📐 Concept diagram

07-08 — Surface Integrals

Phase: 7 — Calculus IV: Vector Calculus Subject: 07-08 Prerequisites: 07-07 — Curl and Divergence, double integrals (Phase 6), parametric curves (Phase 5) Next subject: 07-09 — Stokes' Theorem

Learning Objectives

By the end of this subject, you will be able to:

1. Parametrize common surfaces (planes, spheres, cylinders, graphs) and compute surface area
2. Set up and evaluate scalar surface integrals ∫∫_S f dS representing mass, charge, or other accumulated quantities over a surface
3. Compute surface integrals of vector fields (flux integrals) ∫∫_S F · dS, representing the flow of a vector field through a surface
4. Determine the orientation of a surface and its effect on the sign of flux integrals
5. Apply surface integrals to physical problems: flux of electric fields, fluid flow through surfaces, and heat transfer

Core Content

1. Parametric Surfaces and Surface Area

⚠️ CRITICAL FOUNDATION: Surface integrals ∫∫_S F · dS compute flux — the amount of a vector field flowing through a surface. The key skill is parametrizing the surface, computing the normal vector r_u × r_v, and setting up the double integral over the parameter domain.

A surface S in ℝ³ can be described parametrically as:

$r(u, v) = ⟨x(u, v), y(u, v), z(u, v)⟩,   (u, v) ∈ D
$

Examples of parametric representations:

Surface area element dS:

Given a parametrization r(u, v), the tangent vectors are:

$r_u = ⟨∂x/∂u, ∂y/∂u, ∂z/∂u⟩
r_v = ⟨∂x/∂v, ∂y/∂v, ∂z/∂v⟩
$

The area of the small parallelogram spanned by r_u du and r_v dv is:

$dS = |r_u × r_v| du dv
$

Surface area formula:

$Area(S) = ∬_D |r_u × r_v| du dv
$

For a graph z = g(x, y):

$r(x, y) = ⟨x, y, g(x, y)⟩
r_x = ⟨1, 0, g_x⟩
r_y = ⟨0, 1, g_y⟩
r_x × r_y = ⟨−g_x, −g_y, 1⟩
|r_x × r_y| = √(1 + g_x² + g_y²)

dS = √(1 + (∂g/∂x)² + (∂g/∂y)²) dx dy
$

2. Scalar Surface Integrals

Given a scalar function f(x, y, z) defined on the surface S:

$∬_S f(x, y, z) dS = ∬_D f(r(u, v)) · |r_u × r_v| du dv
$

Interpretations: - If f = 1: ∬_S dS = surface area of S - If f = ρ (density): ∬_S ρ dS = mass of a thin shell - If f = temperature: average temperature = (1/Area) ∬_S T dS - If f = charge density: ∬_S σ dS = total charge

Example 1: Evaluate ∬_S (x + y + z) dS where S is the portion of the plane z = 2 − x − y in the first octant.

$Parametrize using x and y: r(x, y) = ⟨x, y, 2−x−y⟩.
Domain D: x ≥ 0, y ≥ 0, and z = 2−x−y ≥ 0 → x + y ≤ 2. (Triangle.)

r_x = ⟨1, 0, −1⟩, r_y = ⟨0, 1, −1⟩.
r_x × r_y = ⟨1, 1, 1⟩.
|r_x × r_y| = √(1 + 1 + 1) = √3.

f(x, y, z) = x + y + (2−x−y) = 2.

∬_S f dS = ∬_D 2 · √3 dx dy = 2√3 · Area(D) = 2√3 · (1/2)(2)(2) = 2√3 · 2 = 4√3.
$

Example 2: Find the surface area of the portion of the paraboloid z = x² + y² below z = 4.

$z = g(x, y) = x² + y², domain D: x² + y² ≤ 4.

dS = √(1 + g_x² + g_y²) dA = √(1 + 4x² + 4y²) dA
   = √(1 + 4(x²+y²)) dA.

Switch to polar: x²+y² = r², dA = r dr dθ.

