Difficulty: Intermediate to Advanced | Prerequisites: single-variable integration techniques (substitution, integration by parts), polar coordinates from precalculus, basic 3D geometry.
Multiple integrals extend the definite integral to two and three dimensions. Instead of finding the area under a curve, you find the volume under a surface or the mass of a solid. The challenge is choosing the right coordinate system (Cartesian, polar, cylindrical, spherical) and the right order of integration to make the computation manageable. Change-of-variables with a Jacobian is the general tool that ties all coordinate transforms together. This is the computational engine of the second half of Calculus III.
Double and triple integrals are evaluated as iterated (nested) single integrals. Polar coordinates simplify circular regions; spherical coordinates simplify spheres and balls. The volume element in spherical is ρ² sin φ dρ dφ dθ. When density is inversely proportional to distance, the 1/r factor cancels the r in the polar Jacobian, leaving a simpler integrand. The Jacobian measures how areas or volumes stretch under a change of variables.
Iterated integral
A multiple integral written as nested single integrals, evaluated from the inside out. In simple terms, you integrate one variable at a time.
Reversing the order of integration
Rewriting a double integral so that the inner and outer variables swap. Requires re-describing the region geometrically and rewriting the bounds. You cannot simply swap limits.
Polar coordinates
x = r cos θ, y = r sin θ. The area element becomes dA = r dr dθ. Think of it as: instead of a grid of rectangles, you tile the plane with wedge-shaped pieces.
Spherical coordinates
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 dφ dθ. Here ρ is distance from the origin, φ is the angle from the positive z-axis, θ is the angle in the xy-plane.
Jacobian
The absolute value of the determinant of the matrix of partial derivatives when changing variables. It scales the area or volume element to account for the distortion introduced by the change of variables.
Density function
A function δ(x,y) or δ(x,y,z) that gives mass per unit area or volume at each point. Total mass = ∫∫ δ dA or ∫∫∫ δ dV.
Change of variables
Substituting new variables (u, v) for (x, y) to simplify the integrand or the region. The integral picks up a factor of |∂(x,y)/∂(u,v)|, the absolute Jacobian.
Evaluating ∫₀¹ ∫₀¹ ye^x dx dy
Inner integral (with respect to x): ∫₀¹ ye^x dx = y[e^x]₀¹ = y(e - 1).
Outer integral: ∫₀¹ y(e - 1) dy = (e - 1)[y²/2]₀¹ = (e - 1)/2.
Reversing order of integration
To reverse the order, you must first describe the region geometrically and then rewrite the bounds for the other variable as the inner integral. The correct multiple-choice answer for what must be done first is (B): describe the region geometrically and rewrite its bounds.
The region outside x² + y² = 1 and inside x² + y² = 2x
The circle x² + y² = 2x rewrites in polar as r² = 2r cos θ, so r = 2 cos θ. The unit circle is r = 1. The region of interest is where 1 ≤ r ≤ 2 cos θ, and 2 cos θ ≥ 1 requires cos θ ≥ 1/2, giving -π/3 ≤ θ ≤ π/3.
Polar bounds: θ ∈ [-π/3, π/3], r ∈ [1, 2 cos θ].
Density inversely proportional to distance, maximum density 1
Density δ = k/r, and the maximum distance in the region is r = 1 (the inner boundary), so max density k/1 = k = 1. Thus δ = 1/r.
Why the density cancels the polar Jacobian:
The mass integral is ∫∫ δ dA = ∫∫ (1/r) · r dr dθ = ∫∫ 1 dr dθ. The 1/r from the density cancels the r from the polar area element, leaving a constant integrand. This is a standard simplification the exam tests.
Polar-coordinate factor in the mass integrand
The density is 1/r. The polar area element is r dr dθ. Together: (1/r)(r) = 1. The correct multiple-choice answer for the polar factor in the integrand is (B) 1.
Spherical volume element
The correct answer is (A): dV = ρ² sin φ dρ dφ dθ.
Converting (0, 0, 1) to spherical
ρ = √(0² + 0² + 1²) = 1.
φ = arccos(z/ρ) = arccos(1) = 0.
θ is undefined (the point is on the z-axis), but conventionally θ = 0.
Spherical: (ρ, φ, θ) = (1, 0, 0).
The unit ball centred at (0, 0, 1)
Cartesian equation: x² + y² + (z - 1)² ≤ 1.
Expanding: x² + y² + z² - 2z + 1 ≤ 1, so ρ² - 2ρ cos φ ≤ 0, giving ρ ≤ 2 cos φ.
Spherical radial bound: 0 ≤ ρ ≤ 2 cos φ, with 0 ≤ φ ≤ π/2 (since ρ must be non-negative, cos φ ≥ 0), 0 ≤ θ ≤ 2π.
The correct multiple-choice answer for the spherical description of the unit ball at (0, 0, 1) is (D): 0 ≤ ρ ≤ 2 cos φ, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
First-octant volume between spheres of radii 1 and 2
The first octant is one-eighth of the full sphere (x ≥ 0, y ≥ 0, z ≥ 0): θ ∈ [0, π/2], φ ∈ [0, π/2], ρ ∈ [1, 2].
Volume = (1/8) · (4/3)π(2³ - 1³) = (1/8)(4/3)π(7) = 7π/6.
Wait, let me recalculate: full shell volume = (4/3)π(8 - 1) = 28π/3. First octant = (1/8)(28π/3) = 28π/24 = 7π/6.
The correct multiple-choice answer is (D) 7π/6.
Density proportional to distance from the origin, max density 2, ball radius 1 at (0,0,1)
The farthest point from the origin in this ball is (0, 0, 2), at distance 2 from the origin. If density is proportional to distance and the maximum is 2, then δ = kρ with k · 2 = 2, so k = 1 and δ = ρ. The maximum-density point is (0, 0, 2).
