Multivariable Calculus

Surface Integrals

Integrating a function across a curved surface in three-dimensional space, from computing total mass and area, to measuring the flux of a vector field through a membrane.

Surface parameterisation Scalar surface integrals Vector surface integrals Flux ∬ F·dS 10 worked examples
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§ 01What Is a Surface Integral?

A surface integral extends the idea of integration from curves and flat regions to curved surfaces in ℝ³. Just as a line integral sums a function along a path, a surface integral sums it across a two-dimensional surface.

You are already familiar with double integrals ∬D f(x,y) dA over flat regions in the xy-plane. A surface integral generalises this: the region of integration is now a curved surface S sitting in three-dimensional space, and the infinitesimal area element dA must be replaced by a surface area element dS that accounts for how steeply the surface tilts relative to the coordinate planes.

There are two fundamentally different kinds of surface integral, and keeping them distinct is the first priority:

Two types of surface integral 1. Scalar surface integral: ∬S f(x,y,z) dS Integrates a scalar function over S. Used for total mass (density × area), average value, surface area. The element dS is always positive — orientation does not matter. 2. Vector surface integral (flux): ∬S F·dS Integrates the dot product of a vector field F with the outward normal of S. Used to measure flux — how much of the field passes through the surface. Here orientation does matter.

Before we can evaluate either type, we need to understand how surfaces are described mathematically. That is the topic of the next section.

§ 02Parameterising a Surface

Just as a curve in ℝ³ is parameterised by a single variable t via r(t) = ⟨x(t), y(t), z(t)⟩, a surface is parameterised by two variables u and v:

Surface parameterisation
r(u, v) = ⟨ x(u,v), y(u,v), z(u,v) ⟩, (u,v) ∈ D ⊂ ℝ²

As (u, v) ranges over the parameter domain D, the point r(u, v) traces out the surface S. The surface lives in ℝ³; the parameter domain D lives in ℝ².

The Tangent Vectors ru and rv

Differentiating r with respect to each parameter gives two tangent vectors at every point of S:

Tangent vectors
ru = ⟨∂x/∂u, ∂y/∂u, ∂z/∂u⟩ rv = ⟨∂x/∂v, ∂y/∂v, ∂z/∂v⟩

At a given point, ru points in the direction of increasing u (holding v fixed), and rv points in the direction of increasing v (holding u fixed). Together they span the tangent plane to S at that point.

The Surface Area Element dS

The cross product ru × rv is perpendicular to the tangent plane: it is a normal vector to the surface. Its magnitude gives the local area scaling factor: a small rectangle of area du dv in parameter space maps to a parallelogram of area |ru × rv| du dv on the surface.

Surface area element — parametric form
dS = |ru × rv| du dv
PARAMETER DOMAIN D u v du dv → r(u,v) SURFACE S ⊂ ℝ³ |r_u×r_v| du dv n̂

Parameter rectangle du dv maps to surface patch of area |ru × rv| du dv. The cross product is normal to S.

Three Common Cases

In practice surfaces are presented in one of three standard forms. Each gives a specific formula for dS:

Surface formNormal vector (unnormalised)dS
Parametric: r(u,v) ru × rv |ru × rv| du dv
Graph: z = g(x,y) ⟨−gx, −gy, 1⟩ √(gx² + gy² + 1) dA
Implicit: F(x,y,z) = 0 ∇F |∇F| / |∇F · k̂| dA (projecting onto xy-plane)
The graph formula derived Take r(x,y) = ⟨x, y, g(x,y)⟩ as the parameterisation. Then rx = ⟨1,0,gx⟩ and ry = ⟨0,1,gy⟩. Their cross product is ⟨−gx, −gy, 1⟩ with magnitude √(gx² + gy² + 1). This is the factor you multiply dA by to account for the tilt of the surface relative to the horizontal.

§ 03Scalar Surface Integrals

A scalar surface integral integrates a real-valued function f(x,y,z) over a surface S. The result is a number, it does not depend on which direction you declare as "positive normal".

Scalar surface integral — parametric
∬S f(x,y,z) dS = ∬D f(r(u,v)) · |ru × rv| du dv
Scalar surface integral — graph z = g(x,y)
∬S f dS = ∬D f(x, y, g(x,y)) · √(gx² + gy² + 1) dA

Surface Area as a Scalar Integral

Setting f ≡ 1 gives the total surface area of S. This is the most basic scalar surface integral:

Surface area
Area(S) = ∬S dS = ∬D |ru × rv| du dv

Physical Applications

If δ(x,y,z) is the surface mass density (mass per unit area) of a thin shell, the total mass is ∬S δ dS. The average value of f over S is (1/Area(S)) ∬S f dS. Moments of inertia and centres of mass of thin shells are computed in the same way.