Surface area = ∬_D √(1 + 4r²) · r dr dθ
             = ∫₀^{2π} dθ ∫₀² r√(1+4r²) dr
             = 2π · (1/8) ∫₀² 8r√(1+4r²) dr
Let u = 1+4r², du = 8r dr. Limits: r=0→u=1, r=2→u=17.
             = 2π · (1/8) ∫₁¹⁷ √u du
             = (π/4) [2u^{3/2}/3]₁¹⁷
             = (π/6)(17^{3/2} − 1)
             = (π/6)(17√17 − 1).
$

3. Vector Surface Integrals (Flux Integrals)

Given a vector field F(x, y, z) and an oriented surface S, the flux of F through S is:

$Flux = ∬_S F · dS = ∬_S F · n dS
$

Where n is the unit normal vector to the surface, pointing in the chosen direction (the orientation).

Interpretation: If F is a fluid velocity field, the flux is the volume of fluid flowing through S per unit time.

Computation using parametrization:

Given r(u, v), the normal vector is r_u × r_v (or its negative, depending on orientation). Then:

$dS = n dS = (r_u × r_v) du dv   (or its negative)

Flux = ∬_S F · dS = ∬_D F(r(u, v)) · (r_u × r_v) du dv
$

The sign depends on whether r_u × r_v points in the chosen orientation direction.

For a graph z = g(x, y) oriented upward:

$r_u × r_v = r_x × r_y = ⟨−g_x, −g_y, 1⟩  (upward, since z-component = +1)

Flux = ∬_D F(x, y, g(x, y)) · ⟨−g_x, −g_y, 1⟩ dx dy
     = ∬_D (−P g_x − Q g_y + R) dx dy
$

Example 3: Find the upward flux of F(x, y, z) = ⟨x, y, z⟩ through the hemisphere x² + y² + z² = 1, z ≥ 0.

Parametrize sphere: r(φ, θ) = ⟨sin φ cos θ, sin φ sin θ, cos φ⟩, φ ∈ [0, π/2], θ ∈ [0, 2π].

r_φ = ⟨cos φ cos θ, cos φ sin θ, −sin φ⟩
r_θ = ⟨−sin φ sin θ, sin φ cos θ, 0⟩

r_φ × r_θ = ⟨sin²φ cos θ, sin²φ sin θ, sin φ cos φ⟩

Check orientation: at the north pole (φ=0), this gives ⟨0, 0, 0⟩ (degenerate).
At φ=π/4: z-component = sin(π/4)cos(π/4) = 1/2 > 0. So this points outward.
For upward flux, outward is correct for the hemisphere.

F(r(φ,θ)) = ⟨sin φ cos θ, sin φ sin θ, cos φ⟩

F · (r_φ × r_θ) = sin³φ cos²θ + sin³φ sin²θ + sin φ cos²φ
                = sin³φ(cos²θ + sin²θ) + sin φ cos²φ
                = sin³φ + sin φ cos²φ
                = sin φ(sin²φ + cos²φ)
                = sin φ.

Flux = ∫₀^{2π} ∫₀^{π/2} sin φ dφ dθ
     = 2π [−cos φ]₀^{π/2} = 2π(0 − (−1)) = 2π.

Check: The divergence of F is 3, so by the divergence theorem (07-10), 
flux through the closed hemisphere (plus the base disk) would be 3·(2π/3) = 2π.
Since the base disk contributes 0 (F·k = z = 0 on z=0), flux through hemisphere = 2π. ✓

4. Orientation of Surfaces

An orientable surface has two sides. The unit normal n picks one side.

• For closed surfaces: outward orientation is standard (n points outward)
• For graphs z = g(x, y): upward orientation means the z-component of n is positive
• Möbius strip: NOT orientable (you can't consistently define a normal)

Effect on flux: Reversing orientation flips the sign of the flux integral.

5. Special Cases and Formulas

Flux through a plane n · (r − r₀) = 0 with constant normal n̂:

If the surface is a flat plane region R with unit normal n̂:

$∬_S F · dS = (∬_S F dS) · n̂  (when F is constant? No...)

