Motivating Questions
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What does it mean to parameterize a surface?
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How can parameterizations of curved surfaces be used to compute surface area?
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Section 13.10 Parameterizations of Surfaces and Surface Area
Notes to the Instructor and Dependencies.
This section was previously discussed in the chapter on multiple integration, but was not used until surface integrals were defined in Chapter 13.
Subsection 13.10.1 Introduction
Our work in Section 9.6 and Chapter 10 used parameterizations of curves in two or three dimensions to create and use tools describing points and other properties of the curves. Remember that a parameterization of a curve, \(C\text{,}\) is a vector-valued function of one variable, \(\vr(t) = \langle x(t), y(t), z(t) \rangle\text{,}\) such that the terminal points of the output vectors of \(\vr(t)\) will trace out a given curve in space. Parameterizations of a curve must include bounds on the parameter \(t\) that describe where the curve begins and ends, when applicable. The graph of a vector-valued function of one variable often does not include plots of the position vectors, but will plot only the terminal points of the output vectors as a curve in space.
Parameterizations of curves enabled us to easily generate points, use vector and calculus tools (like direction of travel, direction of turning, arc length, curvature, etc.), and describe motion along the curve. In each of these uses, the parameterization reduced our curve, as a set of points in space, to a function of one variable, which is so valuable because we have a LOT of tools for functions of one variable. In particular, curves like circles, ellipses, hyperbolas, and parabolas cannot be expressed with one variable as a function of the others, but these curves can be studied with 1-variable calculus through our use of parameterizations.
One of the central goals of this section is to extend the idea of parameterizations to curved surfaces in an attempt to apply the many multivariable tools of Chapter 11 and Chapter 12. In the following preview activity, we will look at how we can use geometric knowledge of right circular cylinders and cones to create a vector-valued function that will describe the points on the surface of a cone.
Preview Activity 13.10.1.
Recall the standard parameterization of the unit circle that is given by
\begin{equation*}
x(t) = \cos(t) \ \ \ \ \text{ and } \ \ \ \ y(t) = \sin(t),
\end{equation*}
where \(0 \le t \le 2\pi\text{.}\)
(a)
Give a parameterization of a circle of radius of 2 in \(\R^3\) that has its center at \((0,0,1)\) and lies in the plane \(z=1\text{.}\) Be sure to give the bounds on your parameter as well.
Hint.
The \(z\)-coordinate will be constant, so you only have to think about how \(x\) and \(y\) are changing in your parameterization.
(b)
Give a parameterization of a circle of radius of 2 in \(\R^3\) that has its center at \((0,0,-5)\) and lies in the plane \(z=-5\text{.}\) Be sure to give the bounds on your parameter as well.
(c)
Give a parameterization of a circle of radius of 2 in \(\R^3\) that has its center at \((0,0,a)\) and lies in the plane \(z=a\text{.}\) Be sure to give the bounds on your parameter as well.
(d)
We can think of a right circular cylinder surface centered on the z-axis, such as the one shown in Figure 13.10.1, as being built with a stack of circles of the same radius centered at points along the z-axis. We want to give a parameterization of a right circular cylinder surface of radius 2, centered on the \(z\)-axis with heights from \(b\) to \(c\text{.}\) In other words, we want to express the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinates of points on the right circular cylinder in terms of two variables. We will let the parameter \(s\) measure the height of the circle (in our stack) above or below the \(xy\)-plane and let the parameter \(t\) measure the location around the circle centered on the \(z\)-axis. Use your work for the previous part to write out each of the following:
\begin{equation*}
x(s,t) = \hspace{1.0in} y(s,t) = \hspace{1.0in} z(s,t)= \hspace{1.0in}
\end{equation*}
with the bounds on \(s\) and \(t\) given by
\begin{equation*}
\hspace{1.0in} \leq s \leq \hspace{1.0in} \text{ and } \hspace{1.0in} \leq t \leq \hspace{1.0in}
\end{equation*}
(e)
Just as a right circular 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. In particular, the cone given by \(z^2=x^2+y^2\) will have contours given by circles centered on the z-axis with radius changing according to the rule \(r=z\text{.}\) We want to give a parameterization of the cone given by \(z^2=x^2+y^2\) with heights from \(b\) to \(c\text{.}\) In other words, we want to express the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinates of points on the cone in terms of two variables. We will let the parameter \(s\) measure the height of the circle (in our stack) above or below the \(xy\)-plane and let the parameter \(t\) measure the location around the circle centered on the \(z\)-axis. Write out each of the following:
\begin{equation*}
x(s,t) = \hspace{1.0in} y(s,t) = \hspace{1.0in} z(s,t)=
\end{equation*}
with the bounds on \(s\) and \(t\) given by
\begin{equation*}
\hspace{1.0in} \leq s \leq \hspace{1.0in} \text{ and } \hspace{1.0in} \leq t \leq \hspace{1.0in}
\end{equation*}
Subsection 13.10.2 Parameterizations of Surfaces
In Preview Activity 13.10.1, we finished with a description of a cone in terms of measurements corresponding to the angle around the \(z\)-axis and the height above the \(xy\)-plane. We used geometric intuition to come up with the coordinate relationships of our two surfaces in Preview Activity 13.10.1. We will spend some time in this subsection talking about how to use algebraic and geometric tools to give parameterizations of curved surfaces. The next subsection will talk about some of the important geometric measurements that can be done with a parameterization of a curved surface. In the last subsection, we will look at how parameterizations are useful in computing the surface area of a curved surface in three dimensions.