The problem: region bounded by y = 3x, xy = 3, xy = 27, with u = xy, v = x/y
The boundaries become:
y = 3x → x/y = 1/3, so v = 1/3.
xy = 3 → u = 3.
xy = 27 → u = 27.
The transformed region in the uv-plane is a rectangle: 3 ≤ u ≤ 27, v = 1/3 (but we also need the other boundary that closes the region in x,y; the original region is in the first quadrant, so v ranges from some lower bound up to 1/3 or vice versa, depending on the geometry).
Computing the Jacobian
From u = xy, v = x/y, solve for x and y:
x = √(uv), y = √(u/v).
∂x/∂u = 1/(2√(uv)) · v^(1/2) ... more cleanly:
|∂(x,y)/∂(u,v)| = 1/(2v). (The standard result for this transformation.)
Alternatively, compute the Jacobian of the forward map (u,v) with respect to (x,y) and invert:
∂u/∂x = y, ∂u/∂y = x.
∂v/∂x = 1/y, ∂v/∂y = -x/y².
Jacobian of (u,v) w.r.t. (x,y): det = y(-x/y²) - x(1/y) = -x/y - x/y = -2x/y = -2v.
So |∂(x,y)/∂(u,v)| = 1/(2|v|) = 1/(2v) in the first quadrant.
Why the change of variables helps:
The original boundaries (y = 3x and two hyperbolas) are curves in Cartesian coordinates. In (u, v) space they become straight lines (constant u or constant v), turning the region into a rectangle. This reversal of complexity is the entire point of the technique.
Polar area element: dA = r dr dθ
Spherical volume element: dV = ρ² sin φ dρ dφ dθ
Polar conversion: x = r cos θ, y = r sin θ, r² = x² + y²
Spherical conversion: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ
Jacobian (change of variables): ∫∫ f dA = ∫∫ f(x(u,v), y(u,v)) |∂(x,y)/∂(u,v)| du dv
Circle x² + y² = 2x in polar: r = 2 cos θ
Ball centred at (0,0,1) in spherical: ρ ≤ 2 cos φ
Mass integrals with variable density arise in structural engineering (finding the centre of mass of a non-uniform beam) and in astrophysics (mass of a star with radially varying density). Spherical coordinates are the natural language for problems with radial symmetry: gravitational fields, antenna radiation patterns, molecular orbitals. Change of variables is used routinely in probability (transforming random variables) and in computational physics (mapping irregular domains to rectangles for numerical integration).
Forgetting the r in dA = r dr dθ is the most common polar-coordinate error. Every polar integral needs that factor.
Confusing φ and θ in spherical coordinates. In the physics convention, φ is the polar angle (from the z-axis) and θ is azimuthal. Some textbooks swap them. Know which convention your course uses.
Believing you can reverse the order of integration by simply swapping the limits. You must re-describe the region first.
Assuming the Jacobian is always positive. Take the absolute value.
⚠️ The spherical volume element ρ² sin φ dρ dφ dθ is a formula-sheet item. Memorise it anyway; you will use it under time pressure.
⚠️ "Explain why the density factor cancels the polar Jacobian" is a favourite short-answer question. The answer: density 1/r times Jacobian r equals 1.
⚠️ The unit ball at (0,0,1) in spherical (ρ ≤ 2 cos φ) is a classic exam trap. Students who do not practise this derivation lose easy marks.
⚠️ For the first-octant volume between two spheres, remember it is 1/8 of the full shell volume.
dV in spherical coordinates is ___. (Answer: ρ² sin φ dρ dφ dθ)
True or false: To reverse the order of integration, you just swap the limits. (Answer: False)
The polar equation of x² + y² = 2x is ___. (Answer: r = 2 cos θ)
∫₀¹ ∫₀¹ ye^x dx dy = ___. (Answer: (e - 1)/2)
True or false: The Jacobian of polar coordinates is r. (Answer: True)
Q: Evaluate ∫₀¹ ∫₀¹ ye^x dx dy.
A: Inner: ∫₀¹ ye^x dx = y(e - 1). Outer: ∫₀¹ y(e - 1) dy = (e - 1)/2.
Q: Convert (0, 0, 1) to spherical coordinates and state the volume element.
A: (ρ, φ, θ) = (1, 0, 0). Volume element: dV = ρ² sin φ dρ dφ dθ.
Q: State the polar bounds for the region outside x² + y² = 1 and inside x² + y² = 2x, and explain why density 1/r cancels the polar Jacobian.
A: θ ∈ [-π/3, π/3], r ∈ [1, 2 cos θ]. The mass integrand is (1/r)(r dr dθ) = dr dθ, since the 1/r density cancels the r from the area element.
Q: For the unit ball at (0,0,1), write the Cartesian equation and its spherical radial bound.
A: x² + y² + (z-1)² ≤ 1. In spherical: ρ ≤ 2 cos φ, with 0 ≤ φ ≤ π/2.
Q: What is the volume of the first-octant region between spheres of radii 1 and 2? (A) 4π/3 (B) 7π/3 (C) 8π/3 (D) 7π/6
A: (D) 7π/6.
Multiple integrals are the foundation for line integrals, surface integrals, and the big vector-calculus theorems (Green's, Stokes', Divergence). The Jacobian reappears in surface-area integrals. Spherical coordinates are essential for Gauss's law in physics and for solving Laplace's equation.
double integral, triple integral, iterated integral, reversing order of integration, Fubini's theorem, polar coordinates, cylindrical coordinates, spherical coordinates, volume element, area element, Jacobian, change of variables, density, mass integral, centre of mass, unit ball, first octant, MATH 215, Calculus III, University of Michigan