Actually: ∬_S F · n̂ dS, where n̂ is constant, so ∬_S F · n̂ dS = n̂ · ∬_S F dS.
$

Surface area of a sphere of radius R: 4πR². Surface area of a cylinder (side only): 2πRh.

Common Misconceptions

1. "dS and dA are the same." dA is the area element in the parameter domain (dx dy or du dv). dS = |r_u × r_v| dA is the area element on the actual surface.

2. "Surface integrals are always hard." Many are simplified by symmetry. For example, ∬_S x dS = 0 over any surface symmetric about the yz-plane.

3. "The normal vector must be a unit vector." In the flux formula dS = (r_u × r_v) du dv, the length of r_u × r_v IS the area scaling factor. No need to normalize separately.

4. "Flux is always positive." Flux can be negative if the field flows against the chosen orientation.

Key Terms

• "Flux is always positive."
• "Surface integrals are always hard."
• "The normal vector must be a unit vector."
• "dS and dA are the same."
• 07 08 Surface Integrals
• Correct: B) The surface area of S
• Correct: B) π
• Correct: B) ⟨−2x, −2y, 1⟩
• Correct: C) Flips the sign
• Correct: C) Outward
• Correct: C) The net flow of F through S
• Correct: C) √(1 + g_x² + g_y²) dx dy

Worked Examples

Example 1: Surface Area of a Cylinder

Problem: Find the surface area of the cylinder x² + y² = 4 between z = 0 and z = 3.

Solution:

$Parametrize: r(θ, z) = ⟨2 cos θ, 2 sin θ, z⟩, θ ∈ [0, 2π], z ∈ [0, 3].

r_θ = ⟨−2 sin θ, 2 cos θ, 0⟩
r_z = ⟨0, 0, 1⟩

r_θ × r_z = ⟨2 cos θ, 2 sin θ, 0⟩
|r_θ × r_z| = √(4 cos²θ + 4 sin²θ) = 2.

Surface area = ∫₀³ ∫₀^{2π} 2 dθ dz = 2 · 2π · 3 = 12π.

Check: 2πrh = 2π·2·3 = 12π. ✓
$

Example 2: Mass of a Spherical Shell

Problem: A spherical shell x² + y² + z² = 4 has surface density f(x, y, z) = x² + y². Find its mass.

Solution:

$Parametrize sphere: r(φ,θ) = ⟨2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ⟩.
φ ∈ [0, π], θ ∈ [0, 2π].

r_φ = ⟨2 cos φ cos θ, 2 cos φ sin θ, −2 sin φ⟩
r_θ = ⟨−2 sin φ sin θ, 2 sin φ cos θ, 0⟩

r_φ × r_θ = ⟨4 sin²φ cos θ, 4 sin²φ sin θ, 4 sin φ cos φ⟩
|r_φ × r_θ| = 4 sin φ (after simplification: = √(16 sin⁴φ cos²θ + 16 sin⁴φ sin²θ + 16 sin²φ cos²φ)
            = 4√(sin⁴φ + sin²φ cos²φ)
            = 4 sin φ √(sin²φ + cos²φ)
            = 4 sin φ).

f = x²+y² = 4 sin²φ cos²θ + 4 sin²φ sin²θ = 4 sin²φ.

Mass = ∫₀^{2π} ∫₀^π 4 sin²φ · 4 sin φ dφ dθ
     = 32π ∫₀^π sin³φ dφ
     = 32π [−cos φ + (1/3)cos³φ]₀^π
     = 32π [(1−1/3) − (−1+1/3)] = 32π (2/3 + 2/3) = 32π · 4/3 = 128π/3.
$

Example 3: Flux Through a Paraboloid

Problem: Find the upward flux of F(x, y, z) = ⟨x, y, z⟩ through the paraboloid z = 1 − x² − y², z ≥ 0.

Solution:

$z = g(x, y) = 1 − x² − y², domain D: x² + y² ≤ 1.
g_x = −2x, g_y = −2y.

