Double Integrals

The definite integral ${\int }_a^{b}f(x)\mathbf{d}x$ answered the area problem: cut $[a,b]$ into strips, sum rectangle areas, take the limit. The double integral answers the volume problem in exactly the same way, with a rectangle in the plane taking the place of an interval on the line and a surface taking the place of a curve. Everything that follows is the single-variable definite integral with one extra dimension: the same limit computes volume, mass, average value, center of mass, and moment of inertia of a flat plate of variable density.

Volume and the double integral

Let $f(x,y)$ be defined on a closed rectangle $R=[a,b]\times [c,d]=\{(x,y):a\le x\le b,c\le y\le d\}$, and suppose first that $f(x,y)\ge 0$. The graph $z=f(x,y)$ is a surface, and above $R$ it caps off a solid $S=\{(x,y,z):0\le z\le f(x,y),(x,y)\in R\}$. To measure the volume of $S$, partition $R$ into subrectangles. Divide $[a,b]$ into $m$ pieces of width $\Delta x=(b-a)/m$ and $[c,d]$ into $n$ pieces of width $\Delta y=(d-c)/n$; the grid lines cut $R$ into $mn$ subrectangles $R_{ij}$, each of area $\Delta A=\Delta x\Delta y$. Choose a sample point$(x_{ij}^{\ast },y_{ij}^{\ast })$ in each $R_{ij}$ and raise a column of height $f(x_{ij}^{\ast },y_{ij}^{\ast })$ over it. The columns approximate $S$, and their total volume is a double Riemann sum,

As $m$ and $n$ grow the columns thin out and the approximation improves. The limit, when it exists, is the double integral.

The precise meaning matches the single-variable case: for every $\epsilon >0$ there is an integer $N$ such that the sum is within $\epsilon$ of the integral whenever $m,n>N$, for any sample points. A sufficient condition for existence is that $f$ be continuous on $R$, or bounded with discontinuities confined to finitely many smooth curves.¹ The midpoint rule is the practical estimate: take each $(x_{ij}^{\ast },y_{ij}^{\ast })$ to be the center of $R_{ij}$.

Two facts carry over verbatim from single integrals:

• Linearity. $\underset{R}{\iint }(cf+dg)\mathbf{d}A=c\underset{R}{\iint }f\mathbf{d}A+d\underset{R}{\iint }g\mathbf{d}A$.
• Additivity over regions. If $R=R_1\cup R_2$ overlap only on a boundary, then $\underset{R}{\iint }f\mathbf{d}A=\underset{R_1}{\iint }f\mathbf{d}A+\underset{R_2}{\iint }f\mathbf{d}A$.
• Average value. The average of $f$ over $R$ is $f_{\text{avg}}=\frac{1}{A(R)}\underset{R}{\iint }f\mathbf{d}A$, the height of a flat box with the same base and the same volume.

Iterated integrals and Fubini's Theorem

The limit definition is unusable for computation; instead, integrate one variable at a time. Fix $x$ and integrate $f(x,y)$ over $y$ from $c$ to $d$; the result $A(x)={\int }_c^{d}f(x,y)\mathbf{d}y$ is a function of $x$ alone (a cross-sectional area). Integrate that over $x$:

This is an iterated integral: work the inner integral first, treating the outer variable as a constant, then integrate the result. The claim that it equals the double integral, and that the order does not matter, is Fubini's Theorem.

The geometric reason is Cavalieri's slicing principle: the volume under the surface is ${\int }_a^{b}A(x)\mathbf{d}x$ where $A(x)$ is the area of the slice at $x$, and also ${\int }_c^{d}A(y)\mathbf{d}y$ slicing the other way. Both recover the same solid.

When the integrand separates as a product $g(x)h(y)$ on a rectangle, the double integral factors into a product of single integrals,

because the inner integral pulls the constant $g(x)$ out and leaves ${\int }_c^{d}h(y)\mathbf{d}y$, itself a constant, to factor out of the outer one.

General regions

Rectangles are the exception; real regions have curved boundaries. To integrate over a general region $D$, enclose it in a rectangle $R$, extend $f$ to be zero outside $D$, and integrate the extension over $R$. Because the extension vanishes off $D$, the outer limits of integration are the flat sides of $R$ but the inner limits follow the boundary of $D$. Two shapes cover almost everything.

