Difficulty: Intermediate | Prerequisites: Single-variable integration, MATH 215 Midterm 1 material (partial derivatives, gradients)

TL;DR

Multivariable integration extends single-variable integration to two and three dimensions. You integrate over regions in the plane (double integrals) or in space (triple integrals) by choosing the coordinate system that matches the geometry of the region: Cartesian for rectangles and simple algebraic boundaries, polar for circles, cylindrical for problems with circular cross-sections and a free z-axis, and spherical for balls and cones. The Jacobian (r for polar/cylindrical, ρ² sin φ for spherical) corrects for the stretching each coordinate change introduces.

Key Terms

Double integral

The integral ∬_D f(x, y) dA, which sums f over a two-dimensional region D. In simple terms, it measures the "signed volume" under a surface over a flat region.

Triple integral

The integral ∭_E f(x, y, z) dV, which sums f over a three-dimensional region E. Think of it as accumulating a quantity (mass, charge, volume) throughout a solid.

Polar coordinates

x = r cos θ, y = r sin θ, with Jacobian r. In simple terms, you describe each point by its distance from the origin and its angle, which turns circles into simple bounds.

Cylindrical coordinates

x = r cos θ, y = r sin θ, z = z, with Jacobian r. Think of it as polar coordinates in the xy-plane with a regular z-axis bolted on.

Spherical coordinates

x = ρ cos θ sin φ, y = ρ sin θ sin φ, z = ρ cos φ, with Jacobian ρ² sin φ. Each point is described by its distance from the origin (ρ), an azimuthal angle (θ), and a polar angle from the positive z-axis (φ).

Jacobian

The scaling factor that corrects area or volume elements when changing coordinates. For polar and cylindrical it is r; for spherical it is ρ² sin φ. In simple terms, when you warp a grid, some cells get bigger and some get smaller, and the Jacobian tracks that stretching.

Iterated integral

A multiple integral written as nested single integrals with specific limits on each variable. Setting up the correct limits is usually the hard part of the problem.

Density function ρ(x, y) or ρ(x, y, z)

A function giving mass per unit area (2D) or mass per unit volume (3D) at each point. Mass = the integral of density over the region.

Change of variables (substitution)

Replacing (x, y) or (x, y, z) with new variables to simplify the region or the integrand. The Jacobian of the transformation must multiply the integrand.

Order of integration

The sequence in which you integrate nested variables. Switching the order can simplify computation but requires redrawing the region's bounds.

Polar Coordinates and Double Integrals

When to use polar

Switch to polar whenever the region or the integrand involves x² + y² or circles centred at the origin. Circles like x² + y² = R² become r = R. Circles passing through the origin, such as x² + y² = 2x, become r = 2 cos θ after dividing both sides by r.

The area element transforms as dA = r dr dθ. Forgetting the extra r is the single most common error.

Density and mass problems

Mass of a lamina with density ρ(x, y) over a region D:

m = \iint_D \rho(x, y) \, dA

In polar, if ρ depends on distance from the origin (e.g. ρ = c / √(x² + y²) = c / r), the r in the Jacobian often cancels part of the integrand, making the integral cleaner.

Worked pattern: region between two circles (Exam Q1)

The region outside x² + y² = 1 and inside x² + y² = 2x is described in polar as:

• Inner boundary: r = 1

• Outer boundary: r = 2 cos θ

• Angular range: where 2 cos θ ≥ 1, i.e. -π/3 ≤ θ ≤ π/3

The density ρ = c/r has its largest value where r is smallest. On D, the smallest r is 1 (the inner circle), so the largest density is c/1 = c. Setting this equal to 1 gives c = 1.

The mass integral becomes:

m = \int_{-\pi/3}^{\pi/3} \int_1^{2\cos\theta} \frac{1}{r} \cdot r \, dr \, d\theta = \int_{-\pi/3}^{\pi/3} \int_1^{2\cos\theta} dr \, d\theta

The inner integral gives (2 cos θ - 1). Then:

m = \int_{-\pi/3}^{\pi/3} (2\cos\theta - 1) \, d\theta = [2\sin\theta - \theta]_{-\pi/3}^{\pi/3} = 2\sqrt{3} - \frac{2\pi}{3}

Real-world application

Density integrals in polar coordinates appear in engineering whenever a circular plate or disc has non-uniform material distribution, for instance computing the total charge on a circular capacitor plate with radially varying charge density.