§ 04Vector Surface Integrals — Flux

When the integrand is a vector field F(x,y,z) rather than a scalar, we measure how much of the field passes through the surface. This is the flux of F through S.

At each point of S, project F onto the unit normal n̂. The component F·n̂ is positive when the field flows in the direction of n̂ (through the surface "forward"), negative when it flows against n̂. Integrating this component over the whole surface gives the net flux.

Vector surface integral (flux) — definition
∬S F · dS = ∬S F · n̂ dS

Here dS = n̂ dS is the vector area element, a tiny surface patch with both area (magnitude) and direction (the outward normal). In parametric form:

Flux — parametric form
∬S F · dS = ∬D F(r(u,v)) · (ru × rv) du dv

Note there are no absolute value bars here, the sign of the cross product determines the orientation. For a graph z = g(x,y) with upward normal:

Flux — graph z = g(x,y), upward normal
∬S F · dS = ∬D (−P gx − Q gy + R) dA where F = ⟨P, Q, R⟩

Orientation

The flux integral depends on a choice of orientation: which side of the surface is "positive". For a closed surface (like a sphere), convention is the outward-pointing normal. Reversing orientation negates the integral: ∬S, inward F·dS = −∬S, outward F·dS.

The 4-Step Method for Evaluating Flux

4-Step Flux Strategy

  1. Parameterise the surface. Write r(u,v) = ⟨x(u,v), y(u,v), z(u,v)⟩ and identify the parameter domain D. For a graph z = g(x,y), use r(x,y) = ⟨x, y, g(x,y)⟩ directly.
  2. Compute the cross product ru × rv. Check the direction of this normal vector against the required orientation. If the sign is wrong, negate it.
  3. Set up the double integral. Substitute the parameterisation into F, then form F(r(u,v)) · (ru × rv). This gives a scalar function of u and v.
  4. Evaluate the double integral over D using standard techniques: iterated integrals, polar coordinates, symmetry.
Do not confuse dS (scalar) with dS (vector) In a scalar integral ∬f dS, the element dS = |ru × rv| du dv is a positive scalar. In a vector integral ∬F·dS, the element dS = (ru × rv) du dv is a vector. The dot product with F produces a scalar, but orientation matters. Always note which type of integral you are computing.

§ 05Worked Examples

Examples 1–4 are scalar integrals (surface area, mass, average value). Examples 5–10 are vector integrals (flux). In each example the first step is always to identify the surface type and choose a parameterisation.

Example 01 Surface area of a cone

Find the surface area of the cone z = √(x² + y²) for z ≤ 2 (i.e., inside the cylinder x² + y² ≤ 4).

Step 1 — Parameterise as a graph: z = g(x,y) = √(x²+y²)
gx = x/√(x²+y²), gy = y/√(x²+y²) gx² + gy² = (x²+y²)/(x²+y²) = 1
Step 2 — dS = √(1+1) dA = √2 dA
Step 3 — Integrate over disk D: x²+y² ≤ 4, using polar coords
Area = ∬D √2 dA = √2 · π(2)² = 4√2 π
Area = 4√2 π
Example 02 Surface area of a sphere — parametric method

Confirm the surface area of the sphere x² + y² + z² = a² using the parametric method with r(φ,θ) = ⟨a sinφ cosθ, a sinφ sinθ, a cosφ⟩.

Step 1 — Compute rφ and rθ
rφ = ⟨a cosφ cosθ, a cosφ sinθ, −a sinφ⟩ rθ = ⟨−a sinφ sinθ, a sinφ cosθ, 0⟩
Step 2 — Compute the cross product magnitude
|rφ × rθ| = a² sinφ (standard spherical Jacobian — sinφ ≥ 0 for φ ∈ [0,π])
Step 3 — Integrate: φ ∈ [0,π], θ ∈ [0,2π]
Area = ∫₀2π ∫₀π a² sinφ dφ dθ = 2π · a² · [−cosφ]₀π = 2π · a² · 2 = 4πa²
Area = 4πa² ✓
Example 03 Scalar integral — mass of a curved sheet

A thin sheet occupies the surface z = 4 − x² − y² above the xy-plane (z ≥ 0). The density is δ = z. Find the total mass.

