We have seen that a space curve can be parametrized by a vector function where ranges over some interval of the -axis. In an analogous manner we can parametrize a surface in space by a vector function where ranges over some region of the -plane.

**Example 1.** (The graph of a function) The graph of a function

can be parametrized by setting

In the same vein the graph of a function

can be parametrized by setting

As
ranges over
,
the tip of
traces out the graph of
.

**Example 2.** (A plane) If two vectors
and
are not parallel, then the set of all linear combinations
generate a plane
that passes through the origin. We can parametrize this plane by setting

The plane
that is parallel to
and passes through the tip of
can be parametrized by setting

Note that the plane contains the lines

**Example 3.** (A sphere) The sphere of radius
centered at the origin can be parametrized by

with
ranging over the rectangle

Derive this parametrization. The points of latitude
form a circle of radius
on the horizontal plane
.
This circle can be parametrized by

This expands to give

Letting
range from
to
,
we obtain the entire sphere. The
-equation
for this same sphere is
It is easy to verify that the parametrization satisfies this equation:

**Example 4.** (A cone) Considers a cone with apex semiangle
and slant height
.
The points of slant height
form a circle of radius
on the horizontal plane
This circle can be parametrized by

Since we can obtain the entire cone by letting
range from
to
,
the cone is parametrized by

with

**Example 5.** (A spiral ramp) A rod of length
initially resting on the
-axis
and attached at one end to the
-axis
sweeps out a surface by rotating about the
-axis
at constant rate
while climbing at a constant rate
.

To parametrize this surface we mark the point of the rod at a distance
from the
-axis
(
) and ask for the position of this point at time
.
At time
the rod will have climbed a distance
and rotated through an angle
.
Thus the point will be found at the tip of the vector

The entire surface can be parametrized by

Let
be a surface parametrized by a differentiable vector function
For simplicity let us suppose that
varies over an open rectangle

Since
is a function of two variables, we can take two partial derivatives:
,
the partial of
with respect to
,
and
,
the partial with respect to
.
Now let
be a point of
for which

The vector function

(here we are keeping
fixed at
)
traces a differentiable curve
that lies on
.
The vector function

(this time we are keeping
fixed at
)
traces a differentiable curve
that also lies on
.
Both curves pass through the tip of

with tangent vector

with tangent vector

The cross product which we have assumed to be different from zero, is thus perpendicular to both curves at the tip of and can be taken as a normal to the surface at that point.

We record the result as follows:

if is a surface given by a differentiable function then the vector is perpendicular to the surface at the tip of and, if different from zero, can be taken as a normal to the surface at that point.

The cross product
is called the *fundamental vector product* of the surface.

**Example 6.** For the plane
we have

, and therefore . The vector is normal to the plane.

**Example 7.** We parametrized the sphere
by setting

with
In this case

and

Thus

As was to be expected, the fundamental vector product of a sphere is parallel
to the radius vector

A linear function

( and not parallel)

parametrizes a plane
.
Horizontal lines from the
-plane,
lines with equations of the form
,
are carried onto lines parallel to
,
and vertical lines,
,
are carried onto lines parallel to
:

Thus a rectangle
in the
-plane
with sides parallel to the
and
axes,

is carried onto a parallelogram on
with sides parallel to
and
.
What is important to us here is that

the area of the parallelogram = (the area of ).

The parallelogram is generated by the vectors

The area of the parallelogram is thus

We can summarize as follows:

Let
be a rectangle in the
-plane
with sides parallel to the coordinate axes. If
and
are not parallel, the linear function

parametrizes a parallelogram with sides parallel to
and
and

the area of the parallelogram (area of ).

More generally, let's suppose that we have a surface
parametrized by a continuously differentiable function

We will assume that
is a basic region in the
-plane
and that
is one-to-one on the interior of
.
Also we will assume that the fundamental vector product
is never zero on the interior of
.
Under these conditions we call
a **continuously differentiable surface** and define

We will try to show the reasoning behind this definition in the case that
is a rectangle
with sides parallel to the coordinate axes. We begin by breaking up
into
little rectangles
.
This induces a decomposition of
into little pieces
Taking
as the center of
,
we have the tip of
in
.
Since the vector
is normal to the surface at the tip of
,
we can parametrize the tangent plane at this point by the linear function

is the portion of
that corresponds to
.
The portion of the tangent plane that corresponds to this same
is a parallelogram with area

Taking this as our estimate for the area of
,
we have

area of area of (area of ).

This is a Riemann sum for

and tends to this integral as the maximum diameter of the
tends to zero.

**Example 8.** (*The surface area of a sphere*) The
function

with
ranging over the set
parametrizes a sphere of radius
.
For this parametrization

and

According to the new formula

area of the sphere

which is known to be correct.