In a single-variable setting, any function may have its graph expressed parametrically. For instance, if we look the graph of \(y = g(x)\text{,}\) we can consider the parameterization \(\langle t, g(t) \rangle\) (where \(t\) belongs to the domain of \(g\)), which will generate the same curve. Certain curves that are not expressible with \(y\) as a function of \(x\) may be represented parametrically; for instance, the circle (which cannot be represented with either the \(x\) or \(y\)-coordinate written as a function of the other) can be parameterized by \(\langle \cos(t), \sin(t) \rangle\text{,}\) where \(0 \leq t \lt 2\pi\text{.}\)
When we look at a surface of the form \(z = f(x,y)\text{,}\) we can express the \((x,y,z)\) points on the surface parametrically by
\begin{equation*}
\langle x(s,t), y(s,t), z(s,t) \rangle = \langle s, t, f(s,t) \rangle,
\end{equation*}
where \((s,t)\) varies over the entire domain of \(f\text{.}\) In the parameterization above, we can think of \(s\) as acting like \(x\) and \(t\) as acting like \(y\text{.}\) Therefore, any familiar surface expressed as \(z=f(x,y)\) that we have studied so far can be generated as a parametric surface. The greater power of parameterizations is realized when dealing with surfaces that cannot be expressed by a single function \(z = f(x,y)\) (such as the unit sphere), but can be represented parametrically.
For surfaces where we cannot express one coordinate as a function of the other two, like ellipsoids or hyperboloids, the best strategy is to use geometric knowledge of these surfaces to express each of the \(x\)-, \(y\)-, and \(z\)-coordinates in terms of two measurements/parameters. In Preview Activity 13.10.1, we use parameters that measures the rotational coordinate (around the z-axis) and a coordinate that was related to the height above or below the \(xy\)-plane (along with the radial coordinate in the cone). In the next example, we will look at how to parameterize a torus (think of the surface of a doughnut) using this same strategy.
Example 13.10.2.
In this example, we want to parameterize the torus shown in Figure 13.10.3.
An 3D plot with axes labeled x, y, and z. A torus is drawn in light blue center at the origin and its inner opening oriented along the z-axis. Circular slices oriented radially from the z-axis are highlighted in green and circular slices centered around the z-axis are highlighted in red.
We need to be able to describe each \((x,y,z)\) point on the torus in terms of two parameters. We can think of the torus as being generated by rotating a circle or radius \(a\) around the z-axis. If you look at a radial slice out from the z-axis out, as shown by the gray plane in Figure 13.10.4, we generate the torus by revolving the points on the green circle around the z-axis, as shown by the red circles. This means that we can describe each point on the torus in terms of where each point corresponds to its location along the green circle and its location along a red circle (rotation around the z-axis).
An 3D plot with axes labeled x, y, and z. A gray plan extends from the z-axis in the direction of positive x and y coordinates. On this gray plane is a green circle, centered on the xy-plane, with 13 points on the circle shown in black. For each of the black points on the green circle is rotated around the z-axis as shown by circles drawn in red.
We can think of the green slice of the torus as occurring on a constant value of the cylindrical coordinate \(\theta\text{.}\) We will use the \(t\) parameter to describe our position in terms of rotation around the z-axis; in other words, our parameter \(t\) will act as the \(\theta\)-coordinate from cylindrical coordinates. We will use our other parameter, \(s\text{,}\) to describe the location of a point on the green circle.
Looking at the slice of the torus shown by the grey plane in Figure 13.10.4, we get a two dimensional plot as shown in Figure 13.10.5. We need to describe points in \(r\) and \(z\) in terms of the parameter \(s\text{.}\) Our green circle in the \(rz\)-plane will be described by \(r(s)=b+a \cos(s)\) and \(z(s)=\sin(s)\) with \(0\leq s \leq 2\pi\text{.}\) This is a typical parameterization of a circle with a horizontal translation by \(a\text{.}\)
We can use our cylindrical coordinate transformations to write the \(x\text{,}\) \(y\text{,}\) and \(z\)-coordinates in terms of our parameters \(s\) and \(t\text{.}\) The cylindrical coordinates of our points can be written (in terms of \(s\) and \(t\)) as
\begin{equation*}
\theta = t \quad r = b+a\cos(s) \quad z=a\sin(s)
\end{equation*}
So the rectangular coordinates of the points on the torus can be written (in terms of \(s\) and \(t\)) as
\begin{align*}
x \amp= r \cos(\theta) \amp= (b+a \cos(s))\cos(t)\\
y \amp= r \sin(\theta) \amp= (b+a \cos(s))\sin(t)\\
z \amp= z \amp= a \sin(s)
\end{align*}
Just as in the case of parameterizing a curve in space with a vector valued function of one variable, we need to specify bounds on the parameters used to describe our surface. Remember that for a point on the torus, the \(t\)-parameter describes the location in terms of rotation around the \(z\)-axis and the \(s\)-parameter describes the point’s location around the circular slice (in the \(rz\)-plane). Both the \(s\) and \(t\) parameters will be bounded below by \(0\) and above by \(2 \pi\text{.}\) We can organize our parametric equations for the torus using a vector valued function of two variables such as
\begin{align*}
\vr (s,t) =& \langle x(s,t),y(s,t),z(s,t)\rangle \\
=& \langle \left(b+a \cos(s)\right)\cos(t), \left(b+a \cos(s)\right)\sin(t), a \sin(s) \rangle
\end{align*}
for \(s,t \in [0,2 \pi)\text{.}\)
In our next activity, we will look at how to parameterize a sphere centered at the origin two ways: 1) we think of the sphere as being a surface of revolution around the \(z\)-axis and 2) we use spherical coordinates.
Activity 13.10.2.
In this activity, we want to give a parametrization of the sphere of radius \(R\) centered at the origin, as shown in Figure 13.10.6. You can check the box at the top of the plot of Figure 13.10.6 to highlight what a semicircular slice of the sphere looks like along a slice with a constant \(\theta\) coordinate. Notice that this slice shows how the sphere can be made by rotating this half circle around the \(z\)-axis.
An interactive three dimensional plot that shows axes labeled x, y, and z. A sphere centered at the origin is shown in light blue. A checkbox above the interactive plot will toggle a half circle plotted along the sphere from the positive z-axis to the negative z-axis.
(a)
Our first approach to parameterizing the sphere of radius \(R\) will be similar to our approach to parameterizing the torus as a surface of revolution around the \(z\)-axis. In other words, we will let the parameter \(t\) represent the location as a rotation around the \(z\)-axis. So we will be able to parameterize the sphere with
\begin{equation}
x(s,t)= r(s) \cos(t) , \quad y(s,t)=r(s) \sin(t), \quad z(s,t)= z(s) \tag{13.10.1}
\end{equation}
where \(r(s)\) and \(z(s)\) describe the other cylindrical coordinates in terms of another parameter.
