Difficulty: Intermediate to Advanced | Prerequisites: Sections 16.5 – 16.6, parametric surfaces, cross product, double integrals
Surface integrals are the 2D analogue of line integrals: instead of integrating along a curve, you integrate over a surface. Section 16.7 covers two flavours: integrating a scalar function over a surface (for quantities like mass or total heat flux) and integrating a vector field over an oriented surface (for flux). Both rely on the parametric surface machinery from 16.6, specifically the cross product r_u × r_v. This section is the bridge to the divergence theorem and Stokes' theorem in 16.8 and 16.9.
To integrate a scalar function f over a parametric surface, compute ∬ f(r(u,v)) |r_u × r_v| dA. To find surface area, set f = 1. The key computation is always finding |r_u × r_v|. For symmetric integrands over symmetric surfaces, terms that are odd functions over a full period integrate to zero, which can simplify messy-looking integrals dramatically.
Surface integral of a scalar function
The integral ∬_S f(x,y,z) dS, where dS = |r_u × r_v| dA. Think of it as adding up the values of f over the surface, weighted by area. When f = 1, this gives the surface area.
Surface integral of a vector field (flux integral)
The integral ∬_S F · dS = ∬_S F · (r_u × r_v) dA. This measures the net flow of the vector field F through the surface. In simple terms, it answers: how much "stuff" passes through this surface per unit time?
Oriented surface
A surface with a consistent choice of normal direction (inward vs outward, upward vs downward). Orientation matters for flux integrals because it determines the sign.
Torus
A doughnut-shaped surface generated by revolving a circle of radius 1 around a circle of radius R (in Problem 10, R = 4). Parametrised by two angles: one around the tube, one around the hole.
The sphere (x − x₀)² + (y − y₀)² + (z − z₀)² = 4 has radius 2 and is parametrised as r(φ, θ) = ⟨2 sin φ cos θ + x₀, 2 sin φ sin θ + y₀, 2 cos φ + z₀⟩.
The tangent vectors are:
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⟩
The cross product r_φ × r_θ has magnitude |r_φ × r_θ| = 4 sin φ.
To compute ∬_S cos(πz) dS, substitute z = 2 cos φ + z₀:
∬ cos(π(2 cos φ + z₀)) · 4 sin φ dφ dθ
Integrating first over φ from 0 to π, then over θ from 0 to 2π, using substitution w = cos φ:
= −(2/π) ∫₀²π [sin(πz₀ + 2π cos φ)]₀π dθ = −(2/π) · 2π · [sin(πz₀ − 2π) − sin(πz₀ + 2π)]
Since sin is 2π-periodic, sin(πz₀ − 2π) = sin(πz₀) and sin(πz₀ + 2π) = sin(πz₀), so the whole integral equals zero.
The answer does not depend on the centre (x₀, y₀, z₀) of the sphere.
Parametrisation: r(φ, θ) = ⟨(4 + cos θ)cos φ, (4 + cos θ)sin φ, sin θ⟩ for 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ 2π.
This is a torus with major radius R = 4 (distance from the centre of the tube to the centre of the torus) and minor radius r = 1 (radius of the tube).
Grid curves: fixing φ (the angle around the hole) gives circles centred on the tube's centreline. Fixing θ (the angle around the tube cross-section) gives circles centred at the origin.
Surface area computation:
Compute r_φ × r_θ. After working through the cross product, its magnitude simplifies neatly:
|r_φ × r_θ| = √((cos θ + 4)²) = cos θ + 4
(This uses the fact that cos θ + 4 > 0 always, so the absolute value just drops.)
Surface area = ∫₀²π ∫₀²π (cos θ + 4) dθ dφ = 2π ∫₀²π (cos θ + 4) dθ = 2π · [sin θ + 4θ]₀²π = 2π · 8π = 16π².
This agrees with the general formula for a torus: surface area = 4π²Rr = 4π² · 4 · 1 = 16π².
Surface integral of f(x,y,z) = 1 + sin y − sin x cos z:
Substituting the parametrisation into f and multiplying by |r_φ × r_θ| = cos θ + 4, the sin y and sin x terms are odd functions when integrated over the full 2π period in φ. They vanish by symmetry.
The surviving integrand is just (cos θ + 4), which gives the same integral as the surface area: 16π².
