Difficulty: Intermediate to Advanced | Prerequisites: Single-variable integration, polar coordinates basics, partial derivatives
Multiple integration extends single-variable integration to two and three dimensions, letting you compute areas, volumes, masses, and other accumulated quantities over regions in the plane or in space. Choosing the right coordinate system (Cartesian, polar, cylindrical, or spherical) can turn a painful integral into a clean one. Fubini's theorem guarantees that for continuous functions on rectangles you can integrate in either order. This material is the computational engine behind many physical applications in MATH 215 and beyond.
Double integrals compute area, mass, and other quantities over 2D regions; triple integrals do the same in 3D. Switch to polar coordinates when boundaries are circles, and to spherical coordinates when boundaries are spheres or cones. Fubini's theorem lets you swap the order of integration on rectangles. The surface area of z = f(x, y) uses the integral of √(1 + fₓ² + f_y²). Integrating an odd function over a symmetric region gives zero.
Double integral (∬_D f dA)
The integral of a function f(x, y) over a 2D region D. It computes the "accumulated value" of f over D. When f = 1, the double integral gives the area of D.
Triple integral (∭_E f dV)
The integral of a function f(x, y, z) over a 3D region E. When f = 1, it gives the volume of E. When f is a density, it gives the mass.
Polar coordinates (r, θ)
A coordinate system for the plane where x = r cos θ, y = r sin θ. The area element is dA = r dr dθ. Think of it as describing position by "how far from the origin" and "at what angle."
Spherical coordinates (ρ, θ, φ)
A coordinate system for 3D space where x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. The volume element is dV = ρ² sin φ dρ dθ dφ. Ideal when boundaries are spheres (constant ρ) or cones (constant φ).
Jacobian
The scaling factor that appears when changing coordinates. For polar coordinates, the Jacobian is r. For spherical coordinates, it is ρ² sin φ. In simple terms, it corrects for how the coordinate grid stretches or compresses in different regions.
Fubini's theorem
For a continuous function on a rectangle, the iterated integrals in either order are equal. In simple terms, you can integrate x first then y, or y first then x, and get the same answer.
Surface area integral
The area of the surface z = f(x, y) over a region D is ∬_D √(1 + fₓ² + f_y²) dA. The expression under the square root accounts for how tilted the surface is relative to the xy-plane.
Convert when boundaries are circles or parts of circles.
x² + y² = 1 becomes r = 1. x² + y² = 2x becomes r = 2 cos θ.
Find intersection: 1 = 2 cos θ gives θ = ±π/3.
The density ρ(x, y) = 1/√(x² + y²) becomes 1/r in polar.
The mass integral: M = ∫ from −π/3 to π/3 ∫ from 1 to 2 cos θ (1/r)(r dr dθ) = ∫∫ dr dθ.
The Jacobian r cancels with the 1/r density, giving a clean integral.
Result: M = 2√3 − 2π/3.
Use spherical when the region is bounded by spheres (constant ρ) and/or cones (constant φ).
First octant (x, y, z ≥ 0): θ from 0 to π/2, φ from 0 to π/2.
Between spheres ρ = 1 and ρ = 2: ρ from 1 to 2.
The integrand z = ρ cos φ. The volume element is ρ² sin φ dρ dθ dφ.
The integral separates into three single integrals (one in each variable) when the limits are all constants.
Example result: ∭_E z dV = (π/2)(15/4)(1/2) = 15π/16.
Circular or annular regions in 2D: polar coordinates.
Spheres and cones in 3D: spherical coordinates.
Cylinders in 3D: cylindrical coordinates.
The condition z ≥ √(x² + y²) describes a cone; combined with spheres, this calls for spherical.
∫₀¹ ∫ₓ¹ f(x, y) dy dx integrates over the triangle where 0 ≤ x ≤ 1 and x ≤ y ≤ 1.
Vertices: (0, 0), (1, 1), (0, 1).
Reversed: ∫₀¹ ∫₀^y f(x, y) dx dy. For fixed y, x runs from 0 to y.
Sketch the region first. Identify which variable is the outer integral and determine the inner limits from the geometry.
For a continuous function on a rectangle R = [a, b] × [c, d], the order of integration does not matter.
This applies specifically to rectangles. For non-rectangular regions, you can still switch the order, but you must re-derive the limits.
The surface area of z = f(x, y) over D is ∬_D √(1 + fₓ² + f_y²) dA.
This is distinct from the double integral of f itself (which gives the volume under the surface).
Note the "1 +" under the square root. Without it, you would be computing something else entirely.
If the integrand is an odd function (changes sign when you negate the variable) and the region is symmetric about the origin, the integral is zero.
Example: ρ(x, y, z) = x + y + z over the unit ball x² + y² + z² ≤ 1. Each of x, y, z is odd, the ball is symmetric, so the mass is zero.