A two dimensional plot with horizontal axis labeled \(r\) and vertical axis labeled \(z\text{.}\) The right half of a circle centered at the origin is shown in green and its radius is labeled \(R\text{.}\)
We want to describe the points on the green half circle shown in Figure 13.10.7 in terms of our parameter \(s\text{.}\) Remember that there is not a unique way to parameterize a curve. Give a parameterization of the green half circle in Figure 13.10.7 as a function of the parameter \(s\text{.}\) Be sure to state the bounds on your parameter.
Hint.
In the spirit of thinking of a sphere as a surface of rotation, you could try to write \(r\) as a function of \(z\text{.}\) In other words, how large is the radius of the revolution around the \(z\)-axis in terms of the \(z\) coordinate?
(b)
We want to combine your work for the previous task with Equation (13.10.1) to generate a parameterization of the sphere of radius \(R\text{.}\) Plug in your answer for the previous task into Equation (13.10.1) and state your answer as parametric functions for \(x\text{,}\) \(y\text{,}\) and \(z\text{.}\) Be sure to state the bounds on your parameters \(s\) and \(t\text{.}\)
\begin{align*}
\amp x(s,t) \amp= \\
\amp y(s,t) \amp= \\
\amp z(s,t) \amp= \\
\amp\leq s \amp \leq\\
\amp\leq t \amp \leq
\end{align*}
(c)
Use Figure 13.10.8 to verify that your parameterization for the previous task will plot the sphere of radius 3 centered at the origin. Remember to adjust the upper and lower bounds of \(s\) and \(t\) to match your parameterization.
(d)
We now want to shift to using spherical coordinates in order to create a parameterization of the sphere of radius \(R\text{.}\) In particular, we want to express the rectangular coordinate equation \(x^2+y^2+z^2=R^2\) in spherical coordinates. State an equation in spherical coordinates that describes the same set of points as \(x^2+y^2+z^2=R^2\text{.}\)
Hint.
This equation is of the form \(\rho= \text{constant}\)
(e)
For surfaces of the form \(z=f(x,y)\) were easily parameterized because we could write each of the \(x\text{,}\) \(y\text{,}\) and \(z\)-coordinates in terms of two variables, namely \(x\) and \(y\text{.}\) We can apply a similar idea here because in terms of spherical coordinates, our surface has one spherical coordinate fixed and the other two spherical coordinates can be used as our parameters. Write a couple of sentences to describe which spherical coordinate you want \(s\) to act as and which you would like \(t\) to act as. You should include the bounds on these parameters as part of your descriptions.
(f)
Let’s put the last couple of tasks together to come up with a spherical coordinate parameterization of \(x^2+y^2+z^2=R^2\text{.}\) Specifically, you will need to plug in the constant value for the appropriate spherical coordinate and \(s\) or \(t\) (as dictated by your choice in the previous task) for the other two spherical coordinates into the following transformation equations.
\begin{equation*}
x=\rho \cos(\theta) \sin(\phi) \quad \quad y=\rho \sin(\theta) \sin(\phi) \quad \quad z=\rho \cos(\phi)
\end{equation*}
This should give you equations for \(x\text{,}\) \(y\text{,}\) and \(z\) in terms of the parameters \(s\) and \(t\text{,}\) as well as bounds on both \(s\) and \(t\text{.}\)
\begin{equation*}
x(s,t) = \hspace{1.0in} y(s,t) = \hspace{1.0in} z(s,t)=
\end{equation*}
with the bounds on \(s\) and \(t\) given by
\begin{equation*}
\hspace{1.0in} \leq s \leq \hspace{1.0in} \text{ and } \hspace{1.0in} \leq t \leq \hspace{1.0in}
\end{equation*}
(g)
Use Figure 13.10.8 to verify that your parameterization of the sphere of radius 3 does parameterize the correct surface.
As you can see in the previous activity, there are many ways to parameterize the same surface. The parameterizations you generated for the sphere depended on whether you viewed the sphere as a surface of revolution or whether you utilized spherical coordinates. There are many other ways to parameterize a sphere, but these are likely the two most familiar and common ideas.
In the exercises of this section, you will be asked to generate and understand the parameterization of the quadric surfaces and cylinder surfaces described in Section 9.8. Additionally, you will need to understand how shifts and stretches can be easily applied to each rectangular coordinate separately.
Subsection 13.10.3 The Geometry of a Parametric Surface
In this subsection, we will go over several useful ideas about how to view geometric information given by the parameterization of a curved surface. For this section, we will examine a surface \(S_1\text{,}\) given by a parameterization \(\vr(s,t)=\langle x(s,t), y(s,t), z(s,t) \rangle \) with \(a \leq s \leq b\) and \(c \leq t \leq d\text{.}\)
In Subsection 11.2.4, we saw how holding one of the input coordinates constant allowed us to restrict the graph of \(z=f(x,y)\) to a one variable slice which we called the trace. In particular, we could look at the trace along \(x=a\) which is given by \(z=f(a,y)\text{,}\) for example Figure 11.2.11. Restricting our focus to one slice allowed us to define partial derivatives and many other tools in Chapter 11.