The inner limits are functions; the outer limits are the constants. For type I, integrate $y$ from the bottom curve to the top curve, then $x$ across the base:

For type II, integrate $x$ from the left curve to the right curve, then $y$:

Changing the order of integration

Some inner integrals have no elementary antiderivative in one order but are routine in the other. The standard example is ${\int }_0^1{\int }_x^1\sin (y^2)\mathbf{d}y\mathbf{d}x$: the inner integral $\int \sin (y^2)\mathbf{d}y$ has no closed form. Read off the region — $0\le x\le 1$ and $x\le y\le 1$, a triangle — and redescribe it as type II: $0\le y\le 1$ and $0\le x\le y$. Now $y$ is the outer variable and the inner integral is trivial:

The procedure is fixed: sketch the region from the given limits, then rewrite it in the other type. Never swap the numbers on the integral signs without redrawing the region — the inner limits change from curves to constants and back.

Double integrals in polar coordinates

When the region is a disk, an annulus, or a sector, Cartesian limits carry ugly square roots. Polar coordinates$x=r\cos \theta$, $y=r\sin \theta$ describe such regions cleanly. The subtlety is the area element. A polar rectangle$R=\{(r,\theta ):a\le r\le b,\alpha \le \theta \le \beta \}$ splits into subregions bounded by circles $r=\text{const}$ and rays $\theta =\text{const}$. A patch between radii $r$ and $r+\Delta r$ and angles $\theta$ and $\theta +\Delta \theta$ is not a rectangle of area $\Delta r\Delta \theta$; it is a curved wedge whose inner and outer arcs have lengths $r_{\text{in}}\Delta \theta$ and $r_{\text{out}}\Delta \theta$, so its area is the difference of two circular-sector areas,

where $r_i^{\ast }$ is the midpoint radius. In the limit the area element is $\mathbf{d}A=r\mathbf{d}r\mathbf{d}\theta$. The extra factor $r$ accounts for the polar change of variables: a patch far from the origin subtends more area than one near it.

Three substitutions convert a Cartesian polar-friendly integral: replace $x$ by $r\cos \theta$, $y$ by $r\sin \theta$, and $\mathbf{d}A$ by $r\mathbf{d}r\mathbf{d}\theta$, and set limits that describe the region in $(r,\theta )$. Forgetting the $r$ is the most common error here.

The factor $r$ that polar coordinates supply is what makes the inner integral elementary; letting $R\to \infty$ gives the value $\pi$, the route to the Gaussian integral ${\int }_{-\infty }^{\infty }{e}^{-x^2}\mathbf{d}x=\sqrt{\pi }$.

Applications: mass, center of mass, and moments

A double integral computes any quantity that is the sum of a density times an area element. Let a lamina (a thin flat plate) occupy a region $D$ with surface density $\rho (x,y)$ in mass per unit area. The mass of a small piece near $(x,y)$ is $\rho (x,y)\mathbf{d}A$, and summing over $D$ gives the total.

The center of mass is the balance point: place a support under $(\bar{x},\bar{y})$ and the plate is in equilibrium. Note the cross labeling — the moment about the $x$-axis weights by $y$, because a mass's tendency to rotate about the $x$-axis grows with its distance $y$ from that axis.

A related quantity measures resistance to rotation rather than location.

The polar moment $I_0$ pairs naturally with polar coordinates, since $x^2+y^2=r^2$.

Each quantity is the same double integral $\underset{D}{\iint }(\cdot )\mathbf{d}A$ with a different integrand:

Choosing the setup

The technique never changes; only the description of the region does. Three decisions settle every problem.

• Coordinate system. Circular boundaries or an integrand in $x^2+y^2$ favor polar; straight boundaries favor Cartesian.
• Region type. Sketch $D$. If vertical strips have simple top and bottom curves, use type I; if horizontal strips are simpler, use type II. Sometimes one order is elementary and the other is impossible.
• Order of integration. Even in Cartesian coordinates, reversing the order can rescue a non-integrable inner integral — redescribe the region, never just the numbers.