Cylindrical Coordinates and Triple Integrals

When to use cylindrical

Use cylindrical coordinates when the projection of the region onto the xy-plane is circular (or has circular symmetry) but the z-bounds are not spherical. The volume element is dV = r dz dr dθ.

Converting from Cartesian to cylindrical

Replace x² + y² with r², x with r cos θ, y with r sin θ, and multiply the integrand by r. The z-limits often stay the same or become functions of r alone.

Worked pattern: integral over a cylinder with a spherical cap (Exam Q4)

The original integral is:

I = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_1^{\sqrt{2 - x^2 - y^2}} \frac{6x^2 z}{(x^2 + y^2)^{3/2}} \, dz \, dy \, dx

The xy-region is the unit disc (x² + y² ≤ 1). The z-bounds run from z = 1 up to z = √(2 - x² - y²), which is the upper hemisphere of the sphere x² + y² + z² = 2.

In cylindrical coordinates:

• x² + y² = r², so (x² + y²)^(3/2) = r³

• x² = r² cos² θ

• z upper bound: √(2 - r²)

• Region: 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, 1 ≤ z ≤ √(2 - r²)

The integral becomes:

I = \int_0^{2\pi} \int_0^1 \int_1^{\sqrt{2 - r^2}} \frac{6r^2 \cos^2\theta \cdot z}{r^3} \cdot r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^1 \int_1^{\sqrt{2-r^2}} 6\cos^2\theta \cdot z \, dz \, dr \, d\theta

Notice how the r terms cancel completely. The z-integral gives (2 - r² - 1)/2 = (1 - r²)/2. Then integrate over r and use the identity cos² θ = (1 + cos 2θ)/2 for the θ-integral.

Evaluating step by step:

• z-integral: ∫₁^√(2-r²) z dz = (2 - r² - 1)/2 = (1 - r²)/2

• r-integral: ∫₀¹ 6 · (1 - r²)/2 dr = 3 ∫₀¹ (1 - r²) dr = 3[r - r³/3]₀¹ = 3(2/3) = 2

• θ-integral: ∫₀^(2π) cos² θ dθ = π

Final answer: I = 2π

Spherical Coordinates and Triple Integrals

When to use spherical

Use spherical coordinates when the region is bounded by spheres (x² + y² + z² = R²) or cones (z = c√(x² + y²)), or when the integrand depends on x² + y² + z². The volume element is dV = ρ² sin φ dρ dφ dθ.

Angle conventions

• θ (theta): the azimuthal angle in the xy-plane, measured from the positive x-axis. Same as in polar/cylindrical.

• φ (phi): the polar angle measured from the positive z-axis downward. φ = 0 is the north pole, φ = π/2 is the xy-plane, φ = π is the south pole.

The first octant (x ≥ 0, y ≥ 0, z ≥ 0) corresponds to 0 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π/2.

Worked pattern: first-octant shell between two spheres (Exam Q6)

Compute the integral of 3 / (1 + (x² + y² + z²)^(3/2)) over E, where E is the first-octant region between the sphere of radius 1 and the sphere of radius 2.

In spherical coordinates, x² + y² + z² = ρ², so (x² + y² + z²)^(3/2) = ρ³. The bounds are:

• 1 ≤ ρ ≤ 2

• 0 ≤ φ ≤ π/2

• 0 ≤ θ ≤ π/2

The integral becomes:

I = \int_0^{\pi/2} \int_0^{\pi/2} \int_1^2 \frac{3}{1 + \rho^3} \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta

The angular integrals separate completely:

• θ-integral: ∫₀^(π/2) dθ = π/2

• φ-integral: ∫₀^(π/2) sin φ dφ = [-cos φ]₀^(π/2) = 1

For the ρ-integral, notice that 3ρ²/(1 + ρ³) is exactly the derivative of ln(1 + ρ³):

\int_1^2 \frac{3\rho^2}{1 + \rho^3} \, d\rho = \ln(1 + \rho^3) \Big|_1^2 = \ln 9 - \ln 2 = \ln\frac{9}{2}

Final answer: I = (π/2) ln(9/2)

Real-world application

Spherical integrals are used in astrophysics to compute the gravitational potential of spherical mass distributions and in electrostatics for charge distributions on spherical shells.