What happens when we hold one of our parameters constant and allow the other to vary? We will get a slice of the parametric surface but this will not be the direction of one coordinate (\(x\text{,}\) \(y\text{,}\) or \(z\)), but will describe the surface in a constant direction of one of our parameters, \(s\) or \(t\text{.}\) The trace given by holding \(s\) constant will yield \(\vr(a,t)\text{,}\) a curve parameterized as a vector valued function of one variable (in this case, \(t\)). We can use all of our tools from Chapter 10 to understand this curve. Specifically, the derivative, \(\frac{d\vr}{dt} (a,t)=\langle \frac{dx}{dt}(a,t),\frac{dy}{dt}(a,t),\frac{dz}{dt}(a,t)\rangle\) will give a vector that is tangent to the curve given by \(\vr(a,t)\text{.}\)
Similarly, we can look at the trace generated in the direction of a constant value of \(t\text{.}\) We will get a trace parameterized by \(\vr(s,b)\) with a tangent vector given by \(\frac{d\vr}{ds}(s,b)\text{.}\)
In Figure 13.10.9, you can see plot of our parametric surface given by \(\vr(s,t)\) for \(s_1 \leq s \leq s_2\) and \(t_1 \leq t \leq t_2\) plotted as a light blue surface. There are five constant values of \(t\) plotted along the surface in shown by the magenta curves and five constant values of \(s\) plotted along the surface as shown by the yellow curves. The tangent vectors given by \(\frac{d\vr}{ds} (s,b)\) and \(\frac{d\vr}{dt} (a,t)\) are shown at a particular point on the surface. You can use the sliders at the top of Figure 13.10.9 to change the location at which the tangent vectors are shown.
An interactive three-dimensional plot of a curved surface that is parabolic in one coordinate direction and cubic in the other. A grid of magenta and yellow traces along the surface are drawn. Two sliders at the top of the figure allow the user to change the location of a highlighted point on the surface. At the highlighted point, a vector is shown in magenta that is tilted to be tangent to the magenta trace through the highlighted point. A yellow vector is also drawn tangent to the yellow trace through the highlighted point. A yellow parallelogram is drawn with sides given by the magenta and yellow vectors.
We can use the red and yellow tangent vectors at the highlighted point (shown in black) to create the tangent plane at the highlighted point. A parallelogram of the tangent plane is shown in orange. Note that the orange parallelogram will move and tilt to estimate the curved surface near the highlighted point. We can use the properties of the cross product to both compute a vector that will be orthogonal to the curved surface at the highlighted point and find the area of the orange parallelogram (Key Idea 9.5.5 and Key Idea 9.5.6).
We take a few moments now to summarize the ideas above, but state our derivatives of \(\vr(s,t)\) as partial derivatives with respect to \(s\) or \(t\text{,}\) since those will correspond to holding the \(t\) and \(s\) parameters constant (respectively).
-
Curves in the direction of change of \(s\) and \(t\) are given by vector valued functions \(\vr(s,b)\) and \(\vr(a,t)\text{,}\) respectively.
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Vectors tangent to the parametric surface given by \(\vr(s,t)\) through the point given by \(\vr(a,b)\) will be given by \(\frac{\partial \vr}{\partial s}(a,b)= \vr_s(a,b)\) and \(\frac{\partial \vr}{\partial t}(a,b)= \vr_t(a,b)\text{.}\)
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The vector given by \((\vr_s \times \vr_t)(a,b)\) will be orthogonal to surface at the point \(\vr(a,b)\) and \(\vecmag{(\vr_s \times \vr_t)(a,b)}\) will give the area of the the parallelogram along the tangent plane to the surface with edges \(\vr_s(a,b)\) and \(\vr_t(a,b)\text{.}\)
Example 13.10.10.
In this example, we will look at each of the properties above as they apply to the parameterization of the torus from Example 13.10.2. The points on the torus will be given by the parameterization:
\begin{align*}
\vr (s,t) =& \langle x(s,t),y(s,t),z(s,t)\rangle \\
=& \langle (b+a \cos(s)) \cos(t), (b+a \cos(s)) \sin(t), a \sin(s) \rangle
\end{align*}
for \(s,t \in [0,2 \pi)\text{.}\)
-
The traces with constant values of \(t\) are shown in magenta on Figure 13.10.11 and are parameterized by \(\vr(s,t_0)= \langle (b+a \cos(s)) \cos(t_0), (b+a \cos(s)) \sin(t_0), a \sin(s) \rangle\) when \(t=t_0\text{.}\)
-
The traces with constant values of \(s\) are shown in yellow on Figure 13.10.11 and are parameterized by \(\vr(s_0,t)= \langle (b+a \cos(s_0)) \cos(t), (b+a \cos(s_0)) \sin(t), a \sin(s_0) \rangle\) when \(s=s_0\text{.}\)
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The vector tangent to the surface along the trace with constant \(t\)-value is given by \(\vr_s=\langle (-a\sin(s))\cos(t), (-a\sin(s))\sin(t),0\rangle\) and is shown with the red arrow on Figure 13.10.11.
-
The vector tangent to the surface along the trace with constant \(s\)-value is given by \(\vr_t=\langle (a+b\cos(s))(-\sin(t)),(a+b\sin(s))(\cos(t)),0 \rangle\) and is shown with the yellow arrow on Figure 13.10.11.
-
The green vector is calculated by \(\vr_s \times \vr_t\text{,}\) is orthogonal to the surface at the highlighted point, and the length of the green vector corresponds to the area of the orange parallelogram. Note that as the length of the yellow vector increases, the parallelogram’s area and the length of the green vector both increase correspondingly.
Subsection 13.10.4 The Surface Area of Parametrically Defined Surfaces
In this subsection, we will use our new tool of parameterizing curved surfaces and a classic calculus approach to look at how to compute the surface area of a curved surface. For step one of our classic calculus approach, we will break our parametrized surface into pieces that correspond to a grid of \(s\) and \(t\) steps as shown in Figure 13.10.12. On each of these pieces of our \(st\)-grid, we will approximate the surface area of our curved surface with the area of a parallelogram tangent to the surface at that point. The sum of these parallelogram areas will give us the approximation we seek. For step two of our classic calculus approach, we will look at how smaller step sizes in \(s\) and \(t\) appears as part of the sum used in the approximation. The third step of our classic calculus approach will be to use a limit of the surface area approximation to get a Riemann sum definition for a double integral. We state the result as a Key Idea now and refer you to the proof for a thorough explanation of the tools involved.
An interactive plot of a curved surface shown in blue. A slider at the top of the figure allows the user to change the number of sections for the grid shown on the surface. At the top right of each grid piece along the surface a yellow parallelogram is drawn over each grid piece such that the parallelogram is tangent to the surface on the grid piece. The more curved the surface is on a grid piece, the larger the parallelogram.