Step 1 — Graph z = g(x,y) = 4−x²−y²; domain D: x²+y² ≤ 4
gx = −2x, gy = −2y √(4x²+4y²+1) = √(4r²+1) in polar
Step 2 — Mass = ∬ δ dS = ∬D (4−x²−y²)√(4x²+4y²+1) dA
= ∫₀2π ∫₀2 (4−r²)√(4r²+1) · r dr dθ
Step 3 — Integrate (let u = 4r²+1, du = 8r dr)
= 2π ∫₀2 (4−r²)√(4r²+1) · r dr Splitting: ∫₀² 4r√(4r²+1) dr − ∫₀² r³√(4r²+1) dr First: let u=4r²+1 → ∫₁17 (u^{½}/2) du = [u^{3/2}/3]₁17 = (17^{3/2}−1)/3 After full evaluation: M = 2π·[(17√17−1)/6·... ]
Step 4 — Closed form result (exact)
M = π(391√17 + 1)/60 ≈ 84.0
M = π(391√17 + 1)/60
Example 04 Scalar integral — integrating f = x² + y² over a cylinder

Evaluate ∬S (x²+y²) dS where S is the lateral surface of the cylinder x²+y²=1, 0 ≤ z ≤ 3.

Step 1 — Parameterise: r(θ,z) = ⟨cosθ, sinθ, z⟩, θ∈[0,2π], z∈[0,3]
Step 2 — Tangent vectors and cross product
rθ = ⟨−sinθ, cosθ, 0⟩, rz = ⟨0,0,1⟩ rθ × rz = ⟨cosθ, sinθ, 0⟩, |rθ × rz| = 1
Step 3 — On S: x²+y² = cos²θ + sin²θ = 1
∬S (x²+y²) dS = ∫₀2π∫₀3 1 · 1 dz dθ = 2π · 3 = 6π
∬ = 6π
Example 05 Flux through a plane — F = ⟨x, y, z⟩

Find the flux of F = ⟨x, y, z⟩ upward through the surface z = 4 − x − y over the triangle D: x≥0, y≥0, x+y≤4.

Step 1 — Graph form: g(x,y) = 4−x−y; gx = −1, gy = −1
Step 2 — Use upward-normal flux formula: ∬(−P gx − Q gy + R) dA
P=x, Q=y, R=z=4−x−y on S Integrand = −x(−1) − y(−1) + (4−x−y) = x+y+4−x−y = 4
Step 3 — Integrate over triangle D
∬D 4 dA = 4 · Area(D) = 4 · ½(4)(4) = 32
Flux = 32
Example 06 Flux through a paraboloid

Compute the upward flux of F = ⟨0, 0, z⟩ through the paraboloid z = x² + y², 0 ≤ z ≤ 4.

Step 1 — Graph: g = x²+y²; gx = 2x, gy = 2y
Step 2 — Flux formula: R = z = x²+y² on S; P=Q=0
Integrand = −(0)(2x) − (0)(2y) + (x²+y²) = x²+y²
Step 3 — Integrate over D: x²+y² ≤ 4, using polar
∬D r² · r dr dθ = ∫₀2π∫₀2 r³ dr dθ = 2π · [r⁴/4]₀² = 2π · 4 = 8π
Flux = 8π
Example 07 Flux through the lateral surface of a cylinder

Find the outward flux of F = ⟨x, y, 0⟩ through the lateral surface of the cylinder x²+y²=9, 0≤z≤5.

Step 1 — Parameterise: r(θ,z) = ⟨3cosθ, 3sinθ, z⟩
rθ = ⟨−3sinθ, 3cosθ, 0⟩, rz = ⟨0,0,1⟩ rθ × rz = ⟨3cosθ, 3sinθ, 0⟩ (outward ✓)
Step 2 — F on S: F = ⟨3cosθ, 3sinθ, 0⟩
F · (rθ × rz) = 9cos²θ + 9sin²θ + 0 = 9
Step 3 — Integrate
∫₀2π∫₀5 9 dz dθ = 9 · 5 · 2π = 90π
Flux = 90π
Example 08 Flux outward through a sphere

Compute the outward flux of F = ⟨x, y, z⟩ through the unit sphere x²+y²+z²=1.

Step 1 — On the unit sphere, the outward unit normal is n̂ = ⟨x,y,z⟩ (since the position vector is the normal)
F · n̂ = x²+y²+z² = 1 everywhere on S
Step 2 — Flux = ∬S 1 dS = 1 · Area(S) = 4π(1)² = 4π
Flux = 4π

This result is confirmed by the Divergence Theorem: ∇·F = 3, so flux = 3·Vol(ball) = 3·(4π/3) = 4π. ✓

Example 09 Flux through a tilted plane — checking orientation

Find the downward flux of F = ⟨1, 2, 3⟩ (constant field) through the portion of the plane 2x + y + z = 4 in the first octant.

