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Calculus III (Multivariable) Unit 5: Double integrals
Double integrals add up a function over a region in the plane, giving volume under a surface or total mass.
Covers: Iterated integrals, general regions, polar coordinates, applications
Study guide
A double integral ∬_R f(x, y) dA adds up f over a region R; if f ≥ 0 it's the volume under the surface. Over a rectangle, Fubini's theorem says you can compute it as an iterated integral in either order: integrate in one variable while holding the other constant, then integrate the result.
For general regions, describe the region with limits. A type I region lies between two curves g₁(x) ≤ y ≤ g₂(x) for a ≤ x ≤ b, so integrate dy first. A type II region lies between curves in x for c ≤ y ≤ d, so integrate dx first. The outer limits must be constants; the inner limits may depend on the outer variable. Sketch the region, and switch the order of integration if the integral is easier the other way, redrawing the region to find new limits.
Polar coordinates simplify integrals over circles, rings and sectors, or integrands involving x² + y². Substitute x = r cos θ, y = r sin θ and replace dA with r dr dθ (don't forget the extra r).
Applications: ∬_R 1 dA is the area of R; the average value of f is (1/area)∬ f dA; with a density function ρ(x, y), the double integral gives mass, and dividing moments ∬ xρ dA and ∬ yρ dA by mass gives the center of mass.
If a boundary changes formula partway across the region, split the region along a horizontal or vertical line and add the iterated integrals over each piece. If the region and the density are both symmetric about the y-axis, then x̄ = 0. An integrand that is odd in x, integrated over a region symmetric about the y-axis, gives 0.
For the disc x² + y² ≤ a², use 0 ≤ r ≤ a and 0 ≤ θ ≤ 2π. For the circle r = 2a cos θ, use 0 ≤ r ≤ 2a cos θ and −π/2 ≤ θ ≤ π/2.
Key ideas
- Fubini's theorem: on a rectangle, integrate in either order.
- On general regions, the inner limits can depend on the outer variable.
- Sketch the region before setting up or switching the order.
- Polar coordinates suit circles and sectors; remember the extra r.
- ∬_R 1 dA is the area of R.
Formulas and rules
Common mistakes
- Forgetting the r in polar dA.
- Putting variables in the outer limits.
Words to know
- dA in polar coordinates
- r dr dθ
- Area of a region R as a double integral
- ∬_R 1 dA
- Fubini's theorem
- On a rectangle you can integrate in either order.
- Limits of a double integral
- Outer limits must be constants; inner limits may depend on the outer variable.
- Polar coordinates (integrals)
- Use for circles or x² + y²; dA = r dr dθ.
- Double integral
- ∬_R f dA: volume under the surface when f ≥ 0.
- Iterated integral
- Integrate in one variable, then the other.
- Type I region
- g₁(x) ≤ y ≤ g₂(x) for a ≤ x ≤ b; integrate dy first.
- Type II region
- h₁(y) ≤ x ≤ h₂(y) for c ≤ y ≤ d; integrate dx first.
- Reversing the order of integration
- Sketch the region and rewrite the limits the other way.
- Average value (2 variables)
- (1/area)∬ f dA
- Center of mass
- Moments ∬ xρ dA and ∬ yρ dA divided by the mass.
Sample practice questions
19 more questions like these, from level 1 to 5, are in the practice set.
∫₀¹ ∫₀² 1 dy dx
- 2
- 3
- 1
- 1/2
Show the answer and steps
Answer: A) 2
- ∫₀² 1 dy = 2, then ∫₀¹ 2 dx = 2.
- (That's the area of the 1 × 2 rectangle.)
∫₀¹ ∫₀¹ xy dy dx
- 1/8
- 1/2
- 1/4
- 1
Show the answer and steps
Answer: C) 1/4
- Inner: ∫₀¹ xy dy = x/2.
- Outer: ∫₀¹ x/2 dx = 1/4.
Area of a disk of radius 3 using polar coordinates.
- 6π
- 3π
- 9π
- 18π
Show the answer and steps
Answer: C) 9π
- Area = ∫₀^2π ∫₀³ r dr dθ.
- = 2π·9/2 = 9π.