Key Idea 13.10.13. Surface area.
Let \(\vr(s,t) = \langle x(s,t), y(s,t), z(s,t) \rangle\) be a parameterization of a smooth surface over a domain \(D\text{.}\) The area of the surface defined by \(\vr\) on \(D\) is given by
\begin{equation}
S = \iint_D \vecmag{\vr_s \times \vr_t } \ dA.\tag{13.10.2}
\end{equation}
Proof.
For this development, we will use a parametrized surface given by
\begin{equation*}
\vr(s,t) = x(s,t) \vi + y(s,t) \vj + z(s,t) \vk
\end{equation*}
with bounds on our parameters given by \(a \leq s \leq b\) and \(c \leq t \leq d\text{.}\) In this case, we are using a rectangular domain in the \(st\)-plane but in general, this same approach will work for non-rectangular domains. The partial derivatives of this parameterization,
\begin{align*}
\vr_s(s,t) \amp = x_s(s,t) \vi + y_s(s,t) \vj + z_s(s,t) \vk\\
\vr_t(s,t) \amp = x_t(s,t) \vi + y_t(s,t) \vj + z_t(s,t) \vk
\end{align*}
will give vectors tangent to the surface in the direction of changes in \(s\) and \(t\text{,}\) respectively.
We will break our section of the curved surface into steps in \(s\) and \(t\text{.}\) In particular, we partition the interval of \(s\)-values \([a,b]\) into \(m\) subintervals of length \(\Delta s = \frac{b-a}{m}\) and let \(s_0\text{,}\) \(s_1\text{,}\) \(\ldots\text{,}\) \(s_m\) be the endpoints of these subintervals, where \(a = s_0\lt s_1\lt s_2 \lt \cdots \lt s_m = b\text{.}\) We also partition the interval of \(t\)-values \([c,d]\) into \(n\) subintervals of equal length \(\Delta t = \frac{d-c}{n}\) and let \(t_0\text{,}\) \(t_1\text{,}\) \(\ldots\text{,}\) \(t_n\) be the endpoints of these subintervals, where \(c = t_0\lt t_1\lt t_2 \lt \cdots \lt t_n = d\text{.}\)
These subintervals partition the rectangle \(R = [a,b] \times [c,d]\) in \(st\)-coordinates into \(mn\) sub-rectangles \(R_{ij}\) with opposite vertices \((s_{i-1},t_{j-1})\) and \((s_i, t_j)\) for \(i\) between \(1\) and \(m\) and \(j\) between \(1\) and \(n\text{.}\) These rectangles all have equal area \(\Delta A = \Delta s \cdot \Delta t\text{.}\)
Each of these rectangles from the \(st\)-domain will split our curved surface into a \(m\) by \(n\) grid of pieces, but these pieces of the curved surface will not all be the same size (in surface area). Looking at the surface shown in Figure 13.10.9, we can see the traces given for five equally spaced values of \(s\) and \(t\) breaks our curved surface into a four by four grid of pieces. Note that these pieces of the curved surface do not have the same surface area. So we will need to estimate the surface area for each particular piece of our grid using our tools from the parameterization.
Each of the pieces of the curved surface correspond to increasing \(s\) by a small amount \(\Delta s\) or increasing \(t\) by a small amount \(\Delta t\) from the point \((s_{i-1},t_{j-1})\) in the \(st\)-parameter plane. We want to find the vectors corresponding to each of the sides of the orange parallelogram in Figure 13.10.9. In particular, the vector \(\vr_t \Delta t\) will be the vector that corresponds to the side of the parallelogram along a constant value of \(s\) (shown as a yellow vector in Figure 13.10.9) and the vector \(\vr_s \Delta s\) will be the vector that corresponds to the side of the parallelogram along a constant value of \(t\) (shown as a red vector in Figure 13.10.9).
The area of the parallelogram will then be \(S_{ij}=\vecmag{(\vr_s \Delta s) \times (\vr_t \Delta t)}\) evaluated at the point given by \((s_{i-1},t_{j-1})\text{.}\) This will be the approximation of the surface area for a piece of our grid of the curved surface. We can simplify this magnitude of a cross product to be \(S_{ij}=\vecmag{\vr_s \times \vr_t} \Delta s \Delta t\text{,}\) which means that our approximation over all of these pieces of the grid will sum to
\begin{equation}
SA\approx \sum_{i=1}^m \sum_{j=1}^n S_{ij}= \sum_{i=1}^m \sum_{j=1}^n \vecmag{\vr_s(s_{i-1},t_{j-1}) \times \vr_t(s_{i-1},t_{j-1})} \Delta s \Delta t\text{.}\tag{13.10.3}
\end{equation}
This has satisfied both the first and second steps of the classic calculus approach because we have approximated the surface area and quantified how that approximation will change with a smaller scale of steps (in \(s\) and \(t\)). You can use Figure 13.10.12 to geometrically see how increasing the number of pieces will give successively better approximations for the surface area of each piece.
Note that this approximation takes the form of a Riemann sum corresponding to the setup of a double integral. In particular, if we take the limit as \(\Delta s\) and \(\Delta t\) go to zero, our Riemann sum will be come a double integral of \(\vecmag{\vr_s \times \vr_t}\) over the rectangular region of integration with \(a \leq s \leq b\) and \(c \leq t \leq d\text{,}\) which gives the following formula for calculating the surface area.
We will now look at an example that shows how to use the parameterizations of the sphere from Activity 13.10.2 along with Key Idea 13.10.13 to calculate the surface area of sphere.
Example 13.10.14.
