Multiple Integrals

We defined multivariable functions and examined their derivatives in §Functions of Several Variables and Partial Derivatives. Now we turn to integration.

Multiple Integrals

Just as the single-variable definite integral sums function values over an interval, in several variables we sum function values over a region. Let us begin with closed boxes in \(\mathbb{R}^n\).

Definition 1 For a closed box \(R = [a_1,b_1]\times\cdots\times[a_n,b_n]\subseteq \mathbb{R}^n\) and a bounded function \(f\) on it, partition \(R\) into small boxes (volume \(\Delta V\)) and choose a sample point \(\mathbf{x}^\ast\) in each piece to form the Riemann sum

\[\sum f(\mathbf{x}^\ast) \Delta V\]

If this converges to a single value as the partition is refined, that value is called the multiple integral of \(f\) and is written as

\[\int_R f \dd{V}\]

When we need to exhibit the variables, we also write

\[\int\cdots\int_R f(x_1,\ldots,x_n) \dd{x_1}\cdots \dd{x_n}\]

Depending on the dimension, this integral computes different quantities. When \(n=1\), it is the familiar definite integral, that is, computing the area under a curve. When \(n=2\) and \(f\geq 0\), the quantity computed is the volume of the solid whose base is the region \(R\) and whose upper surface is the surface \(z=f(x,y)\); the case \(n=2\) is specially called the double integral. The notation \(\dd{V}\) above is a notation with \(n\)-dimensional volume in mind, but since \(2\)-dimensional volume, that is, area, is an object already familiar to us, it is also customary to write it as

\[\iint_R f \dd{A}\]

From this perspective, the meaning of the integral is clearer: each term \(f(x_i^\ast, y_j^\ast) \Delta A_{ij}\) of the Riemann sum is the volume of a thin column with base area \(\Delta A_{ij}\) and height \(f\), and the double integral is precisely the limit obtained by assembling these columns to approximate the solid and then letting the partition become infinitely fine. Raising the dimension by one more, when \(n=3\) it becomes the triple integral

\[\iiint_E f \dd{V}\]

Continuous functions are multiple-integrable, and the integral over a general region \(D\) that is not a box is defined, inside a box containing \(D\), as the integral of the function extended so that outside \(D\) it takes the value \(0\). As long as the boundary of \(D\) consists of smooth surfaces and the function is continuous on it, the integral is well defined, and one can verify that the basic properties of integration hold exactly as in the single-variable case. In this post, rather than enumerating them one by one, we only briefly summarize what is newly introduced.

Fubini’s Theorem

The definition of a multiple integral is an \(n\)-dimensional limit, so it is difficult to compute directly. Fortunately it reduces to an iterated integral, integrating one variable at a time.

Theorem 2 (Fubini) If \(f\) is continuous on the box \(R = [a_1,b_1]\times\cdots\times[a_n,b_n]\), then

\[\int_R f \dd{V} = \int_{a_1}^{b_1}\left(\int_{a_2}^{b_2}\cdots\left(\int_{a_n}^{b_n} f(x_1,\ldots,x_n) \dd{x_n}\right)\cdots \dd{x_2}\right)\dd{x_1}\]

and the order of integration may be changed arbitrarily.

A rigorous proof relies essentially on the analytic properties of continuous functions, so we omit it here. In any case, since the inner integral on the right-hand side of the above theorem is an ordinary definite integral where one variable is held constant and integrated with respect to another, a multiple integral becomes an iteration of single-variable integrals solved by §The Fundamental Theorem of Calculus and techniques of integration. If the integrand separates variables as \(f(x_1,\ldots,x_n) = g_1(x_1)\cdots g_n(x_n)\) and the region is a box, we can go one step further: the multiple integral decomposes into the product of \(n\) single-variable integrals

\[(\int_{a_1}^{b_1} g_1)\cdots(\int_{a_n}^{b_n} g_n)\]

For a general region, the limits of integration may depend on the other variables. Taking \(n=2\) as an example, if

\[D = \{(x,y) \mid a\leq x\leq b,\ g_1(x)\leq y\leq g_2(x)\}\]

then

\[\iint_D f \dd{A} = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \dd{y} \dd{x}\]

and the limits of the inner integral become functions of the outer variable. Hence, to integrate the same region in the opposite order, we must describe the region in terms of \(y\) instead. For instance, the triangle with vertices \((0,0), (1,0), (1,1)\) can be written as

\[0 \leq x \leq 1,\quad 0 \leq y \leq x\]

but to switch the order, it can also be written as

\[0 \leq y \leq 1,\quad y \leq x \leq 1\]

How one chooses the order does not change the answer, but it is common that the inner integral evaluates to an elementary function in one order while failing to do so in the other.

