Topic 17 / 17advanced

Surface Parameterization & Area

Parameterize surfaces $\mathbf r(u, v)$, find a normal vector $\mathbf r_u \times \mathbf r_v$, and compute surface area for parametric and explicit surfaces.

A parametric surface is the image of a 2D parameter region $D$ under a map $r (u, v)$ . Two tangent vectors $r_{u}, r_{v}$ span the tangent plane; their cross product is normal. The magnitude of the cross product gives the surface area element.

Parameterizing surfaces

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

for

(u, v) \in D

. Common cases: - Graph

z = f (x, y)

r (x, y) = ⟨ x, y, f (x, y)⟩

. - Sphere of radius

a

⟨ a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ ⟩

. - Cylinder of radius

a

along

z

⟨ a cos θ, a sin θ, z ⟩

. - Cone:

r (r, θ) = ⟨ r cos θ, r sin θ, r ⟩

. - Surface of revolution about the

z

-axis with profile

r (z)

⟨ r (z) cos θ, r (z) sin θ, z ⟩

Normal vector via cross product

Tangent vectors:

r_{u}, r_{v}

. Normal:

n = r_{u} \times r_{v}

. The direction depends on the order of partials and the parameterization. To match a desired orientation (e.g., outward), reverse the order if needed.

Strategy tips

Always check the sign of the normal against the desired orientation.
Reverse the order $r_{u} \times r_{v} \to r_{v} \times r_{u}$ to flip.

Surface area

A = \iint_{D} ∣ r_{u} \times r_{v} ∣ d u d v

. For graphs

z = f (x, y)

A = \iint_{R} 1 + f_{x}^{2} + f_{y}^{2} d A .

The square root factor comes from

∣ r_{u} \times r_{v} ∣

for the parameterization

r (x, y) = ⟨ x, y, f (x, y)⟩

Choosing parameter bounds

D

should match the **piece** of surface you want, no more, no less. Sphere:

θ \in [0, 2 π), ϕ \in [0, π]

. Cylinder slice: pick the

z

range you want. Surface of revolution: pick the angular sweep.

Worked examples

Example 1

Find the area of the part of the surface

z = x^{2} + 2 y

above the triangle with vertices

(0, 0), (1, 0), (1, 2)

1
Use the explicit formula.
$A = \iint_{R} 1 + f_{x}^{2} + f_{y}^{2} d A = \iint_{R} 1 + 4 x^{2} + 4 d A = \iint_{R} 5 + 4 x^{2} d A$
2
Region: $0 \leq x \leq 1, 0 \leq y \leq 2 x$ .
$A = \int_{0}^{1} \int_{0}^{2 x} 5 + 4 x^{2} d y d x = \int_{0}^{1} 2 x 5 + 4 x^{2} d x$
3
$u = 5 + 4 x^{2}$ , $d u = 8 x d x$ .
$= \frac{1}{4} \int_{5}^{9} u d u = \frac{1}{4} \cdot \frac{2}{3} [u^{3/2}]_{5}^{9} = \frac{1}{6} (27 - 55)$

Answer

A = \frac{27 - 5 5}{6}

Example 2

Parameterize the upper hemisphere

x^{2} + y^{2} + z^{2} = 1, z \geq 0

and find a normal vector.

1
Use spherical with $ρ = 1$ .
$r (θ, ϕ) = ⟨ sin ϕ cos θ, sin ϕ sin θ, cos ϕ ⟩, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π /2$
2
Partials.
$r_{θ} = ⟨ - sin ϕ sin θ, sin ϕ cos θ, 0 ⟩, r_{ϕ} = ⟨ cos ϕ cos θ, cos ϕ sin θ, - sin ϕ ⟩$
3
Cross product.
$r_{θ} \times r_{ϕ} = - sin ϕ \cdot ⟨ sin ϕ cos θ, sin ϕ sin θ, cos ϕ ⟩$
4
This points inward; flip to outward by reversing.
$r_{ϕ} \times r_{θ} = sin ϕ \cdot r$

Answer

Normal

= sin ϕ \cdot r

, magnitude

sin ϕ

— surface area element

= sin ϕ d θ d ϕ

Example 3

Find the surface area of the part of the paraboloid

z = x^{2} + y^{2}

below

z = 4

1
$f_{x} = 2 x, f_{y} = 2 y$ .
$A = \iint_{R} 1 + 4 x^{2} + 4 y^{2} d A$
2
Polar: $R$ is the disk $x^{2} + y^{2} \leq 4$ .
$= \int_{0}^{2 π} \int_{0}^{2} 1 + 4 r^{2} r d r d θ$
3
$u = 1 + 4 r^{2}, d u = 8 r d r$ .
$= 2 π \cdot \frac{1}{8} \int_{1}^{17} u d u = \frac{π}{4} \cdot \frac{2}{3} (1 7^{3/2} - 1) = \frac{π}{6} (1 7^{3/2} - 1)$

Answer

A = \frac{π ( 17 17 - 1 )}{6}

Interactive visualizations

Loading visualization…

u3.14

v1.57

Sphere: r(θ, φ) = ⟨sin φ cos θ, sin φ sin θ, cos φ⟩

r_{u} \times r_{v} = ⟨ 1.00, 0.00, 0.00 ⟩, ∣ r_{u} \times r_{v} ∣ = 1.00

Drag

u

and

v

: red dot moves on the sphere; red vector is the normal

r_{u} \times r_{v}

(or its negative). Sign depends on orientation choice.

Loading visualization…

u3.14

v1.00

Cone: r(θ, r) = ⟨r cos θ, r sin θ, r⟩

r_{u} \times r_{v} = ⟨ - 1.00, - 0.00, - 1.00 ⟩, ∣ r_{u} \times r_{v} ∣ = 1.41

Upper cone parameterization. Vary

r

along the slant,

θ

around.

Formulas in this topic

Surface area (parametric)

A = \iint_{D} ∣ r_{u} \times r_{v} ∣ d u d v

Magnitude of normal integrated over parameter domain.

Surface area (graph)

A = \iint_{R} 1 + f_{x}^{2} + f_{y}^{2} d A

For surfaces z = f(x,y).

Normal vector

n = r_{u} \times r_{v}

Direction depends on order of partials.

Sphere parametrization

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

Use spherical coordinates with fixed ρ = a.

Cylinder parametrization

r (θ, z) = ⟨ a cos θ, a sin θ, z ⟩

Polar in xy plus free z.

Parameterizing surfaces

Normal vector via cross product

Surface area

Choosing parameter bounds

Worked examples

Interactive visualizations

Formulas in this topic

Related topics