\iint_S f(x,y,z)\, dS = \iint_D f(\mathbf{r}(u,v))\, |\mathbf{r}_u \times \mathbf{r}_v|\, dA\text{Surface area} = \iint_D |\mathbf{r}_u \times \mathbf{r}_v|\, dAFor a sphere of radius ρ: |r_φ × r_θ| = ρ² sin φ
For a torus with major radius R and minor radius r, parametrised as r(φ, θ) = ⟨(R + r cos θ)cos φ, (R + r cos θ)sin φ, r sin θ⟩:
|r_φ × r_θ| = r(R + r cos θ)
Surface area of a torus = 4π²Rr
Students often forget the |r_u × r_v| factor when setting up a surface integral. The dS is not simply dA; the cross-product magnitude accounts for how the parametrisation stretches area.
A common mistake with the torus is losing track of which angle is which. In the standard parametrisation, φ goes around the hole (the big circle) and θ goes around the tube (the small circle). Mixing them up gives wrong bounds or wrong cross products.
Students sometimes attempt to simplify |r_φ × r_θ| for the torus before checking whether the expression under the square root is a perfect square. For the standard torus it is, which makes the integral much cleaner.
When symmetry cancels terms, students sometimes do not trust the result and spend time computing integrals that are obviously zero over a full period. If sin(φ) is integrated from 0 to 2π, it is zero. Use this.
⚠️ Computing |r_u × r_v| for standard surfaces (sphere, cylinder, torus) is almost certainly on the exam. Memorise the sphere result (ρ² sin φ) and be able to derive the torus result.
⚠️ Symmetry arguments that eliminate terms are expected on exams. If you can explain why an integral vanishes because the integrand is odd over a symmetric interval, you save time and show understanding.
⚠️ The surface area of a torus (16π² for R = 4, r = 1) is a common "check your answer" benchmark. If you get something wildly different, recheck.
⚠️ Problem 9's result (the integral of cos(πz) over any sphere of radius 2 is zero, regardless of centre) illustrates a deeper symmetry. This type of centre-independence argument appears in exam problems.
True or false: the surface area of a parametric surface is ∬ |r_u × r_v| dA. True.
Fill in the blank: for a sphere of radius 3, |r_φ × r_θ| = ___. 9 sin φ.
True or false: ∫₀²π sin θ dθ = 0. True. This is the symmetry fact that eliminates many terms in torus integrals.
Fill in the blank: the surface area of a torus with R = 5 and r = 2 is ___. 4π²(5)(2) = 40π².
True or false: the surface integral ∬_S f dS depends on the orientation of the surface. False – scalar surface integrals do not depend on orientation (flux integrals do).
Q: Set up (but do not evaluate) the surface integral ∬_S z² dS where S is the sphere x² + y² + z² = 9.
A: Parametrise: r(φ, θ) = ⟨3 sin φ cos θ, 3 sin φ sin θ, 3 cos φ⟩. Then z = 3 cos φ and |r_φ × r_θ| = 9 sin φ. The integral is ∫₀²π ∫₀π (3 cos φ)² · 9 sin φ dφ dθ = ∫₀²π ∫₀π 81 cos²φ sin φ dφ dθ.
Q: Why is the surface area of a torus with major radius R and minor radius r equal to 4π²Rr?
A: |r_φ × r_θ| = r(R + r cos θ). Integrating over 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ 2π: ∫₀²π ∫₀²π r(R + r cos θ) dθ dφ = 2π · r · [Rθ + r sin θ]₀²π = 2π · r · 2πR = 4π²Rr.
Q: Explain why ∬_S cos(πz) dS = 0 over any sphere of radius 2, regardless of where it is centred.
A: After parametrising and substituting, the integral reduces to a constant times [sin(πz₀ − 2π) − sin(πz₀ + 2π)]. Since sin has period 2π, both terms equal sin(πz₀), so the difference is zero.
Q: For the torus in Problem 10, why do the sin(x) and sin(y) terms in f(x,y,z) = 1 + sin y − sin x cos z drop out?
A: After substituting the parametrisation, sin(x) and sin(y) contain factors of sin φ or cos φ that are odd functions over the full period [0, 2π] in φ. Integrating an odd function over a full period gives zero.
Surface integrals of scalar functions generalise the arc-length integral from line integrals (Section 16.2): replace ds = |r'(t)| dt with dS = |r_u × r_v| dA.
Flux integrals (the vector version, covered later in 16.7) lead directly to the divergence theorem (Section 16.9), which relates the flux of F through a closed surface to ∭ div F dV.
The torus parametrisation is a concrete example of a surface that is not simply connected, which matters for Stokes' theorem (Section 16.8). Understanding its geometry helps when deciding whether a vector field on a torus is conservative.
surface integral, scalar surface integral, flux integral, surface area, parametric surface, torus, torus surface area, sphere integral, cross product magnitude, dS, area element, oriented surface, symmetry in integrals, Stewart 16.7, Calculus III, multivariable calculus, MATH 215, University of Michigan, Homework 9