Setting Up Iterated Integrals and Changing Order

The general approach

To write a triple integral ∭_E f dV as an iterated integral ∫∫∫ f dz dy dx:

1. Project E onto the xy-plane to find the shadow region D.

1. For each (x, y) in D, determine the z-bounds: h₁(x, y) ≤ z ≤ h₂(x, y).

1. Describe D itself with x-bounds (constants a₁ ≤ x ≤ a₂) and y-bounds that depend on x: g₁(x) ≤ y ≤ g₂(x).

Worked pattern: parabolic cylinder region (Exam Q8)

F is the region above the xy-plane and below z = 1 - x². E is the part of F where x ≥ 0 and |y| ≤ 2x.

Setting up the bounds:

• x ranges from 0 to 1 (since z = 1 - x² ≥ 0 requires x² ≤ 1, and x ≥ 0)

• For each x: y ranges from -2x to 2x (the constraint |y| ≤ 2x)

• For each (x, y): z ranges from 0 (xy-plane) to 1 - x² (parabolic cylinder)

So: a₁ = 0, a₂ = 1, g₁(x) = -2x, g₂(x) = 2x, h₁(x, y) = 0, h₂(x, y) = 1 - x².

Computing the volume of E

Volume = ∭_E 1 dV. Since h₂ - h₁ = 1 - x² does not depend on y:

V = \int_0^1 \int_{-2x}^{2x} (1 - x^2) \, dy \, dx = \int_0^1 4x(1 - x^2) \, dx

Expanding: 4x - 4x³. Integrating: [2x² - x⁴]₀¹ = 2 - 1 = 1.

Using symmetry to simplify

For the integral ∭_E (2 + sin y) dV over the same region, note that E is symmetric about y = 0 (it spans -2x ≤ y ≤ 2x). Since sin y is odd in y, the sin y part integrates to zero over this symmetric interval. The remaining integral is just 2V = 2.

Changing order of integration in 2D

To switch ∫₀¹ ∫₀^(x²) f(x, y) dy dx to dx dy order:

1. Sketch the region: 0 ≤ y ≤ x², 0 ≤ x ≤ 1.

1. Rewrite bounds solving for x: y = x² means x = √y, so for a given y, x runs from √y to 1.

1. y ranges from 0 to 1.

1. Result: ∫₀¹ ∫_(√y)¹ f(x, y) dx dy.

Note the exam asked whether ∫₀¹ ∫_(y²)¹ f dx dy equals the original. The correct reversed form has x from √y to 1, so the proposed form with y² is wrong.

Formulas and Coordinate Conversion Reference

Coordinate conversions

Key identities used in integration

\cos^2\theta = \frac{1 + \cos 2\theta}{2}, \quad \sin^2\theta = \frac{1 - \cos 2\theta}{2}

\int_0^{2\pi} \cos^2\theta \, d\theta = \pi, \quad \int_0^{2\pi} \sin^2\theta \, d\theta = \pi

\int_0^{\pi/2} \sin\phi \, d\phi = 1, \quad \int_0^{\pi} \sin\phi \, d\phi = 2

Common region descriptions in polar

• Circle centred at origin, radius R: r = R

• Circle x² + y² = 2ax (passes through origin, centre (a, 0)): r = 2a cos θ

• Circle x² + y² = 2by (passes through origin, centre (0, b)): r = 2b sin θ

Volume and area formulas

• Disc of radius a: area = πa²

• Sphere of radius a: volume = 4πa³/3, surface area = 4πa²

• Cylinder of radius a, height h: volume = πa²h

• Cone with base radius a, height b: volume = πa²b/3

Surface area of a graph

\text{SA} = \iint_D \sqrt{1 + f_x^2 + f_y^2} \, dA

where z = f(x, y) and D is the projection onto the xy-plane.

Common Misconceptions

• Students often forget the Jacobian when converting to polar, cylindrical, or spherical coordinates. dA = r dr dθ, not dr dθ. dV = ρ² sin φ dρ dφ dθ, not dρ dφ dθ. This is the most frequent source of lost marks.