Upward normal: n dS = ⟨−g_x, −g_y, 1⟩ dx dy = ⟨2x, 2y, 1⟩ dx dy.

F(x, y, g(x,y)) = ⟨x, y, 1−x²−y²⟩.
F · n dS = (x)(2x) + (y)(2y) + (1−x²−y²)(1) dx dy
         = (2x² + 2y² + 1 − x² − y²) dx dy
         = (x² + y² + 1) dx dy.

Flux = ∬_D (r² + 1) r dr dθ  (polar)
     = ∫₀^{2π} ∫₀¹ (r³ + r) dr dθ
     = 2π [r⁴/4 + r²/2]₀¹
     = 2π (1/4 + 1/2) = 2π (3/4) = 3π/2.
$

Quiz

Q1: A surface integral ∬_S f(x, y, z) dS computes:

A) The volume of the solid bounded by S B) The surface area (when f = 1) or total mass of a thin shell (when f is density) C) The line integral around the boundary of S D) The flux across the surface

Correct: B)

• If you chose B: ∬_S f dS generalizes scalar line integrals to surfaces. With f = 1, it gives surface area; with density, it gives mass. Correct!
• If you chose A: Volume is a triple integral, not a surface integral.
• If you chose C: Line integrals are along curves, not surfaces.
• If you chose D: Flux uses ∬_S F · n dS (vector surface integral), not ∬_S f dS (scalar).

Q2: To evaluate ∬_S f(x, y, z) dS for a surface z = g(x, y), the surface element dS is:

A) dx dy B) √(1 + g_x² + g_y²) dx dy C) g_x g_y dx dy D) ‖∇g‖ dx dy

Correct: B)

• If you chose B: dS = √(1 + (∂g/∂x)² + (∂g/∂y)²) dA — the area scaling factor for a parametric surface. Correct!
• If you chose A: This is dA (area in the xy-plane), which doesn't account for surface tilt.
• If you chose C: This is not the correct formula.
• If you chose D: ‖∇g‖ = √(g_x² + g_y²), missing the 1 under the square root.

Q3: The flux of a vector field F through a surface S is computed as:

A) ∬_S F dS B) ∬_S F · n dS C) ∬_S ∇ × F dS D) ∬_S (∇ · F) dS

Correct: B)

• If you chose B: Flux = ∬_S F · n dS — the surface integral of the normal component of F. It measures how much fluid flows through S. Correct!
• If you chose A: F dS doesn't make sense dimensionally; you need the dot product with the normal.
• If you chose C: That's related to circulation via Stokes' theorem.
• If you chose D: This would integrate divergence over area, but flux uses the normal component F · n.

Q4: For a surface parametrized as r(u, v), the surface element dS equals:

A) du dv B) ‖r_u × r_v‖ du dv C) r_u · r_v du dv D) ‖r_u‖ ‖r_v‖ du dv

Correct: B)

• If you chose B: dS = ‖r_u × r_v‖ du dv — the magnitude of the cross product gives the area of the parallelogram spanned by the tangent vectors. Correct!
• If you chose A: Forgetting the area scaling (only true for a flat square in the uv-plane).
• If you chose C: The dot product doesn't give area — it gives cos of the angle times product of lengths.
• If you chose D: Close but misses the sin factor between the tangent vectors.

Q5: The normal vector n to a parametrized surface r(u, v) is proportional to:

A) r_u + r_v B) r_u × r_v C) r_u · r_v D) ‖r_u‖ + ‖r_v‖

Correct: B)

• If you chose B: The cross product r_u × r_v is perpendicular to both tangent vectors — it's the normal vector to the surface. Correct!
• If you chose A: The sum is also a tangent vector, not a normal.
• If you chose C: The dot product is a scalar, not a vector.
• If you chose D: This is a scalar sum of magnitudes, not a vector.

