Calculus
Double Integrals
How does Pixar compute the light hitting a single pixel? A double integral over the visible solid angle, evaluated billions of times per frame.
- Pixar RenderMan: each pixel is a double integral of incoming radiance over a 2D solid angle
- Boeing aerodynamics: lift is computed as a double integral of pressure over wing surface
- MRI image reconstruction: signal intensity integrated over disks and annuli inside the patient
- Catastrophe insurance pricing: expected loss is a double integral of damage density over a risk region
Предварительные знания
- Single-variable definite integration
- Partial derivatives and multivariable regions
- Polar coordinates
Riemann Sums in 2D and Iterated Integrals
Pixar's RenderMan averages light incoming over a 2D solid angle for every pixel. That is a double integral evaluated billions of times per frame. The math behind Toy Story 4 is the same construction we cover here: partition a region, sum heights times areas, take the limit.
Setting f = 1 gives the area of D. Setting f = rho (surface density) gives total mass. Setting f = h(x,y) (height above the plane) gives the volume under the graph z = h(x,y) above D.
Sketch D before integrating. The picture tells you which variable to integrate first and what the inner-integral limits are. Algebra without a sketch is the #1 source of bound-of-integration errors.
Compute the double integral of f(x,y) = x + y over R = [0,1] x [0,2].
integral_0^1 integral_0^2 (x+y) dy dx = integral_0^1 (2x + 2) dx = 1 + 2 = 3.
Type I, Type II, and Changing the Order of Integration
AutoCAD's hatching algorithm needs to compute the area of arbitrary closed regions. It decomposes them into Type I and Type II pieces and runs a double integral on each. The choice of order can turn a 12-second computation into a 0.4-second one. Same answer, different path.
Swapping order only works for absolutely integrable f (Fubini). For oscillating integrals with cancellation, the order can matter. In intro courses, all our f satisfy the hypothesis, but be aware of the assumption.
Heuristic: if the inner integral is hard, draw D and try the other order. Roughly one in three textbook integrals collapses to an elementary one after swapping.
Region D is bounded by y = x and y = x^2 with 0 <= x <= 1. Which iterated integral correctly computes the area of D?
On [0,1], x >= x^2, so y goes from x^2 (parabola, below) up to x (line, above). The area is integral_0^1 (x - x^2) dx = 1/2 - 1/3 = 1/6.
Double Integrals in Polar Coordinates
MRI scanners reconstruct an image by integrating signal intensity over disks and annuli inside the patient. Doing this in Cartesian coordinates triples the runtime versus polar. Polar coordinates are the natural language for any problem with circular symmetry, and the Jacobian r is the price of admission.
The famous Gaussian integral integral over R^2 of e^{-(x^2+y^2)} = pi is the polar-coordinate parlor trick that proves integral over R of e^{-x^2} dx = sqrt(pi). Every probability textbook reproduces it.
Switch to polar whenever the region or the integrand involves x^2 + y^2, circles, sectors, or annuli. The Jacobian r often cancels a pesky 1/r in the integrand, simplifying things further.
What replaces dA when changing from Cartesian to polar coordinates?
The Jacobian determinant of the polar map (r, theta) -> (r cos theta, r sin theta) is r. So dA = dx dy = r dr d theta.
Volume, Mass, and Average Value
Boeing's wing aerodynamics team integrates pressure over the wing surface to compute lift. NOAA integrates precipitation over watersheds to forecast floods. Insurance actuaries integrate loss density over risk regions to price catastrophe bonds. Three industries, same construction: a double integral of a density over a region.
Moment of inertia about an axis is also a double integral: I_x = double integral of y^2 * rho dA. Mechanical engineers use it to compute the bending stiffness of arbitrary cross-sections.
Whenever a real-world quantity is described as 'total something, distributed over a region', you are almost always looking at a double integral. Identify the density, identify the region, then choose coordinates.
A lamina occupies the unit disk x^2 + y^2 <= 1 with density rho(x,y) = x^2 + y^2. What is its total mass?
Switch to polar: rho * dA = r^2 * r dr d theta = r^3 dr d theta. Mass = integral_0^{2pi} d theta * integral_0^1 r^3 dr = 2 pi * 1/4 = pi/2.
Where double integrals lead next
Double integrals are the entry point to all of multivariable integration: triple integrals, surface integrals, and the great theorems of vector calculus all build on them.
- Triple integrals — Next step: integration over solid regions
- Change of variables — Generalises the polar Jacobian trick to arbitrary coordinates
- Green's theorem — Converts a double integral over a region into a line integral over its boundary
- Probability density functions — Joint distributions are integrated using double integrals
Итоги
- Double integral = limit of Riemann sums: double integral of f dA over R = lim sum f(x*, y*) Delta x Delta y
- Fubini's theorem: on a rectangle (or absolutely integrable f) the iterated integral equals the double integral and order does not matter
- Type I region: y between curves g_1(x) and g_2(x); Type II: x between h_1(y) and h_2(y). Swap order if the inner integral is intractable
- Polar coordinates: dA = r dr d theta. Use for circular symmetry; turns the Gaussian over R^2 into a one-line proof
- Applications: volume = double integral of h dA; mass = double integral of rho dA; average value = (1/area) * double integral of f dA
Связанные уроки
- calc-20-extrema-multi — Partial derivatives and multivariable regions set the stage for integration in 2D
- calc-22-triple-integrals — Triple integrals extend the same construction one dimension higher
- calc-23-coordinate-changes — Change of variables lets us tackle integrals that look ugly in Cartesian coordinates