• Students sometimes confuse the spherical coordinate conventions. In MATH 215, φ is measured from the positive z-axis (not the xy-plane). The first octant has 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ π/2.

• When a circle passes through the origin (e.g. x² + y² = 2x), students often set r = 2x instead of properly dividing by r to get r = 2 cos θ. Remember: substitute x = r cos θ and divide the entire equation by r.

• Students frequently set up bounds for double integrals in the wrong order, putting the constant limits on the inner integral and the variable limits on the outer. The outer integral always has constant limits; the inner integral's limits can depend on the outer variable.

• When using symmetry to evaluate integrals, students sometimes claim an integral is zero because "the function is odd" without checking that the region is symmetric about the relevant axis. Both the function and the region must have the right symmetry.

Why It Matters / Exam Flags

⚠️ Choosing the right coordinate system is worth more than computing the integral. A problem that takes two pages in Cartesian may take four lines in polar or spherical. If the region involves circles or spheres, switch coordinates.

⚠️ The Jacobian is tested on every coordinate-change problem. Write dA = r dr dθ or dV = ρ² sin φ dρ dφ dθ explicitly as your first step.

⚠️ Density and mass problems nearly always appear on this exam. The pattern is: find c (often by identifying where density is largest or by a given condition), then integrate ρ · dA or ρ · dV.

⚠️ Setting up limits of integration (without evaluating) is a standalone question type. You can earn full marks just for correctly identifying the six bounds a₁, a₂, g₁, g₂, h₁, h₂.

⚠️ Symmetry arguments (odd functions over symmetric domains integrate to zero) save significant time and are heavily tested in true/false questions.

⚠️ The integral of an odd function of one variable over a region symmetric in that variable is zero. This applies to terms like sin(y) over -2x ≤ y ≤ 2x, or xy³z⁵ over the full unit ball (the term xy³z⁵ is odd in x, y, and z, and the unit ball is symmetric in all three).

Quick Self-Test

1. True or false: The Jacobian for spherical coordinates is ρ sin φ. False (it is ρ² sin φ).

1. Fill in the blank: The circle x² + y² = 2x in polar coordinates is r = ___. 2 cos θ.

1. True or false: ∫₀^(2π) cos²θ dθ = 2π. False (it equals π).

1. Fill in the blank: In spherical coordinates, the first octant requires θ ∈ [0, ___] and φ ∈ [0, ___]. π/2 and π/2.

1. True or false: If f(x, y) is odd in y and the region D is symmetric about the x-axis, then ∬_D f dA = 0. True.

Practice Q&A

Q: Convert the integral ∫∫∫_E 3/(1 + ρ³) · ρ² sin φ dρ dφ dθ, where E is the first octant between spheres of radius 1 and 2, into a product of three single integrals. What is the value?

A: The integral separates as (∫₀^(π/2) dθ)(∫₀^(π/2) sin φ dφ)(∫₁² 3ρ²/(1+ρ³) dρ) = (π/2)(1)(ln(9/2)) = (π/2) ln(9/2).

Q: A lamina occupies the region between r = 1 and r = 2cos θ. Its density is ρ = 1/r. Set up and evaluate the mass integral.

A: The curves intersect where 1 = 2cos θ, so θ = ±π/3. Mass = ∫{-π/3}^{π/3} ∫₁^{2cosθ} (1/r)·r dr dθ = ∫{-π/3}^{π/3} (2cos θ - 1) dθ = 2√3 - 2π/3.

Q: Write the triple integral ∭_E f dV as an iterated integral where E is the region x ≥ 0, |y| ≤ 2x, 0 ≤ z ≤ 1 - x².

A: ∫₀¹ ∫_{-2x}^{2x} ∫₀^{1-x²} f(x,y,z) dz dy dx.

Q: Compute the volume of the region in the previous question.

A: V = ∫₀¹ 4x(1-x²) dx = [2x² - x⁴]₀¹ = 1.

Q: If g(x,y,z) = 2 + sin(y) and the region E is symmetric in y (i.e. -2x ≤ y ≤ 2x), what is ∭_E g dV?

A: The sin(y) term integrates to 0 by symmetry (odd function, symmetric interval). So the answer is 2V = 2(1) = 2.

Related Terms / Search Tags

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