Q6: For the sphere x² + y² + z² = R², the outward unit normal at (x, y, z) is:

A) ⟨x, y, z⟩ B) ⟨x/R, y/R, z/R⟩ C) ⟨R, R, R⟩ D) ⟨x/R², y/R², z/R²⟩

Correct: B)

• If you chose B: The gradient ∇(x²+y²+z²) = ⟨2x, 2y, 2z⟩ gives the normal direction. Unit normal: ⟨x, y, z⟩/R = ⟨x/R, y/R, z/R⟩. Correct!
• If you chose A: This is the unnormalized normal vector (magnitude R, not 1).
• If you chose C: This is constant, not the position-dependent normal of a sphere.
• If you chose D: Wrong normalization — divides by R² instead of R.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Find the surface area of the portion of the plane 2x + 3y + z = 6 that lies in the first octant.

Problem 2: Evaluate ∬_S z dS where S is the portion of the paraboloid z = x² + y² below z = 1.

Problem 3: Compute the flux of F(x, y, z) = ⟨z, y, x⟩ downward through the surface z = 1 − x² − y², z ≥ 0.

Problem 4: Find the upward flux of F(x, y, z) = ⟨0, 0, z⟩ through the hemisphere x² + y² + z² = 4, z ≥ 0.

Problem 5: Evaluate ∬_S (x² + y²) dS where S is the cylinder x² + y² = 1 between z = 0 and z = 2.

Problem 6: Compute the flux of F = ⟨x, 0, 0⟩ through the unit sphere x² + y² + z² = 1 (outward orientation). Use symmetry.

Problem 7: Find the surface area of the torus obtained by revolving the circle (x−R)² + z² = r² (with R > r) around the z-axis.

Summary

1. Surface integrals extend double integrals to curved surfaces: scalar surface integrals ∬ f dS accumulate quantities like mass or charge; vector surface integrals ∬ F · dS compute flux through oriented surfaces
2. The surface area element dS = |r_u × r_v| du dv is the magnitude of the cross product of tangent vectors; for graphs z = g(x,y), it simplifies to √(1 + g_x² + g_y²) dx dy
3. Flux integrals require choosing an orientation (side) of the surface; reversing orientation flips the sign of the integral; for closed surfaces, the outward orientation is standard
4. Parametrization is the key skill: common surfaces (planes, spheres, cylinders, paraboloids, graphs) have standard parametrizations that simplify computation
5. Symmetry can dramatically simplify surface integrals: odd functions over symmetric domains integrate to zero; constant divergence leads to flux proportional to enclosed volume (preview of the Divergence Theorem)

Pitfalls

1. Forgetting the Jacobian when switching to polar/cylindrical coordinates. When the domain D is a disk and you switch to polar coordinates to evaluate ∬ |r_u × r_v| du dv, don't forget the extra r factor: dA = r dr dθ. The surface area element dS already has a built-in scaling factor; adding polar conversion on top of that requires careful bookkeeping.

2. Confusing the surface normal direction with the orientation required by the problem. For graphs z = g(x, y), r_x × r_y = ⟨−g_x, −g_y, 1⟩ gives the upward normal. If the problem asks for downward flux, you must negate this. Getting the sign wrong flips the flux result.

3. Using the wrong parametrization for the surface type. The cylinder x² + y² = R² should be parametrized as r(θ, z) = ⟨R cos θ, R sin θ, z⟩, not as r(x, y). Using the graph parametrization for a cylinder is messy and often wrong. Each standard surface has its natural parametrization.

4. Neglecting that dS = |r_u × r_v| du dv already includes the scaling. The magnitude of the cross product gives the area scaling factor. Some students erroneously multiply by an additional factor, or forget to take the magnitude when computing scalar surface integrals.

5. Misidentifying which parametrization goes with which surface. Confusing spherical (r(φ, θ)) with cylindrical (r(θ, z)) parametrizations, or using the wrong parameter ranges (e.g., φ ∈ [0, 2π] instead of φ ∈ [0, π] for a sphere), leads to incorrect surface areas and flux values.

Next Steps

Move on to 07-09 — Stokes' Theorem to learn the grand generalization that connects the circulation of a vector field around a closed curve to the flux of its curl through any surface bounded by that curve — the ultimate unification of line and surface integrals.