The area of a region D in the xy-plane equals ∬_D 1 dA. Integrating the constant function 1 over D simply accumulates the area.
dA = r\, dr\, d\theta \quad \text{(polar area element)}dV = \rho^2 \sin\phi\, d\rho\, d\theta\, d\phi \quad \text{(spherical volume element)}\text{Surface area} = \iint_D \sqrt{1 + f_x^2 + f_y^2}\, dA\text{Mass} = \iint_D \rho(x,y)\, dA \quad \text{or} \quad \iiint_E \rho(x,y,z)\, dV\text{Area of } D = \iint_D 1\, dADouble and triple integrals compute physical quantities like mass, centre of mass, and moments of inertia for objects with varying density. Engineers use them to calculate the weight distribution of structural components. Polar and spherical integrals arise in antenna design, gravitational field calculations, and heat conduction in circular or spherical geometries.
Students often forget the Jacobian when converting to polar or spherical. In polar, dA = r dr dθ (not just dr dθ). In spherical, dV = ρ² sin φ dρ dθ dφ.
When reversing the order of integration, students sometimes swap limits without re-deriving them from the region's geometry. Always sketch the region first.
Students confuse the surface area formula (with the "1 +" under the square root) with √(fₓ² + f_y²), which is something else. The "1" accounts for the flat base area.
Students sometimes assume Fubini's theorem applies to non-rectangular regions without modification. On non-rectangles, you can still switch order, but the limits change.
Students sometimes forget that the integral of an odd function over a symmetric region is zero, and compute a long integral unnecessarily.
⚠️ Always include the Jacobian. This is the single most common error in coordinate-change problems.
⚠️ Know when to use each coordinate system. Spheres + cones = spherical. Circles in the plane = polar.
⚠️ The surface area formula ∬ √(1 + fₓ² + f_y²) dA is a standard multiple-choice question.
⚠️ Reversing integration order: sketch the region, identify the triangle or other shape, and re-derive limits from the geometry.
⚠️ Symmetry shortcuts: if the integrand is odd and the region is symmetric, the answer is zero. State this reasoning clearly for full marks.
Fill in the blank: the area element in polar coordinates is dA = ______ dr dθ. (r)
Fill in the blank: the volume element in spherical coordinates is dV = ______ dρ dθ dφ. (ρ² sin φ)
True or false: Fubini's theorem allows swapping integration order for any continuous function on a rectangle. (True)
True or false: the surface area of z = f(x, y) is ∬_D f(x, y) dA. (False; it is ∬_D √(1 + fₓ² + f_y²) dA)
True or false: ∬_D 1 dA gives the area of D. (True)
Q: Which coordinate system is best for a region bounded by 1 ≤ x² + y² + z² ≤ 9 and z ≥ √(x² + y²)?
A: Spherical coordinates. The region is bounded by spheres (constant ρ) and a cone (constant φ = π/4).
Q: Evaluate ∭_E z dV where E is the first octant between the spheres ρ = 1 and ρ = 2.
A: In spherical: ∫₀^{π/2} ∫₀^{π/2} ∫₁² (ρ cos φ)(ρ² sin φ) dρ dφ dθ. Separates into (π/2)(15/4)(1/2) = 15π/16.
Q: Reverse the order of integration for ∫₀¹ ∫ₓ¹ f(x, y) dy dx.
A: The region is the triangle with 0 ≤ x ≤ y ≤ 1. Reversed: ∫₀¹ ∫₀^y f(x, y) dx dy.
Q: Which formula gives the surface area of z = f(x, y) over D?
A: ∬_D √(1 + fₓ² + f_y²) dA.
Q: True or false: the mass of a solid with density ρ(x, y, z) = x + y + z over the unit ball is zero.
A: True. Each of x, y, z is an odd function, the unit ball is symmetric about the origin, and the integral of an odd function over a symmetric region is zero.
Q: True or false: Fubini's theorem says the order of integration does not matter for continuous functions on a rectangle.
A: True.
Q: In spherical coordinates, is the volume element dV = ρ² sin φ dρ dθ dφ?
A: True. The ρ² sin φ is the Jacobian for the spherical coordinate transformation.
Double integrals in polar coordinates connect back to the geometry of circles and the polar form of curves from precalculus. Triple integrals in spherical coordinates connect to the gradient and level surfaces topic (spheres are level surfaces of x² + y² + z²). The surface area formula uses partial derivatives, tying integration to differentiation. The symmetry shortcut (odd function, symmetric region) is a special case of a broader principle that also applies in probability and physics.
double integral, triple integral, iterated integral, polar coordinates, cylindrical coordinates, spherical coordinates, Jacobian, change of variables, Fubini's theorem, order of integration, reversing integration order, surface area, lamina, mass, density function, volume element, area element, odd function, symmetry, first octant, unit ball, annular region, MATH 215, Calculus III, multivariable calculus, University of Michigan