Example 3 (Changing the order of integration) The double integral

\[\int_0^1 \int_x^1 e^{y^2} \dd{y} \dd{x}\]

cannot be evaluated from the inside because the antiderivative of \(e^{y^2}\) is not an elementary function. However, the region of integration is \(\{0 \leq x \leq y \leq 1\}\), so switching the order brings the outer variable into the limits of the region and introduces a new factor \(y\), so it evaluates as

\[\int_0^1 \int_0^y e^{y^2} \dd{x} \dd{y} = \int_0^1 y e^{y^2} \dd{y} = \frac{1}{2}(e - 1)\]

Change of Variables

The multivariable analogue of single-variable substitution is change of variables, where the Jacobian determinant takes the place of the length ratio \(g'\).

Theorem 4 (Change of variables) If a \(C^1\) map \(\mathbf{x} = \mathbf{T}(\mathbf{u})\) sends a region \(D'\subseteq \mathbb{R}^n\) to \(D\), is one-to-one on the interior of \(D'\), and its Jacobian determinant is not \(0\), then on \(D\), for a continuous function \(f\),

\[\int_D f(\mathbf{x}) \dd{V} = \int_{D'} f(\mathbf{T}(\mathbf{u})) \lvert \det J_{\mathbf{T}}(\mathbf{u})\rvert \dd{V'}\]

Here \(J_{\mathbf{T}}\) is the Jacobi matrix collecting the partial derivatives of \(\mathbf{T}\),

\[J_{\mathbf{T}}=\begin{pmatrix} \partial x_1/\partial u_1 & \cdots & \partial x_1/\partial u_n \\ \vdots & \ddots & \vdots \\ \partial x_n/\partial u_1 & \cdots & \partial x_n/\partial u_n\end{pmatrix}\]

Again, since we have agreed to treat linear algebra as a black box, we defer the proof to analysis. In any case, what matters is the intuition: the Jacobian determinant \(\lvert\det J_{\mathbf{T}}\rvert\) is the local scaling factor by which \(\mathbf{T}\) expands volume. That is, a small box in \(\mathbf{u}\)-space is carried by \(\mathbf{T}\) to a small parallelepiped in \(\mathbf{x}\)-space whose volume is the original box’s volume multiplied by \(\lvert\det J_{\mathbf{T}}\rvert\), so the volume element transforms as \(\dd{V} = \lvert\det J_{\mathbf{T}}(\mathbf{u})\rvert \dd{V'}\). For \(n=2\) the parallelepiped becomes a parallelogram, and the area formed by the two edges \(\mathbf{T}_u \Delta u\) and \(\mathbf{T}_v \Delta v\) (where \(\mathbf{T}_u, \mathbf{T}_v\) are the partial derivative vectors of \(\mathbf{T}\)) is exactly \(\lvert\det J_{\mathbf{T}}\rvert \Delta u \Delta v\). Substituting this into the Riemann sum and taking the limit yields Theorem 4 (Change of variables); the reason for taking the absolute value is that volume is always positive.

The most common use of change of variables is to change coordinate systems, and since we have only introduced determinants of \(2\times 2\) and \(3\times 3\) matrices, these are all of our examples.

Example 5 (Change of variables in double integrals, polar coordinates) The most common change of variables for double integrals is polar coordinates \(x = r\cos\theta\), \(y = r\sin\theta\). Then its Jacobian determinant is

\[\det J = \det\begin{pmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix} = r\]

so \(\dd{A} = r \dd{r} \dd{\theta}\). This extra factor \(r\) often resolves integrals that were stuck in rectangular coordinates. For example, the double integral over the unit disk \(D\) is

\[\iint_D e^{-(x^2+y^2)} \dd{A} = \int_0^{2\pi} \int_0^1 e^{-r^2} r \dd{r}\dd{\theta} = \pi(1 - e^{-1})\]

and this computation can be completed because an antiderivative of \(e^{-r^2} r\) is explicitly given by \(-e^{-r^2}/2\).

