting a known, simple domain for the new integral. (This idea was irrelevant with functions one variable, where the domain of integration is always equally simple: an interval.)
Integrals over Intervals Considered as Domains.
To mimic the notation for multiple integrals over domains, it is convenient to express an definite interval as being over an interval \(I=[a,b]\text{.}\) Thus,
\begin{equation*}
\int\limits_{[a,b]} f(x) \, dx \text{ means }
\int_a^b f(x) \, dx, \text{ where we must have } a \leq b.
\end{equation*}
If function \(g\) is increasing, the Substitution Rule can be written as
Integrals over Intervals with Decreasing \(g\) (\(g' \lt 0\)).
However, if function \(g\) is decreasing, \(a \lt b\) leads to \(c \gt d\text{,}\) so the new interval is \(J=[d,c]\text{,}\) not \([c,d]\text{.}\) Thus
\begin{equation*}
\int\limits_{[a,b]} f(x) \, dx
= \int_a^b f(x) \, dx
= \int_c^d f(g(u)) g'(u) \, du
= -\int_d^c f(g(u)) g'(u) \, du
= -\int\limits_{[d,c]} f(g(u)) g'(u) \, du
\end{equation*}
Since \(g\) is decreasing, \(g'(u) \lt 0 \) and so \(-g'(u) = |g'(u)|\text{,}\) giving
A function like \(T\) that takes points in the plane \(\mathbb{R}^2\) to other points in the plane is called a transformation. We now consider more general transformations to new variables \(u\) and \(v\text{,}\)
where we want both \(g\) and \(h\) to be differentiable, (so all four first partial derivatives exist and are continuous): these are called \(C^1\) transformations.
Images and One-to-One Transformations.
If \(T(u_1,v_1)=(x_1,y_1)\text{,}\) the point \((x_1,y_1)\) is called the image of \((u_1,v_1)\text{.}\) The set of all such images is the range, also called the image of the domain of the transformation.
With polar coordinates, we tried to have unique values of \(r\) and \(\theta\) for each point \((x,y)\) in the plane.
A transformation that achieves this uniqueness is one-to-one: no two points in its domain have the same image. For example, the transformation from polar coordinates to cartesian coordinates is one-to-one on any domain on which \(r \gt 0\) and \(-\pi \lt \theta \leq \pi\text{,}\) which is the same as excluding the origin.
(To include the origin too, one could allow only \(\theta=0\) when \(r=0\text{.}\))
The Inverse of a One-to-One Transformation.
If a transformation is one-to-one, then for any point \((x,y)\) in its range, there is only only point \((u,v)\) with \(T(u,v)=(x,y)\text{,}\) so this point \((u,v)\) is the inverse of \((x,y)\) under \(T\), and there is an inverse transformation \(T^{-1}\text{,}\)
For \(T\) the transformation in Equation (5.7.7) from polar to cartesian coordinates, and on a domain with \(r>0\) and \(-\pi \lt \theta \leq \pi\text{,}\)
Subsection5.7.3Transformations and Double Integals
Consider a double integral \(\iint\limits_R f(x,y)\ dA\) and a transformation \((x,y)=T(u,v)\) such that the domain \(R\) in the \((x,y)\) plane is the image of a domain \(S\) in coordinates \((u,v)\text{.}\) This is sometimes expressed as \(R=T(S)\text{.}\) One example we have seen is when the new coordinates are polar coordinates \(r\) and \(\theta\text{,}\)\(R\) is a polar rectangle, so \(S\) is the rectangular domain \(a \leq r \leq b\text{,}\)\(\alpha \leq \theta \leq \beta\text{.}\)
We want to get an expression for this integral as an integral in the new variables \(u\) and \(v\text{,}\) such as
\begin{equation}
\iint\limits_D f(x,y) dA = \int_a^b \int_\alpha^\beta f(r\cos\theta,r\sin\theta)\, r \, d\theta\, dr\tag{5.7.8}
\end{equation}
for that case.