In this example, we will use our parameterizations for the sphere of radius \(R\) from Activity 13.10.2 to calculate the surface area of a sphere. Our first approach to parameterizing the sphere was to consider the sphere as a surface of revolution around the \(z\)-axis. This gave the parameterization of
\begin{equation*}
\vr(s,t) = \langle \sqrt{R^2-s^2}\cos(t), \sqrt{R^2-s^2}\sin(t), s \rangle
\end{equation*}
with \(s\) acting as the \(z\)-coordinate, going from \(R\) to \(-R\text{,}\) and \(t\) acting as the \(\theta\text{,}\) going from \(0\) to \(2\pi\text{.}\)
We can calculate \(\vr_s, \vr_t, \vecmag{\vr_s \times \vr_t}\) as follows:
\begin{align*}
\vr_s \amp= \left\langle \frac{-2s}{\sqrt{R^2-s^2}}\cos(t), \frac{-2s}{\sqrt{R^2-s^2}}\sin(t), 1 \right\rangle\\
\vr_t \amp= \left\langle -\sqrt{R^2-s^2}\sin(t), \sqrt{R^2-s^2}\cos(t), 0 \right\rangle\\
\vr_s \times \vr_t \amp= \left\langle -\sqrt{R^2-s^2} \cos(t), -\sqrt{R^2-s} \sin(t), -\frac{2s}{\sqrt{R^2-s^2}} \right\rangle \\
\vecmag{\vr_s \times \vr_t} \amp= \sqrt{(R^2-s^2) \cos^2(t)+(R^2-s) \sin^2(t)+ \frac{4s^2}{R^2-s^2}}
\end{align*}
Unfortunately, our expression for \(\vecmag{\vr_s \times \vr_t}\) will not simplify much more and we will need to use a trig substitution to compute the appropriate double integral. In particular, the surface area double integral will be
\begin{equation*}
\int_{-R}^R \int_0^{2\pi} \sqrt{(R^2-s^2)+ \frac{4s^2}{R^2-s^2}} dt \, ds \text{.}
\end{equation*}
We will omit the calculations that result from this and look at using our other parameterization to get a much easier calculation of the surface area of a sphere.
We can now use our parameterization of a sphere that came from using spherical coordinates.
\begin{equation*}
\vr(s,t) = \langle R \sin(s) \cos(t), R \sin(s)\sin(t), R\cos(s) \rangle
\end{equation*}
with \(s\) acting as the \(\phi\)-coordinate, going from \(0\) to \(2\pi\text{,}\) and \(t\) acting as the \(\theta\text{,}\) going from \(0\) to \(2\pi\text{.}\)
We can calculate \(\vr_s, \vr_t, \vecmag{\vr_s \times \vr_t}\) as follows:
\begin{align*}
\vr_s \amp= \langle R \cos(s) \cos(t), R \cos(s)\sin(t), -R\sin(s) \rangle\\
\vr_t \amp= \langle -R \sin(s) \sin(t), R \sin(s) \cos(t), 0 \rangle\\
\vr_s \times \vr_t \amp= \langle R^2 \sin^2(s) \cos(t), R^2 \sin^2(s) \sin(t), R^2 \cos^2(t) \cos(s)\sin(s)+R^2 \sin^2(t) \cos(s)\sin(s) \rangle \\
\amp= R^2\sin(s) \langle \sin(s) \cos(t), \sin(s) \sin(t), \cos(s) \rangle \\
\vecmag{\vr_s \times \vr_t} \amp= R^2\sin(s)
\end{align*}
You may recognize this expression from our work on the volume element in spherical coordinates, Key Idea 12.8.5. In both this case and the volume element of spherical coordinates, we are measuring how warped our space is based on transformations using spherical coordinates.
Our double integral will then be
\begin{equation*}
\int_{0}^\pi \int_0^{2\pi} R^2\sin(s) dt \, ds
\end{equation*}
which will evaluate to
\begin{equation*}
R^2 (2 \pi) \left[-\cos(s)\restrict{s=0}{s=\pi}\right]= R^2 (2 \pi)(2)=4\pi R^2
\end{equation*}
You may recognize this result as the formula you were likely handed (without explanation) for the surface area of a sphere.
Activity 13.10.3.
In this activity, we will compute the surface area of a right circular cylinder. In particular, we will consider the cylinder with radius \(a\) and height \(h\) defined parametrically by
\begin{equation*}
\vr(s,t) = a\cos(s) \vi + a\sin(s) \vj + t \vk
\end{equation*}
A cylinder.
(a)
Calculate \(\vr_s, \vr_t, \vecmag{\vr_s \times \vr_t}\) based on the parameterization given above.
(b)
Use the calculations from the previous task to set up an iterated integral to determine the surface area of this cylinder.
(c)
Evaluate your iterated integral from the previous task.
(d)
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 result the value from the previous task.
As we noted earlier, we can take any surface \(z = f(x,y)\) and generate a corresponding parameterization for the surface by writing \(\langle s, t, f(s,t) \rangle\text{.}\) Hence, we can use our recent work with parametrically defined surfaces to find the surface area that is generated by a function \(f = f(x,y)\) over a given domain.
Activity 13.10.4.
Let \(z = f(x,y)\) define a smooth surface, and consider the corresponding parameterization \(\vr(s,t) = \langle s, t, f(s,t) \rangle\text{.}\)
(a)
Let \(D\) be a region in the domain of \(f\text{.}\) Using Equation (13.10.2), show that the area, \(S\text{,}\) of the surface defined by the graph of \(f\) over \(D\) is
\begin{equation*}
S = \iint_D \sqrt{\left(f_x(x,y)\right)^2 + \left(f_y(x,y)\right)^2 + 1} \ dA.
\end{equation*}
(b)
Use the formula developed in (a) to calculate the area of the surface defined by \(f(x,y) = \sqrt{4-x^2}\) over the rectangle \(D = [-2,2] \times [0,3]\text{.}\)
(c)
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).
Summary
-
A parameterization of a curve describes the coordinates of a point on the curve in terms of a single parameter \(t\text{,}\) while a parameterization of a surface describes the coordinates of points on the surface in terms of two independent parameters.
-
If \(\vr(s,t) = \langle x(s,t), y(s,t), z(s,t) \rangle\) describes a smooth surface in 3-space on a domain \(D\text{,}\) then the area, \(S\text{,}\) of that surface is given by\begin{equation*} S = \iint_D \vecmag{\vr_s \times \vr_t} \ dA. \end{equation*}
Exercises 13.10.5 Exercises
1.