Using the example above, we can find the value of the Gaussian integral, which could not be solved with a single variable.

Example 6 (Gaussian integral) Let us find the value of the integral \(I = \int_{-\infty}^{\infty} e^{-x^2} \dd{x}\). For \(x \geq 1\), we have \(e^{-x^2} \leq e^{-x}\), and since \(\int_1^\infty e^{-x} \dd{x}\) converges, by §Improper Integrals, ⁋Proposition 3 (Comparison test) the integral \(\int_0^\infty e^{-x^2} \dd{x}\) converges; since \(e^{-x^2}\) takes the same value at \(x\) and \(-x\), \(\int_{-\infty}^0 e^{-x^2} \dd{x}\) also converges. Hence this improper integral converges. Now, on the square \(Q_t = [-t,t]\times[-t,t]\) the integrand separates variables, so by Theorem 2 (Fubini)

\[\left(\int_{-t}^{t} e^{-x^2} \dd{x}\right)^2 = \left(\int_{-t}^{t} e^{-x^2} \dd{x}\right)\left(\int_{-t}^{t} e^{-y^2} \dd{y}\right) = \iint_{Q_t} e^{-(x^2+y^2)} \dd{A}\]

Since both improper integrals converge, as \(t \rightarrow \infty\) the integral \(\int_{-t}^{t} e^{-x^2} \dd{x}\) approaches \(I\), and therefore the left-hand side of the above equation goes to \(I^2\). Meanwhile, over the radius-\(s\) disk \(B_s\), the calculation of Example 5 (Change of variables in double integrals, polar coordinates) repeats verbatim to give

\[\iint_{B_s} e^{-(x^2+y^2)} \dd{A} = \int_0^{2\pi} \int_0^s e^{-r^2} r \dd{r} \dd{\theta} = \pi(1 - e^{-s^2})\]

Now \(B_t \subseteq Q_t \subseteq B_{\sqrt{2}t}\) and the integrand is positive, so the integral is monotonically increasing with respect to the region, yielding

\[\pi(1 - e^{-t^2}) \leq \left(\int_{-t}^{t} e^{-x^2} \dd{x}\right)^2 \leq \pi(1 - e^{-2t^2})\]

and since as \(t \rightarrow \infty\) both ends approach \(\pi\), we obtain \(I^2 = \pi\), that is, \(I = \sqrt{\pi}\).

In \(2\) dimensions the only classical example is polar coordinates, but in \(3\) dimensions there are two substitution methods, spherical coordinates and cylindrical coordinates, and calculating them in an appropriate manner is helpful when evaluating many integrals.

Example 7 (Change of variables in triple integrals, spherical and cylindrical coordinates) First, cylindrical coordinates \((r, \theta, z)\) are obtained by adding the height \(z\) to polar coordinates in the plane:

\[x = r\cos\theta,\quad y = r\sin\theta, \quad z = z\]

and the Jacobian determinant is

\[\det J = \det\begin{pmatrix}\cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix} = r\]

so \(\dd{V} = r \dd{r} \dd{\theta} \dd{z}\).

For spherical coordinates \((\rho, \phi, \theta)\), a point is represented by the distance \(\rho\) from the origin, the angle with the positive \(z\)-axis \(\phi\), and the azimuthal angle \(\theta\):

\[x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi\]

Computing its Jacobian determinant gives

\[\det J = \det\begin{pmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \\ \cos\phi & -\rho\sin\phi & 0 \end{pmatrix} = \rho^2\sin\phi\]

so \(\dd{V} = \rho^2\sin\phi \dd{\rho} \dd{\phi} \dd{\theta}\). For example, the volume of a ball of radius \(R\) can be derived in spherical coordinates because the three variables separate:

\[\iiint_{B_R} \dd{V} = \left(\int_0^{2\pi} \dd{\theta}\right)\left(\int_0^\pi \sin\phi \dd{\phi}\right)\left(\int_0^R \rho^2 \dd{\rho}\right) = 2\pi\cdot 2\cdot \frac{R^3}{3} = \frac{4\pi R^3}{3}\]

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.