Math 314 Lecture #33
§16.6: Parametric Surfaces and their Areas
Outcome A: Find a parametric representation of a surface.
Just as a curve C in the xy-plane is not always the graph of y = f (x), a surface S in
xyz-space may not be the graph of z = f (x, y).
However, just as every “nice” enough curve can be parameterized, so can every “nice”
enough surface.
Points (x, y, z) on a surface S in R3 are represented by a position vector whose components
are functions of two parameters, says u and v:
S : ~r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k
for (u, v) belonging to some domain D in the uv-plane.
We are dealing with functions from a subset D of R2 to a subset S of R3 , i.e., two numbers
in with three numbers out. [Technically, the “graph” of ~r lives in five dimensional space.]
We call ~r(u, v), (u, v) ∈ D, a parametric representation of the surface S.
Example. Let S be graph of z = f (x, y), (x, y) ∈ D.
A standard parametric representation of the surface S is
S : ~r(x, y) = x~i + y~j + f (x, y)~k, (x, y) ∈ D.
Example. The cylinder x2 + y 2 = 1 between z = −2 and z = 3, is not a graph over a
subset of the xy-plane.
Using a standard trigonometric identity, here is a parametric representation of this cylinder:
S : ~r(u, v) = cos u~i + sin u ~j + v ~k, 0 ≤ u < 2π, −2 < v < 3.
Here is a geometric rendering of this function from D (the purple shaded region on the
left) to S (the cylinder on the right).
Outcome B: Identify a surface from its parametric representation.
Let ~r(u, v), (u, v) ∈ D, be a parametric representation of a surface S.
Fixing the value v = v0 results in a curve ~r(u, v0 ) that lies on S.
Instead, fixing the value u = u0 results in a curve ~r(u0 , v) that lies on S.
These give grid curves on the surface S, and understanding these curves helps us identify
the surface S.
Examples. Here are parametric representations for four surfaces:
1.
2.
3.
4.
~r1 (u, v) = hv cos u, v sin u, vi, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2,
~r2 (u, v) = hu2 − v 2 , u, vi, −2 ≤ u ≤ 2, −2 ≤ v ≤ 2,
~r3 (u, v) = hsin u cos v, sin u sin v, cos ui, 0 ≤ u ≤ π/2, 0 ≤ v ≤ 2π,
~r4 (u, v) = hsin u sin v, cos u cos v, sin ui, 0 ≤ u ≤ π/2, 0 ≤ v ≤ 2π.
Use grid curves to match these with the following four surfaces.
Outcome C: Find the tangent plane at a point on a parametric surface.
Passing through the point ~r(u0 , v0 ) on S are the grid curves ~r(u, v0 ) and ~r(u0 , v).
Lying in the tangent plane of S at the point ~r(u0 , v0 ), the tangent vectors of these curves
are
~ru (u0 , v0 ) = xu (u0 , v0 )~i + yu (u0 , v0 ) ~j + zu (u0 , v0 )~k,
and
~rv (u0 , v0 ) = xv (u0 , v0 )~i + yv (u0 , v0 ) ~j + zv (u0 , v0 )~k.
Here is a illustration of what these tangent vectors look like.
A vector normal to the tangent plane of S at the point ~r(u0 , v0 ) is the cross product of
these two tangent vectors:
~ru × ~rv ,
provided this cross product is not ~0.
The surface S is called smooth when ~ru × ~rv is never ~0, so that there is a tangent plane
at each point of S with ~ru × ~rv as its normal.
Example. Let the surface S have the parametric representation
~r(u, v) = hu2 , v 2 , uvi.
Then
~ru = h2u, 0, vi, ~rv = h0, 2v, ui,
and so a vector normal to the tangent plane of S at the point ~r(u, v) is
~i ~j ~k ~ru × ~rv = 2u 0 v = h−2v 2 , −2u2 , 4uvi.
0 2v u
At the point ~r(1, 1) = h1, 1, 1i on S, a vector normal to the tangent plane of S is
h−2, −2, 4i, and so an equation of this tangent plane is
−2(x − 1) − 2(y − 1) + 4(z − 1) = 0.
Outcome D: Compute the area of a parametric surface, including graphs.
We can approximate a small piece of the surface S : ~r(u, v) near the point ~r(u0 , v0 ) by
the part of the tangent plane with sides ~ru (u0 , v0 )∆u and ~rv (u0 , v0 )∆v.
When ∆u, ∆v > 0, the area of this approximating piece of the tangent plane is
k~ru (u0 , v0 )∆u × ~rv (u0 , v0 )∆vk = k~ru (u0 , v0 ) × ~rv (u0 , v0 )k∆u∆v.
The surface area of S : ~r(u, v), (u, v) ∈ D, is
ZZ
k~ru × ~rv k dA.
A(S) =
D
For a surface S that is the graph z = f (x, y), (x, y) ∈ D, the standard parameterization
of the surface, ~r(x, y) = hx, y, f (x, y)i, gives
~i ~j ~k ~rx × ~ry = 1 0 fx = h−fx , −fy , 1i,
0 1 fy and so the surface area of S becomes
ZZ q
ZZ p
A(S) =
(fx )2 + (fy )2 + 1 dA =
1 + k∇f k2 dA.
D
D
Example. Let S be the graph of z = f (x, y) = (2/3)(x3/2 + y 3/2 ), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
Here is a picture of S.
Since ∇f = hx1/2 , y 1/2 i, the area of the surface S is
Z 1Z 1p
A(S) =
1 + x + y dydx
0
0
Z
y=1
2 1
=
(1 + x + y)3/2 y=0 dx
3 0
Z
2 1
=
(2 + x)3/2 − (1 + x)3/2 dx
3 0
1
4
=
(2 + x)5/2 − (1 + x)5/2 0
15
4(35/2 − 25/2 − 25/2 + 1)
=
15
5/2
7/2
4(3 − 2 + 1)
=
.
15