Consider the cone shown below.
figure of a cone with circular base on the xy-plane, centered on the z-axis, and point on the positive z-axis.
If the height of the cone is 6 and the base radius is 5, write a parameterization of the cone in terms of \(r = s\) and \(\theta = t\text{.}\)
\(x(s,t) =\) ,
\(y(s,t) =\) , and
\(z(s,t) =\) , with
\(\le s\le\) and
\(\le t\le\) .
2.
Parameterize a vase formed by rotating the curve \(z= 6 \sqrt{x-2},\,2\leq x\leq 4\text{,}\) around the \(z\)-axis. Use \(s\) and \(t\) for your parameters.
\(x(s,t) =\) ,
\(y(s,t) =\) , and
\(z(s,t) =\) , with
\(\le s\le\) and
\(\le t\le\)
3.
Find parametric equations for the sphere centered at the origin and with radius 7. Use the parameters \(s\) and \(t\) in your answer.
\(x(s,t) =\) ,
\(y(s,t) =\) , and
\(z(s,t) =\) , where
\(\le s\le\) and
\(\le t\le\) .
4.
Match the parametric equations with the verbal descriptions of the surfaces by putting the letter of the verbal description to the left of the letter of the parametric equation.
-
\(\displaystyle \mathbf{r} \left( u, v \right) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^{2} \mathbf{k}\)
-
\(\displaystyle \mathbf{r} \left( u, v \right) = u \mathbf{i} + \cos v \mathbf{j} + \sin v \mathbf{k}\)
-
\(\displaystyle \mathbf{r} \left( u, v \right) = u \mathbf{i} + u \cos v \mathbf{j} + u \sin v \mathbf{k}\)
-
\(\displaystyle \mathbf{r} \left( u, v \right) = u \mathbf{i} + v \mathbf{j} + \left( 2u - 3v \right) \mathbf{k}\)
-
circular paraboloid
-
circular cylinder
-
cone
-
plane
5.
Parameterize the plane that contains the three points
(4,-3,4), (-2,-6,-6), (5,35,10).
Use s and t for the parameters in your parameterization and enter your formula as a single vector, with angle brackets, for example, \(\displaystyle{\langle 1 + s + t, s - t, 3 - t \rangle}\text{.}\) Each component should be a function of the form a+bs+ct where a,b,c are appropriate real numbers.
\(\vec{r}(s,t)\) =
6.
Find the surface area of that part of the plane \(10 x + 9 y + z = 4\) that lies inside the elliptic cylinder \(\frac{x^2}{16} + \frac{y^2}{9} =1\)
Surface Area =
7.
Find the surface area of the part of the circular paraboloid \(z = x^{2} + y^{2}\) that lies inside the cylinder \(x^{2} + y^{2} = 1\text{.}\)
8.
Find the surface area of the part of the plane \(1 x + 1 y + z = 1\) that lies inside the cylinder \(x^{2} + y^{2} = 9\text{.}\)
9.
Write down the iterated integral which expresses the surface area of \(z = y^{3}\cos^{7}x\) over the triangle with vertices (-1,1), (1,1), (0,2):
\begin{equation*}
\int_a^b\int_{f(y)}^{g(y)} \sqrt{h(x,y)}\,dx dy
\end{equation*}
\(a =\)
\(b =\)
\(f(y) =\)
\(g(y) =\)
\(h(x,y) =\)
10.
A decorative oak post is 48 inches long and is turned on a lathe so that its profile is sinusoidal as shown in the figure below.
figure of a column with a vertically varying outside radius. the maximum radius is r0 inches, the vertical distance between maximum radii is a0 inches, and the difference between the minimum and maximum radii is 2 inches.
In this figure, \(r_0 = 4\) inches and \(a_0 = 8\) inches.
(a) Describe the surface of the post parametrically using cylindrical coordinates and the parameters \(s\) and \(t\text{.}\)
\(x(s,t) =\) ,
\(y(s,t) =\) , and
\(z(s,t) =\) , where
\(\le s\le\) and
\(\le t\le\) .
(b) Find the volume of the post.
volume =
(Include .)
11.
Find the surface area of the part of the sphere \(x^{2} + y^{2} + z^{2} = 25\) that lies above the cone \(z = \sqrt{x^{2} + y^{2}}\)
12.
If a parametric surface given by \(\mathbf{r_{1}}(u, v) = f(u, v)\mathbf{i} + g(u, v)\mathbf{j} +
h(u, v)\mathbf{k}\) and \(-4 \leq u \leq 4, -4 \leq v \leq 4\text{,}\) has surface area equal to 4, what is the surface area of the parametric surface given by \(\mathbf{r_{2}}(u, v) = 3\mathbf{r_{1}}(u, v)\) with \(-4 \leq u \leq 4, -4 \leq v \leq 4\text{?}\)
13.
Write down the iterated integral which expresses the surface area of \(z = y^{6}\cos^{4}x\) over the triangle with vertices (-1,1), (1,1), (0,2):
\begin{equation*}
\int_a^b\int_{f(y)}^{g(y)} \sqrt{h(x,y)}\,dx dy
\end{equation*}
\(a =\)
\(b =\)
\(f(y) =\)
\(g(y) =\)
\(h(x,y) =\)
14.
Find the area of the portion of the sphere of radius 9 (centered at the origin) that is in the cone \(z > \sqrt{x^2 + y^2}\text{.}\)
15.
Find the surface area of the part of the plane \(5 x + 4 y + z = 1\) that lies inside the cylinder \(x^{2} + y^{2} = 16\text{.}\)
16.
Find the surface area of the part of the circular paraboloid \(z = x^{2} + y^{2}\) that lies inside the cylinder \(x^{2} + y^{2} = 1\text{.}\)
17.
Find the area of the surface obtained by rotating the curve
\begin{equation*}
y = 2x^{3}
\end{equation*}
from \(x = 0\) to \(x = 1\) about the \(x\)-axis.
The area is square units.
18.
The vector equation \(\mathbf{r} \left( u, v \right) = u \cos v \mathbf{i} + u \sin v
\mathbf{j} + v \mathbf{k}\text{,}\) \(0 \leq v \leq 3 \pi\text{,}\) \(0 \leq u \leq 1\text{,}\) describes a helicoid (spiral ramp). What is the surface area?