**Example 9.** (*The area of a region*) If
is a plane region
,
then
can be parametrized by setting

Here
and
In this case we have the familiar formula

**Example 10.** (*The area of a surface of revolution*)
Let
be the surface generated by revolving the graph of a function

about the
-axis.
We will assume that
is positive and continuously differentiable.

We can parametrize
by setting

with
ranging over the set
In this case

Therefore
and

area of

**Example 11.** (*Spiral ramp*) One turn of the spiral
ramp of Example 5 is the surface

with
ranging over the set
In this case

Therefore

and

Thus

area of

The integral can be evaluated by setting

Let
be the surface of a function

We are to show that if
is continuously differentiable, then

**Derivation.** We can parametrize
by setting

We may just as well use
and
and write

Clearly

Thus

Therefore
and the formula is verified.

**Problem 12.** Find the surface area of that part of the
parabolic cylinder
that lies over the triangle with vertices
in the
-plane.

*Solution.* Here
so that

The base triangle can be expressed by writing

The surface has area

**Problem 13.** Find the surface area of that part of the
hyperbolic paraboloid
that lies inside the cylinder

*Solution.* Here
so that

The formula gives

In polar coordinates the base region takes the form

Thus we have

There is an elegant version of this last area formula that is geometrically
vivid. We know that the vector

is normal to the surface at the point
The unit vector in that direction, the vector

is called the **upper unit normal** (It is the unit normal with a
nonnegative
-component.)

Now let
be the angle between
and
Since
and
are both unit vectors,

Taking reciprocals we have

The area formula can therefore be written

Imagine a thin distribution of matter spread out over a surface S. We call
this a *material surface.*

If the mass density (the mass per unit area) is constant throughout, then the total mass of the material surface is the density times the area of

(area of ).

If, however, the mass density varies continuously from point to point, then the total mass must be calculated by integration.

To develop the appropriate integral we suppose that

is a continuously differentiable surface, a surface that meets the conditions
for area formula. Our first step is to break up
into
little basic regions
This decomposes the surface into
little pieces
.
The area of
is given by the integral

By the mean-value theorem for double integrals, there exists a point
in
for which

It follows that

area of = (area of ).

Since the point
is in
,
the tip of
is on
The mass density at this point is
.
If
is small (which we can guarantee by choosing
,
small), then the mass density on
is approximately the same throughout. Thus we can estimate
,
the mass contribution of
,
by writing

Adding up these estimates we have an estimate for the total mass of the
surface:

This last expression is a Riemann sum for

The above double integral can be calculated not only for a mass density
function
but for any scalar field
continuous over
.
We call this integral the **surface integral** of
over
and write

Note that, if
is identically
,
then the right-hand side gives the area of
.
Thus

**Problem 1. **Calculate

*Solution.* Call the parameter set
.
Then

By a previous calculation
.
Thus

To find
and
we need the
and
components of
We can get these as follows:

Therefore
and
We can now write

**Problem 2.** Calculate

where
is the ramp of Example 11 of the previous section:

**Solution.** Call the parameter set
.
By a previous calculation

Therefore

Like the other integrals we have studied, the surface integral satisfies a
mean-value condition; namely, if
is continuous, then there is a point
on
for which

We call
the **average value** of
on
.
We can thus write

We can also take weighted averages: if
and
are continuous on
,
then there is a point
on
for which

is called the
**-weighted
average** of
on
.

The coordinates of the centroid
of a surface are simple averages over the surface: for a surface
of area

In the case of a material surface of mass density
the coordinates of the center of mass
are density-weighted averages: for a surface
of total mass

**Problem 3.** Locate the center of mass of a material surface in
the form of a hemisphere
with
given that the mass density varies directly as the distance from the
-plane.

*Solution.* The surface
can be parametrized by

Call the parameter set
and recall that
.
The density function can be written
We can calculate the mass as follows:

By symmetry
and
.
To find
we write

Since
we see that
The center of mass is the point

Suppose that a material surface
rotates about an axis. The moment of inertia of the surface about that axis is
given by the formula

where
is the mass density function and
is the distance from the axis to the point
.
(As with other configurations the moments of inertia about the
axes are denoted by
)

**Problem 4.** Calculate the moment of inertia about the
-axis
of a material sphere

Assume that the surface is homogeneous and has constant mass density
.

*Solution.* We parametrize
by setting

Call the parameter set
and recall that
.
We can calculate the moment of inertia as follows:

Since the surface has mass
we can write

A surface

can be parametrized by

Since
where
is the angle between
and the upper unit normal. Therefore

In evaluating this last integral we use the fact that

**Problem 5.** Calculate

**Solution.** The base region
is the unit disc. The function
has partial derivatives
.
Therefore

and

We evaluate this last integral by changing to polar coordinates.
is the set of all
with polar coordinates
in the set

Therefore