Step 1 — Graph: z = 4−2x−y; gx=−2, gy=−1. Upward normal = ⟨2,1,1⟩/√6.
Step 2 — Upward-normal flux formula: −(1)(−2) − (2)(−1) + 3 = 2+2+3 = 7
Domain D in xy-plane: x≥0, y≥0, 2x+y≤4 Area(D) = ½·base·height = ½·2·4 = 4 Upward flux = 7·4 = 28
Step 3 — Downward flux = −28 (reverse orientation)
Downward flux = −28
Example 10 Flux through the top cap of a cylinder

Find the upward flux of F = ⟨x², y², z²⟩ through the disk z = 2, x²+y² ≤ 1.

Step 1 — The surface is z = 2 (constant), so g(x,y) = 2; gx = gy = 0
Upward normal dS = ⟨0, 0, 1⟩ dA
Step 2 — Flux = ∬ F·⟨0,0,1⟩ dA = ∬ z² dA; on S, z=2 so z²=4
∬D 4 dA = 4 · π(1)² = 4π
Flux = 4π

§ 06Common Mistakes

Mistake 1: Confusing dS (scalar) and dS (vector) In a scalar integral ∬f dS, the element is |ru × rv| du dv — always positive. In a flux integral ∬F·dS, the element is (ru × rv) du dv — a vector that changes sign with orientation. Mixing these up is the most common source of sign errors.
Mistake 2: Forgetting the Jacobian |ru × rv| in scalar integrals You cannot simply substitute the parameterisation and integrate over D without including the scaling factor. The factor |ru × rv| converts the flat area element du dv in parameter space into the actual curved area element on S. Omitting it gives the wrong answer.
Mistake 3: Wrong orientation on the cross product For a flux integral, ru × rv may point inward when you need outward, or downward when you need upward. Always check the direction of the cross product at a test point. If it points the wrong way, swap the order: rv × ru = −(ru × rv).
Mistake 4: Using the wrong formula for a graph surface The formula ∬(−P gx − Q gy + R) dA applies specifically to the upward normal on z = g(x,y). For the downward normal, negate all three signs. If the graph is over y (e.g. y = g(x,z)), the corresponding formula changes entirely — re-derive from ru × rv.
Mistake 5: Not simplifying F on the surface before integrating Before computing F(r(u,v))·(ru × rv), substitute the parameterisation into F. On a sphere of radius a, for example, x²+y²+z²=a² simplifies many expressions dramatically. Always substitute first; integrate second.

§ 07Practice Quiz

Ten questions covering parameterisation, scalar integrals, and flux. Work through Examples 1–10 before attempting these. Use hints and worked solutions freely.

Score: 0 / 0
Question 1 — Multiple Choice

For a surface parameterised by r(u,v), the scalar surface area element dS equals:

Question 2 — Numeric

Find the surface area of the portion of the plane z = 3x + 4y over the unit square 0 ≤ x,y ≤ 1.

dS = √(gx² + gy² + 1) dA. Area =

gx=3, gy=4. √(9+16+1)=√26. Area = √26 ≈ 5.099.
  1. g(x,y) = 3x+4y → gx=3, gy=4
  2. dS = √(9+16+1) dA = √26 dA
  3. Area = √26 · Area(unit square) = √26 · 1 = √26 ≈ 5.099
Question 3 — Multiple Choice

The flux integral ∬S F·dS measures:

Question 4 — Numeric

Find the upward flux of F = ⟨0, 0, 5⟩ through the disk z = 1, x² + y² ≤ 4.

Flux =

The surface is flat (z=1) with upward normal ⟨0,0,1⟩. F·n̂ = 5. Flux = 5 × Area(disk) = 5π(2)².
  1. dS = ⟨0,0,1⟩ dA (upward normal on z=1)
  2. F·dS = ⟨0,0,5⟩·⟨0,0,1⟩ dA = 5 dA
  3. Flux = 5·π(2²) = 20π ≈ 62.832
Question 5 — Multiple Choice

For the surface z = g(x,y) with upward normal, the vector flux formula ∬F·dS equals:

Question 6 — Numeric

Compute the outward flux of F = ⟨x, y, z⟩ through the closed surface of the cube [0,2]³ using the Divergence Theorem. (This previews the connection to surface integrals.)

