16.6 Parametric Surfaces and their Areas

Can define surfaces similarly to spacecurves: need two parameters u, v instead of just t.

Definition. Let x, y, z be functions of two variables u, v, all with the same domain D. The parametric surfacedefined by the co-ordinate functions x, y, z is the collection S of position vectors

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, for all (u, v) ∈ Dv

u

D

(u, v)

r

We think of a parameterized surface as a function r : D → R3.

Example The surface S parameterized by

r(u, v) = u cos vi + u sin vj + u2 cos 2vk

is drawn for 0 ≤ u ≤ 1 and 0 ≤ v < 2π. Since

x2 − y2 = u2 cos2 v− u2 sin2 v= u2 cos 2v = z

we recognize S as the saddle surface z = x2 − y2

Definition. The grid lines of a parametric surface S are the two families of curves for which either u or v isconstant.

Important: surface distinct from parameterization! All surfaces can be parameterized in infinite manyways. Below are two different parameterizations of the saddle surface with the grid lines marked.Note how the grid lines look like a map for the surface. If you rotate the images to look at the (x, y)-plane, the grid lines should look familiar.

r1(u, v) =

u cos vu sin vu2 cos 2v

r2(u, v) = uv

u2 − v2

1

Standard Parameterized Surfaces

Planes The plane through a point with position vector r0 and containing the fixed vectors u and vhas parametric equation

r(u, v) = r0 + uu + vv, u, v ∈ R

The grid lines are parallel to u, v. Equivalent to (r− r0) · n = 0 where n = u×v|u×v| is the unit normalvector.

The blue lines in the picture are thegrid lines with u = 0, u = 1 andu = 2 respectively.The orange lines are v = 0, v = 1and v = 2.

Any other choice of r0, u, v leads toa new parameterization.

Cylinders A cylinder is a parametric surface

r(u, v) = p(u) + vv, u0 ≤ u ≤ u1, v ∈ R,

where p(u) is a spacecurve and v is a constant vec-tor.

The grid lines come in two families: copies of theoriginal spacecurve translated in the v-direction,and lines parallel to v.

A right-circular cylinder is a cylinder where p(u) isa circle and v is perpendicular to the plane of thecircle.

Graphs of Functions The graph of any function z = f (x, y) is automatically parameterized: take(u, v) = (x, y) to obtain

r(u, v) = ui + vj + f (u, v)k

We can also consider graphs parameterized using polar-co-ordinates (u, v) = (r, θ)

r(u, v) = u cos vi + u sin vj + f (u, v)k

2

The saddle surface z = x2 − y2 was parameterized earlier using both these methods.

It is often easiest to think of a surface as a graph: for example. . .

Parameterize the surface S, being the part of the paraboloid z = 10− x2− y2 lying inside the cylinder(x− 1)2 + y2 = 4.

Viewing S as a graph, we first project onto the xy-plane toobtain the domain of the function: the disk D of radius 2 cen-tered at (1, 0).Now parameterize D using modified polar co-ordinates:

(x, y) = (1 + u cos v, u sin v)

Finally, compute z using the formula for the paraboloid toobtain the parameterization

r(u, v) =

1 + u cos vu sin v10− (1 + u cos v)2 − u2 sin2 v

=

1 + u cos vu sin v9− u2 − 2u cos v

where 0 ≤ u ≤ 2, 0 ≤ v < 2π.

Spheres The sphere of radius R has three common parameterizations:

• As two graphs in Cartesian co-ordinates: z = ± f (x, y) = ±√

R2 − x2 − y2.

• As two graphs in polar co-ordinates: z = ± f (r, θ) = ±√

R2 − r2.

• Using spherical co-ordinates: (u, v) = (φ, θ) =⇒ r(u, v) = R

sin u cos vsin u sin vcos u

The grid lines for each are drawn below.

3

Surfaces of revolution Rotate a curve y = f (x)for a ≤ x ≤ b around the x-axis.Letting u measure the distance along the x-axisand v the polar angle in the yz-plane, we obtain

r(u, v) = ui + f (u)(cos vj + sin vk)

The plot is the result when f (x) = 2 + cos x

The grid lines come in two families: copies of theoriginal curve rotated around the x-axis, and cir-cles centered on the x-axis.

Curves can be rotated around any axis for alterna-tive versions.

Tangent Planes

Definition. Let r(u, v) be a parameterization of S. A point r(û, v̂) is regular if (ru × rv)(û, v̂) 6= 0.The tangent plane at a regular point r(û, v̂) is the plane through said point spanned by the tangent vectorsru(û, v̂) and rv(û, v̂).The unit normal vector is n(û, v̂) = ru×rv|ru×rv| (û, v̂).

At r(1, π4 ) = (1√2, 1√

2, 1), the paraboloid

r(u, v) = u cos vi + u sin vj + (2− u2)k

has tangent vectors

ru =

cos vsin v−2u

=

1√2

1√2−2

rv = u

− sin vcos v0

=−

1√2

1√2

0

The unit normal vector can be easily found:

ru × rv =

√

2√2

1

=⇒ n = 1√5

√

2√2

1

The tangent plane therefore has equation

√2(x− 1√

2) +√

2(y− 1√2) + (z− 1) = 0 ⇐⇒

√2x +

√2y + z = 3

4

Surface area

Let S be parameterized by r : D → R3 and consider a point P0 = r(a, b) on S. Increase each of theco-ordinates u, v by small quantities ∆u, ∆v to give three new points

P1 = r(a + ∆u, b), P2 = r(a + ∆u, b + ∆v), P3 = r(a, b + ∆v).