The key to deriving this formula for polar coordinates was looking at a small rectangle of values of \(r\) and \(\theta\text{,}\) of dimensions \(\Delta r\) and \(\Delta \theta\text{,}\) and computing the area of the corresponding region of \((x,y)\) values.
Look at a small rectangle \(S\) of \((u,v)\) values with one corner \((u_0,v_0)\) and adjacent corners \((u_0+\Delta u,v_0)\) and \((u_0,v_0+\Delta v)\text{.}\) Its image under transformation \(T\) is roughly quadrilateral, with one corner \(A\) of coordinates \((x_0,y_0)=T(u_0,v_0) = (g(u_0,v_0),h(u_0,v_0))\text{.}\)
The image \(B\) of the adjacent corner \((u_0+\Delta u,v_0)\) has coordinates \((g(u_0+\Delta u,v_0),h(u_0+\Delta u,v_0))\) and these are given approximately by linearizations of \(g\) and \(h\) respectively:
where we have ignored far smaller terms proportional to \(\Delta u \Delta v\text{.}\)
So the image \(R\) of a small rectangle \(S\) is roughly a parallelogram.
The Area Approximation and the Jacobian.
The area \(\Delta A\) of small region \(R\) can thus be approximated by the cross product formula for parallelogram area:
\begin{align*}
\Delta A \amp\approx
\left| \overrightarrow{AB} \times \overrightarrow{AC} \right|\\
\amp\approx
\left| [g_u(u_0,v_0) h_v(u_0,v_0) - g_v(u_0,v_0) h_u(u_0,v_0)] \Delta u \Delta v \veck \right|\\
\amp=
\left| g_u(u_0,v_0) h_v(u_0,v_0) - g_v(u_0,v_0) h_u(u_0,v_0) \right| \Delta u \Delta v
\end{align*}
The quantity inside the absolute value is the determinant of a \(2 \times 2\) matrix, and is known as the Jacobian of the transformation given by \(x=g(u,v)\text{,}\)\(y=h(u,v)\text{,}\) denoted
Beware the unfortunate double meaning of vertical bars: absolute value and determinant!
With this notation the image \(R=T(S)\) of small rectangle \(S\) is approximately a parallelogram of area
\begin{equation}
\Delta A \approx \left| \frac{\partial(x,y)}{\partial(u,v)} \right | \Delta u \Delta v\tag{5.7.10}
\end{equation}
(c.f. Equation (2.4.7) for the area of a parallelogram in Subsection 2.4.3), and for the "infinitesimal rectangle" of dimensions \(du\) by \(dv\) appearing in double integrals we get the change of coordinates formula using
\begin{equation}
dA = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| du\ dv.\tag{5.7.11}
\end{equation}
Note: The vertical bar notation is used here in two different ways: for the determinant in Equation (5.7.9) defining the Jacobian, and the for the absolute value of that in \(\left| \frac{\partial(x,y)}{\partial(u,v)} \right|\text{.}\) To avoid possible confusion, the expression is sometimes written as
For \(T\) a one-to-one \(C^1\) transformation and \(R\) the image of region \(S\) under \(T\) (i.e. \(R=T(S)\)),
\begin{align}
\amp\iiint\limits_R f(x,y,z) \, dV\notag\\
\amp= \iiint\limits_S f(x(u,v,w),y(u,v,w),z(u,v,w))
\left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right | du \, dv \, dw\tag{5.7.13}
\end{align}
Exercise5.7.5.Cylindrical and Spherical Coordinates.
Use the above formula to derive the formulas (5.5.3) and (5.5.5) for integrals in cylindrical and spherical coordinates seen in Section 5.5.
Note the importance of the absolute value in Equation (5.7.13) for the case of spherical coordinates \((\rho, \theta, \phi)_s\text{:}\) this is because they are left-handed whereas the caretesion coordinates \((x, y, z)_s\) are right-handed.
All the Definitions, Theorems, and Examples (and Checkpoints).
One or several exercises from each of the following ranges and pairs: 365–361, 362–367, 368–373, 374–377, 378–387, 388 & 389, 390–393, 399 & 400, 404 & 405.
