We have now studied at length how curves in space can be defined parametrically by functions of the form , and surfaces can be represented by functions . In what follows, we will see how we can also define surfaces parametrically. A one-dimensional curve in space results from a vector function that relies upon one parameter, so a two-dimensional surface naturally involves the use of two parameters. If ,, and are functions of independent parameters and , then the terminal points of all vectors of the form
form a surface in space. The equations ,, and are the parametric equations for the surface, or a parametrization of the surface. In Preview Activity 11.6.1 we investigate how to parameterize a cylinder and a cone.
Determine a parameterization of the circle of radius 1 in that has its center at and lies in the plane .
Determine a parameterization of the circle of radius 1 in 3-space that has its center at and lies in the plane .
Determine a parameterization of the circle of radius 1 in 3-space that has its center at and lies in the plane .
Taking into account your responses in (a), (b), and (c), describe the graph that results from the set of parametric equations
and
where and . Explain your thinking.
Just as a cylinder can be viewed as a “stack” of circles of constant radius, a cone can be viewed as a stack of circles with varying radius. Modify the parametrizations of the circles above in order to construct the parameterization of a cone whose vertex lies at the origin, whose base radius is 4, and whose height is 3, where the base of the cone lies in the plane . Use appropriate technology to plot the parametric equations you develop. (Hint: The cross sections parallel to the -plane are circles, with the radii varying linearly as increases.)
In a single-variable setting, any function may have its graph expressed parametrically. For instance, given , by considering the parameterization (where belongs to the domain of ), we generate the same curve. What is more important is that certain curves that are not functions may be represented parametrically; for instance, the circle (which cannot be represented by a single function) can be parameterized by , where .
where varies over the entire domain of . Therefore, any familiar surface that we have studied so far can be generated as a parametric surface. But what is more powerful is that there are surfaces that cannot be generated by a single function (such as the unit sphere), but that can be represented parametrically. We now consider an important example.
To find a parametrization of this torus, we recall our work in Preview Activity 11.6.1. There, we saw that a circle of radius that has its center at the point and is contained in the horizontal plane , as shown in Figure 11.6.3, can be parametrized using the vector-valued function defined by
Figure11.6.3.A circle in a horizontal plane centered at .
To obtain the torus in Figure 11.6.2, we begin with a circle of radius in the -plane centered at , as shown on the left of Figure 11.6.4. We may parametrize the points on this circle, using the parameter , by using the equations
Figure11.6.4.Revolving a circle to obtain a torus.
Let’s focus our attention on one point on this circle, such as the indicated point, which has coordinates for a fixed value of the parameter . When this point is revolved about the -axis, we obtain a circle contained in a horizontal plane centered at and having radius , as shown on the right of Figure 11.6.4. If we let be the new parameter that generates the circle for the rotation about the -axis, this circle may be parametrized by
In this activity, we seek a parametrization of the sphere of radius centered at the origin, as shown on the left in Figure 11.6.5. Notice that this sphere may be obtained by revolving a half-circle contained in the -plane about the -axis, as shown on the right.
Figure11.6.5.A sphere obtained by revolving a half-circle.
Begin by writing a parametrization of this half-circle using the parameter :
Be sure to state the domain of the parameter .
By revolving the points on this half-circle about the -axis, obtain a parametrization of the points on the sphere of radius . Be sure to include the domain of both parameters and . (Hint: What is the radius of the circle obtained when revolving a point on the half-circle around the axis?)
Draw the surface defined by your parameterization with appropriate technology.
Subsection11.6.2The Surface Area of Parametrically Defined Surfaces
Recall that a differentiable function is locally linear — that is, if we zoom in on the surface around a point, the surface looks like its tangent plane. We now exploit this idea in order to determine the surface area generated by a parametrization . The basic idea is a familiar one: we will subdivide the surface into small pieces, in the approximate shape of small parallelograms, and thus estimate the entire the surface area by adding the areas of these approximation parallelograms. Ultimately, we use an integral to sum these approximations and determine the exact surface area.
define a surface over a rectangular domain and . As a function of two variables, and , it is natural to consider the two partial derivatives of the vector-valued function , which we define by
In the usual way, we slice the domain into small rectangles. In particular, we partition the interval into subintervals of length and let ,,, be the endpoints of these subintervals, where . Also partition the interval into subintervals of equal length and let ,,, be the endpoints of these subintervals, where . These two partitions create a partition of the rectangle in -coordinates into sub-rectangles with opposite vertices and for between and and between and . These rectangles all have equal area .
Now we want to think about the small piece of area on the surface itself that lies above one of these small rectangles in the domain. Observe that if we increase by a small amount from the point in the domain, then changes by approximately . Similarly, if we increase by a small amount from the point , then changes by approximately . So we can approximate the surface defined by on the -rectangle with the parallelogram determined by the vectors and , as seen in Figure 11.6.6.
Say that the small parallelogram has area . If we can find its area, then all that remains is to sum the areas of all of the generated parallelograms and take a limit. Recall from our earlier work in the course that given two vectors and , the area of the parallelogram spanned by and is given by the magnitude of their cross product, . In the present context, it follows that the area, , of the parallelogram determined by the vectors and is
Set up an iterated integral to determine the surface area of this cylinder.
Evaluate the iterated integral.
Recall that one way to think about the surface area of a cylinder is to cut the cylinder horizontally and find the perimeter of the resulting cross sectional circle, then multiply by the height. Calculate the surface area of the given cylinder using this alternate approach, and compare your work in (b).
As we noted earlier, we can take any surface and generate a corresponding parameterization for the surface by writing . Hence, we can use our recent work with parametrically defined surfaces to find the surface area that is generated by a function over a given domain.
Let be a region in the domain of . Using Equation (11.6.2), show that the area, , of the surface defined by the graph of over is
Use the formula developed in (a) to calculate the area of the surface defined by over the rectangle .
Observe that the surface of the solid describe in (b) is half of a circular cylinder. Use the standard formula for the surface area of a cylinder to calculate the surface area in a different way, and compare your result from (b).
A parameterization of a curve describes the coordinates of a point on the curve in terms of a single parameter , while a parameterization of a surface describes the coordinates of points on the surface in terms of two independent parameters.
If describes a smooth surface in 3-space on a domain , then the area, , of that surface is given by
(Use and for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e.g., as \lt 1 + s + t, s - t, 3 - t \gt.)
Let be a point on the ellipsoid and let ,, and . Show that lies on the sphere . Hence, find a parameterization of in terms of ,, and as functions of and .
Use the result of part (a) to find a parameterization of the ellipse in terms of ,, and as functions of and . Check your parametrization by substituting ,, and into the equation of the ellipsoid. Then check your work by plotting the surface defined by your parameterization.
Set up an iterated integral whose value is the portion of the surface area of a sphere of radius that lies in the first octant (see the parameterization you developed in Activity 11.6.2).
Then, evaluate the integral to calculate the surface area of this portion of the sphere.
By what constant must you multiply the value determined in (b) in order to find the total surface area of the entire sphere.
Finally, compare your result to the standard formula for the surface area of sphere.
Sketch a picture of the overall solid generated by the plane over the given domain.
Determine a parameterization for the plane over the domain .
Use Equation (11.6.2) to determine the surface area generated by over the domain .
Observe that the vector points from to along one side of the surface generated by the plane over . Find the vector such that and together span the parallelogram that represents the surface defined by over , and hence compute . What do you observe about the value you find?