Difficulty: Intermediate | Prerequisites: Chapters 12–14, single-variable integration, polar coordinates from MATH 125
Chapter 15 extends integration to two dimensions. Where single-variable integrals compute areas under curves, double integrals compute volumes under surfaces. The key skill is setting up the correct limits of integration, either in Cartesian (rectangular) or polar coordinates.
Double integrals compute the "signed volume" between a surface z = f(x, y) and the xy-plane over a region R. You evaluate them as iterated integrals, choosing whether to integrate in x first or y first depending on which is simpler. When the region is circular or has radial symmetry, converting to polar coordinates (replacing dA with r dr dθ) often makes the integral far easier.
Double integral ∬_R f(x, y) dA
The limit of Riemann sums over a 2D region R. It gives the signed volume under the surface z = f(x, y): positive when f is above the xy-plane, negative when below.
Type 1 region (top/bottom)
A region where, for each x in [a, b], y runs from a lower curve g₁(x) to an upper curve g₂(x). The iterated integral is ∫ₐᵇ ∫_{g₁(x)}^{g₂(x)} f(x, y) dy dx.
Type 2 region (left/right)
A region where, for each y in [c, d], x runs from a left curve h₁(y) to a right curve h₂(y). The iterated integral is ∫c^d ∫{h₁(y)}^{h₂(y)} f(x, y) dx dy.
Polar coordinates
A coordinate system where a point is described by r (distance from the origin) and θ (angle from the positive x-axis). x = r cos θ, y = r sin θ, x² + y² = r².
z = f(x, y) is the height of the surface above (or below) the xy-plane
R is the region in the xy-plane over which you integrate
Area element: dA = dx dy (or dy dx)
Volume of one small box: f(xᵢⱼ, yᵢⱼ) ΔA
Units of ∬_R f(x, y) dA: (units of f) × (units of x) × (units of y)
∬_R 1 dA = area of R
For each x between a and b, y ranges from g₁(x) (bottom) to g₂(x) (top):
∫ₐᵇ ∫_{g₁(x)}^{g₂(x)} f(x, y) dy dx
Integrate with respect to y first (inner), then x (outer).
For each y between c and d, x ranges from h₁(y) (left) to h₂(y) (right):
∫c^d ∫{h₁(y)}^{h₂(y)} f(x, y) dx dy
Integrate with respect to x first (inner), then y (outer).
The choice between Type 1 and Type 2 depends on which makes the boundaries simpler to express.
When the region R involves circles or radial symmetry, convert to polar:
Describe the region in terms of r and θ
Replace x with r cos θ
Replace y with r sin θ
Replace dA with r dr dθ (the extra factor of r is essential)
The factor r in "r dr dθ" accounts for the fact that area elements in polar coordinates are not rectangles; they are wedge-shaped, and their area grows with distance from the origin.
Topic | Formula |
|---|---|
Type 1 iterated integral | ∫ₐᵇ ∫_{g₁(x)}^{g₂(x)} f(x, y) dy dx |
Type 2 iterated integral | ∫c^d ∫{h₁(y)}^{h₂(y)} f(x, y) dx dy |
Area of region R | ∬_R 1 dA |
Polar conversion | x = r cos θ, y = r sin θ, dA = r dr dθ |
Polar identity | x² + y² = r² |
Students often forget the extra r when converting to polar. The area element is r dr dθ, not just dr dθ. Missing the r makes every polar integral wrong.
Students sometimes mix up the order of integration and the limits. The inner integral's variable must match the inner limits (which can be functions of the outer variable). The outer limits must be constants.
Students occasionally set up a Type 1 integral when Type 2 would be simpler (or vice versa). Before computing, sketch the region and decide which orientation gives the simplest boundary expressions.
⚠️ Setting up the limits of integration is where most marks are won or lost. Practise sketching regions and writing the bounds.
⚠️ Converting to polar coordinates: know the substitution by heart. The most common error is forgetting the r in r dr dθ.
⚠️ "Reverse the order of integration" problems appear frequently. These require re-describing the region from Type 1 to Type 2 (or vice versa).
⚠️ ∬_R 1 dA = area of R is a useful sanity check and sometimes a standalone exam question.
Fill in the blank: in polar coordinates, dA = ______.
r dr dθ.
True or false: ∬_R f(x, y) dA is always positive.
False. It is the signed volume; it is negative when f < 0.
Fill in the blank: in a Type 1 region, the outer integral is with respect to ______.
x.
True or false: the outer limits of an iterated integral can be functions of the inner variable.
False. The outer limits must be constants.
Q: Set up (but do not evaluate) the double integral of f(x, y) = x over the region bounded by y = 0, y = x, and x = 1, as a Type 1 integral.
A: For each x in [0, 1], y runs from 0 to x. The integral is ∫₀¹ ∫₀ˣ x dy dx.
Q: Convert the integral ∬_R (x² + y²) dA to polar, where R is the disk x² + y² ≤ 4.
A: In polar, x² + y² = r², the disk is 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, and dA = r dr dθ. The integral becomes ∫₀^{2π} ∫₀² r² · r dr dθ = ∫₀^{2π} ∫₀² r³ dr dθ.
Q: Compute ∬_R 1 dA where R is the rectangle [0, 3] × [0, 2].
A: ∫₀³ ∫₀² 1 dy dx = ∫₀³ 2 dx = 6. This equals the area of the rectangle, as expected.
Double integrals are the natural continuation of Chapter 14's work with surfaces. Where partial derivatives measure local behaviour, double integrals aggregate quantities over regions. They connect back to single-variable integration (MATH 125): an iterated integral is, at each step, just a single-variable integral.
Polar coordinates were first introduced in MATH 125 for area calculations and parametric curves. Here they extend to double integrals, which is essential for any problem involving circular regions.
Related Terms / Search Tags
double integral, iterated integral, Type 1 region, Type 2 region, top bottom, left right, order of integration, polar coordinates, polar double integral, r dr dθ, area element, signed volume, Fubini's theorem, MATH 126, Calculus III, UW, multivariable integration