Double and triple integrals
Integrating in more than one variable means summing over a region of the plane or of space: areas, volumes, masses and distributed charges.
01Summing over a region
The single integral summed over an interval. The double integral sums over a region of the plane: the region is cut into small rectangles, the function is evaluated in each one, multiplied by its area, and all of them are added up.
As before, nobody computes the limit: it is converted into two chained single integrals, the iterated integrals. You integrate first in one variable, treating the other as a constant, and then in the one that is left.
02The limits, the only hard part
For a region where, for each x between a and b, y runs between two curves:
The same region can be swept the other way around: for each y, x between two curves. The result is the same —guaranteed by Fubini’s theorem— but one of the two forms is usually much easier, and sometimes the other is simply impossible to solve by hand. Drawing the region is mandatory: it is the only way to read the limits without making mistakes.
The grid shows the small rectangles of the sum; each one contributes f·ΔA. Increase the resolution and watch how it converges.
03Polar coordinates: when the region is round
If the region is a circle or a sector, the limits in Cartesian coordinates come out with awkward roots. In polar coordinates they become constants:
The area of a circle of radius R: ∬dA = ∫₀2π∫₀R ρ dρ dθ = ∫₀2π (R²/2) dθ = πR². Forgetting the ρ would have given 2πR, which does not even have units of area.
In three dimensions the same approach gives cylindrical coordinates (dV = ρ dρ dθ dz), convenient for tubes and coils, and spherical coordinates (dV = r² sin φ dr dφ dθ), convenient for spheres and for radial fields.
04What they are used for
| Quantity | Integral | In electronics |
|---|---|---|
| Area | \( \iint \, dA \) | Surface of a heat sink or of a trace |
| Volume | \( \iiint \, dV \text{ or } \iint f \, dA \) | Volume of a package, of a magnetic core |
| Mass | \( \iint \delta (x,y) dA \) | With variable density; the same for charge with surface density \( \sigma \) |
| Center of mass | \( \dfrac{1}{m}\iint x\cdot \delta \, dA \) | Mechanical balance of an assembly, center of gravity of a plate |
| Moment of inertia | \( \iint r^{2}\cdot \delta \, dA \) | Rotors, disks, shafts: it determines the angular acceleration |
| Average value | \( \dfrac{1}{A}\iint f \, dA \) | Average temperature of a plate, average illuminance of a room |
And one that appears as soon as fields are studied: the total charge of a surface distribution is the double integral of the density, and from there the field follows by applying Gauss’s theorem, which is seen in the theorems of vector calculus.
05In the lab
Compute ∬ xy dA over the triangle with vertices (0,0), (2,0) and (2,2), first integrating in y and then in x, and then the other way around. Verify that the result is the same and compare the work involved in each path.
Compute ∬ (x² + y²) dA over the circle of radius 2 in both systems. The goal is to get a feel for why in polar coordinates it takes three lines.
Cut an irregularly shaped plate out of cardboard, compute its center of mass by dividing it into simple shapes, and check it by hanging it from two different points: the plumb lines cross at the center of gravity.
06Common mistakes
- Putting variable limits on the outer integral. The outer ones are numbers; if a loose variable is left, the setup is wrong.
- Forgetting the Jacobian ρ when switching to polar coordinates, or the r² sin φ in spherical coordinates.
- Not drawing the region and deducing the limits by heart.
- Changing the order without redoing the limits: it is not enough to swap dx and dy.
- Integrating a density and calling the result “area”: it pays to check the units of the result.
07Self-assessment
Compute ∫₀¹∫₀² xy dy dx.
The inner one: ∫₀² xy dy = x·[y²/2]₀² = 2x. Then ∫₀¹ 2x dx = 1.
What does ∬ dA represent over a region?
The area of the region.
Why does the factor ρ appear in polar coordinates?
Because the area element in polar coordinates is not dρ dθ but ρ dρ dθ: the arc grows with the radius. It is the Jacobian of the change of variables.
Area of the circle of radius 3 using a double integral in polar coordinates.
∫₀2π∫₀³ ρ dρ dθ = ∫₀2π (9/2) dθ = 9π ≈ 28.3.
How is the charge of a plate with density σ(x,y) computed?
By integrating the density over the surface: Q = ∬ σ(x,y) dA.
08Further reading
- James Stewart. Multivariable Calculus. 7th ed. (Spanish edition, “Cálculo de varias variables. Trascendentes tempranas”), Cengage Learning, 2012. Chapter 15 covers double and triple integrals, changing the order, polar coordinates and physical applications.
- Jerrold E. Marsden and Anthony J. Tromba. Vector Calculus. 5th ed. (Spanish edition, “Cálculo vectorial”), W. H. Freeman, 2003. Carefully develops the change of variables and the Jacobian in general.
- Claudio Pita Ruiz. Cálculo vectorial. Prentice Hall (in Spanish). Many worked examples of setting up the limits, which is where almost everyone gets stuck.