These four points are the corners of a small piece of surface

∆S = {r(u, v) : a ≤ u ≤ a + ∆u, b ≤ v ≤ b + ∆v}

By the definition of partial derivative, we see that

r(a + ∆u, b)− r(a, b)∆u

≈ drdu

(a, b) =⇒ r(a + ∆u, b)− r(a, b) ≈ ru(a, b)∆u

Similarly r(a, b + ∆v)− r(a, b) ≈ rv(a, b)∆v.It follows that the small piece of surface∆S is approximately the parallelogramin the tangent plane at P0 with sidesru∆u and rv∆v. Since cross-productscompute the area of a parallelogram, wesee that the area of ∆S is

A(∆S) ≈ |ru × rv|∆u∆v

We can approximate the surface area of the entire surface S by summing the areas of small approxi-mating parallelograms. Explicitly, if the domain D is a rectangle, subdivided into mn subrectangles,each containing a sample point (ui, vj),, then the area of S is approximately the Riemann sum

Area(S) ≈m

∑i=1

n

∑j=1

∣∣ru(ui, vj)× rv(ui, vj)∣∣∆u∆v

Torus approximated by 192 and 1728 parallelograms resp.

Taking limits as m, n→ ∞, we obtain. . .

Theorem. If a surface S is defined by the vector-valued function r(u, v) where u, v are co-ordinates on somedomain D ⊆ R2, then the surface area of S is given by

A(S) =∫∫

D|ru × rv| du dv

5

Definition. dS = |ru × rv| du dv is the area element of a surface.

Lemma. dS is independent of co-ordinates. Indeed, if u = u(α, β) and v = v(α, β) for some alternativeco-ordinates (α, β), then

|ru × rv| =∣∣∣ ∂(α,β)∂(u,v) ∣∣∣ ∣∣rα × rβ∣∣

du dv =∣∣∣ ∂(u,v)∂(α,β) ∣∣∣ dα dβ

=⇒ |ru × rv| du dv = ∣∣rα × rβ∣∣ dα dβSurface Area of a Graph If S is the graph of z = f (x, y) for (x, y) ∈ D, then we may parameterizewith u = x and v = y:

r(u, v) = ui + vj + f (u, v)k

Therefore

|ru × rv| =∣∣rx × ry∣∣ =

∣∣∣∣∣∣ 10

fx

× 01

fy

∣∣∣∣∣∣ =∣∣∣∣∣∣− fx− fy

1

∣∣∣∣∣∣=√

f 2x + f 2y + 1

Hence the surface area is

A(S) =∫∫

D

√f 2x + f 2y + 1 dx dy

Example The plane z = y− 2x + 2 for 0 ≤ x, y ≤ 1 has surfacearea

A(S) =∫ 1

0

∫ 10

√(−2)2 + 12 + 1 dx dy =

√6

Surfaces of Revolution If S is parameterized by u = x and v = θ, then

r(u, v) = ui + f (v)(cos vj + sin vk)

Supposing that a ≤ x ≤ b and 0 ≤ θ < 2π, we have

|ru × rv| = |rx × rθ |∣∣∣∣( 1f ′(x) cos θf ′(x) sin θ

)× f (x)

( 0− sin θcos θ

)∣∣∣∣ = ∣∣∣∣ f (x)( f ′(x)− cos θ− sin θ)∣∣∣∣

= f (x)√

f ′(x)2 + 1 (supposing f (x) ≥ 0)

The surface area is therefore

A(S) =∫ 2π

0

∫ ba

f (x)√

f ′(x)2 + 1 dx dθ = 2π∫ b

af (x)

√f ′(x)2 + 1 dx

6

Example If f (x) =√

x for 0 ≤ x ≤ 1 then its surface area is

A = 2π∫ 1

0x1/2

√(1

2√

x

)2+ 1 dx

= π∫ 1

0(4x + 1)1/2 dx =

π

6(5√

5− 1)

More General Surfaces Expect hard calculations where you have parameterize the surface and cal-culate ru × rv yourself! For example. . .

A helicoid is parameterized using polar co-ordinates (u, v) = (r, θ):

r(u, v) = u cos v i + u sin v j + v k

for −1 ≤ u ≤ 1 and −π ≤ v ≤ π. Its surface area is

A(S) =∫ π−π

∫ 1−1|ru × rv| du dv

=∫ π−π

∫ 1−1

∣∣∣( cos vsin v0

)×( −r sin v

u cos v1

)∣∣∣ du dv=∫ π−π

∫ 1−1

∣∣∣( sin v− cos vu

)∣∣∣ du dv=∫ π−π

∫ 1−1

√1 + u2 du dv

= 2π(u√

1 + u2 + ln(u +√

1 + u2))∣∣∣10

= 2π(√

2 + ln(1 +√

2))A torus has parametric equation

r(u, v) =(

(R+r cos v) cos u(R+r cos v) sin u

r sin v

)where 0 ≤ u, v ≤ 2π

|ru × rv| =∣∣∣∣( −(R+r cos v) sin u(R+r cos v) cos u

0

)×( −r sin v cos u−r sin v sin u

r cos v

)∣∣∣∣= r(R + r cos v)

Hence the surface area is∫ 2π0

∫ 2π0

r(R + r cos v)du dv = 4π2rR

7

Parametric Surfaces and their Areas

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