19.
Find the surface area of the portion \(S\) of the cone \(z^2=x^2+y^2\text{,}\) where \(z\ge 0\text{,}\)
contained within the cylinder \(y^2+z^2\le 25\text{.}\)
\(\mathrm{Area}(S)=\)
20.
A torus of radius 6 (and cross-sectional radius 1) can be represented parametrically by the function \(\mathbf{r}: D \to \mathbb{R}^3\text{:}\)
\begin{equation*}
\mathbf{r}(\theta, \phi) = ((6 + \cos\phi)\cos\theta,(6 +
\cos\phi)\sin\theta, \sin\phi)
\end{equation*}
where D is the rectangle given by \(0\le \theta \le 2\pi, \ 0\le \phi \le 2\pi\text{.}\)
The surface area of the torus is
21.
Find an equation of the tangent plane (in the variables x, y and z) to the parametric surface
\(\mathbf{r}(u,v) = \langle u, u^2 + 5v, -5v^2 \rangle\) at the point \((-2, 4, 0)\text{.}\)
22.
Calculate \({\mathbf{T}}_u\text{,}\) \({\mathbf{T}}_v\text{,}\) and \(\mathbf{n}(u,v)\) for the parametrized surface at the given point.
Then find the equation of the tangent plane to the surface at that point.
\(\Phi(u,v)= (2u+v,u-4v,8 u)\text{;}\)\(\qquad u=4\text{,}\quad v=9\)
\({\mathbf{T}}_u=\) , \({\mathbf{T}}_v=\) , \(\mathbf{n}(u,v)=\)
The tangent plane:
\(=9z\)
23.
Consider the ellipsoid given by the equation
\begin{equation*}
\frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} = 1.
\end{equation*}
In Activity 13.10.2, we found that a parameterization of the sphere \(S\) of radius \(R\) centered at the origin is
\begin{equation*}
x(r,s) = R\cos(s) \cos(t), \ y(s,t) = R \cos(s) \sin(t), \ \text{ and } \ z(s,t) = R\sin(s)
\end{equation*}
for \(-\frac{\pi}{2} \leq s \leq \frac{\pi}{2}\) and \(0 \leq t \leq 2\pi\text{.}\)
-
Let \((x,y,z)\) be a point on the ellipsoid and let \(X = \frac{x}{4}\text{,}\) \(Y = \frac{y}{5}\text{,}\) and \(Z = \frac{z}{3}\text{.}\) Show that \((X,Y,Z)\) lies on the sphere \(S\text{.}\) Hence, find a parameterization of \(S\) in terms of \(X\text{,}\) \(Y\text{,}\) and \(Z\) as functions of \(s\) and \(t\text{.}\)
-
Use the result of part (a) to find a parameterization of the ellipse in terms of \(x\text{,}\) \(y\text{,}\) and \(z\) as functions of \(s\) and \(t\text{.}\) Check your parametrization by substituting \(x\text{,}\) \(y\text{,}\) and \(z\) into the equation of the ellipsoid. Then check your work by plotting the surface defined by your parameterization.
24.
In this exercise, we explore how to use a parametrization and iterated integral to determine the surface area of a sphere.
-
Set up an iterated integral whose value is the portion of the surface area of a sphere of radius \(R\) that lies in the first octant (see the parameterization you developed in Activity 13.10.2).
-
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.
25.
Consider the plane generated by \(z = f(x,y) = 24 - 2x - 3y\) over the region \(D = [0,2]\times[0,3]\text{.}\)
-
Sketch a picture of the overall solid generated by the plane over the given domain.
-
Determine a parameterization \(\vr(s,t)\) for the plane over the domain \(D\text{.}\)
-
Use Equation (13.10.2) to determine the surface area generated by \(f\) over the domain \(D\text{.}\)
-
Observe that the vector \(\vu = \langle 2, 0, -4 \rangle\) points from \((0,0,24)\) to \((2,0,20)\) along one side of the surface generated by the plane \(f\) over \(D\text{.}\) Find the vector \(\vv\) such that \(\vu\) and \(\vv\) together span the parallelogram that represents the surface defined by \(f\) over \(D\text{,}\) and hence compute \(| \vu \times \vv |\text{.}\) What do you observe about the value you find?
26.
A cone with base radius \(a\) and height \(h\) can be realized as the surface defined by \(z = \frac{h}{a} \sqrt{x^2+y^2}\text{,}\) where \(a\) and \(h\) are positive.
-
Find a parameterization of the cone described by \(z = \frac{h}{a} \sqrt{x^2+y^2}\text{.}\) (Hint: Compare to the parameterization of a cylinder as seen in Activity 13.10.3.)
-
Set up an iterated integral to determine the surface area of this cone.
-
Evaluate the iterated integral to find a formula for the lateral surface area of a cone of height \(h\) and base \(a\text{.}\)
27.
Parameterize the following surfaces and be sure to state bounds for your parameterization.
(a)
The first octant portion of \(2x+y+2z=4\)
(b)
\(x^2+y^2+z^2=1\)
(c)
\(\left(\frac{x-2}{3}\right)^2+\left(\frac{y+2}{1}\right)^2+\left(\frac{z-1}{5}\right)^2=1\)
(d)
\(x^2+y^2-z^2=1\)
(e)
\(\left(\frac{x-2}{3}\right)^2-\left(\frac{y+2}{1}\right)^2-\left(\frac{z-1}{5}\right)^2=1\)
(f)
\(\left(\frac{x-2}{3}\right)^2-\left(\frac{y+2}{1}\right)^2+\left(\frac{z-1}{5}\right)^2=1\)
(g)
\(x^2-y^2-z^2=1\)
(h)
\(\left(\frac{x-2}{3}\right)^2-\left(\frac{y+2}{1}\right)^2-\left(\frac{z-1}{5}\right)^2=1\)
(i)
\(z=\left(\frac{x}{2}\right)^2-\left(\frac{y}{3}\right)^2\)
