# PARAMETRIZED SURFACES; SURFACE AREA

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

## The Fundamental Vector Product

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

## The Area of a Parametrized Surface

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

## The Area of a Surface

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

# SURFACE INTEGRALS

## The Mass of a Material Surface

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

## Surface Integrals

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

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