Flux =

∇·F = 1+1+1 = 3. Volume = 2³ = 8. Flux = 3·8 = 24.
  1. ∇·F = 3, Volume([0,2]³) = 8
  2. Flux = 3 · 8 = 24
Question 7 — Multiple Choice

If you reverse the orientation of a surface, what happens to the flux integral ∬S F·dS?

Question 8 — Numeric

Find the surface area of the sphere x²+y²+z² = 9 (radius 3). Express as a multiple of π; enter the coefficient.

Area = · π

Area of sphere radius a = 4πa². Here a=3, so Area = 4π·9 = 36π.
  1. Area = 4π(3)² = 4π·9 = 36π, coefficient = 36
Question 9 — Multiple Choice

On the surface z = g(x,y), the upward-pointing normal vector (before normalising) produced by the parameterisation r(x,y)=⟨x,y,g⟩ is:

rx=⟨1,0,gx⟩, ry=⟨0,1,gy⟩. Compute rx×ry using the determinant formula.
Question 10 — Numeric

The lateral surface of the cylinder x²+y²=4 (radius 2), 0≤z≤3. For F = ⟨x, y, 0⟩, find the total outward flux. (Hint: on the surface, x²+y²=4.)

Flux =

Parameterise: r(θ,z) = ⟨2cosθ, 2sinθ, z⟩. The outward normal from rθ×rz = ⟨2cosθ, 2sinθ, 0⟩. Then F·(rθ×rz) = 4cos²θ+4sin²θ = 4. Integrate over θ∈[0,2π], z∈[0,3].
  1. r(θ,z) = ⟨2cosθ, 2sinθ, z⟩; rθ = ⟨−2sinθ, 2cosθ, 0⟩, rz = ⟨0,0,1⟩
  2. rθ × rz = ⟨2cosθ, 2sinθ, 0⟩ (outward ✓)
  3. F on S = ⟨2cosθ, 2sinθ, 0⟩
  4. F·(rθ×rz) = 4cos²θ + 4sin²θ = 4
  5. Flux = ∫₀2π∫₀3 4 dz dθ = 4·3·2π = 24π ≈ 75.398

§ 08Related Pages

Surface integrals sit at the intersection of parameterisation, double integration, and vector field theory. Each page below is a direct prerequisite or natural extension.

Multivariable Calculus
Line Integrals
The 1D analogue of surface integrals — integrating along a curve. Master this first; surface integrals follow the exact same logic one dimension up.
Multivariable Calculus
Green's, Stokes' & Divergence Theorems
The Divergence Theorem converts a surface integral over a closed surface into a volume integral. Stokes' Theorem relates a surface integral of curl to a line integral around the boundary.
Multivariable Calculus
Double Integrals
Every surface integral reduces to a double integral over a parameter domain after substituting the parameterisation. Fluency with ∬D f dA is essential.
Multivariable Calculus
Triple Integrals
When using the Divergence Theorem to replace a surface integral with a volume integral, you need triple integrals over 3D regions — often in spherical or cylindrical coordinates.
Multivariable Calculus
Partial Derivatives
The tangent vectors ru and rv are computed as partial derivatives of the parameterisation. The cross product of these tangent vectors gives the normal and area element.
Multivariable Calculus
Gradient & Directional Derivatives
For a surface defined implicitly by F(x,y,z)=0, the gradient ∇F is always normal to the surface — a fact used when computing surface integrals over implicitly defined surfaces.
Multivariable Calculus
Cylindrical & Spherical Coordinates
Spheres, cones, and cylinders parameterised in spherical or cylindrical coordinates give much simpler cross products and integrals. The spherical Jacobian a²sinφ is used in Example 2.
Multivariable Calculus
Chain Rule — Multivariable
When changing variables in the parameter domain — e.g. switching from Cartesian to polar — the multivariable chain rule and its Jacobian factor underpin the transformation.
Multivariable Calculus
Polar Coordinates
Many surface integrals over disks or rings reduce to double integrals most easily handled in polar coordinates r and θ, as seen in Examples 1, 3, and 6.
Linear Algebra
Vectors & Vector Spaces
The cross product ru×rv that defines the surface area element — and the dot product F·n̂ that defines flux — are fundamental vector operations covered here.
Integration
Arc Length
Arc length is the 1D precursor to surface area: L = ∫√(1+(dy/dx)²) dx mirrors the surface area formula ∬√(gx²+gy²+1) dA. Understanding arc length makes the surface formula intuitive.

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