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Parameterizations of Surfaces and Surface Area
In the first edition, this section was in the chapter on multiple integration, but it felt like a bit of an afterthought until it was used after surface integrals were defined in the vector calculus chapter.
Parameterizations of Surfaces and Surface Area
In the first edition, this section was in the chapter on multiple integration, but it felt like a bit of an afterthought until it was used after surface integrals were defined in the vector calculus chapter. The second edition moves this material to a natural place in the flow of the text overall. Parameterizations of surfaces is a required topic for the remainder of this chapter. While surface area can be treated lightly, the ideas of approximating patches of a surface by parallelograms with sides determined by tangent vectors to the parametric coordinate traces will be used again when studying flux integrals.
Introduction
In and , we used parameterizations of curves as a means to use the many tools of single-variable calculus to analyze the curve. One of the central goals of this section is to extend the idea of parameterizations to surfaces in an attempt to apply the many multivariable tools of and . In the preview activity, we look at how geometric knowledge of right circular cylinders and cones allows us to create a vector-valued function that describes the points on the surface of a cone.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Parameterizations of Surfaces
concluded with a description of a cone in terms of measurements corresponding to the angle around the \(z\)-axis and the position relative to the \(xy\)-plane. Geometric intuition described the coordinate relationships of the two surfaces in . This subsection discusses how to use algebraic and geometric tools to find parameterizations of curved surfaces. The next subsection describes some important geometric measurements of a surface that can be done using a parameterization. The section concludes by examining how parameterizations can be used to compute the surface area of a surface in three dimensions.
In a single-variable setting, any function may have its graph expressed parametrically. For instance, the graph of \(y = g(x)\) can be parameterized as \(\langle t, g(t) \rangle\), where \(t\) belongs to the domain of \(g\). Certain curves that are not expressible with \(y\) as a function of \(x\) can also be represented parametrically. For instance, the unit circle cannot be represented with either the \(x\)- or \(y\)-coordinate written as a function of the other. However, it can be parameterized as \(\langle \cos(t), \sin(t) \rangle\), where \(0 \leq t \lt 2\pi\).
A surface of the form \(z = f(x,y)\) can be expressed parametrically by \[\begin{aligned}\end{aligned}\] where \((s,t)\) varies over the entire domain of \(f\). In the parameterization above, we can think of \(s\) as acting like \(x\) and \(t\) as acting like \(y\). Therefore, any familiar surface expressed as \(z=f(x,y)\) can be written as a parametric surface. The greater power of parameterizations is realized when dealing with surfaces such as a sphere that cannot be expressed by a single function \(z = f(x,y)\) but can be represented parametrically.
For surfaces such as ellipsoids or hyperboloids where one coordinate cannot be expressed as a function of the other two, a good 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 , the parameters measured the rotational coordinate around the \(z\)-axis and 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 using this same strategy.
The next activity finds two ways to parameterize a sphere centered at the origin. The first views the sphere as a surface of revolution around the \(z\)-axis, similar to the preceding example. The second uses spherical coordinates.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
The Geometry of a Parametric Surface
The parameterization of a surface allows us to access important geometric information about the surface. In this subsection, we consider a surface \(S_1\), 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 \(c \leq t \leq d\).
In , we considered a function \(f(x,y)\) and saw how holding one of the input coordinates constant allowed us to restrict the graph of \(z=f(x,y)\) to a single-variable slice, which we called the trace. For example, we looked at the trace along \(x=a\), which is given by \(z=f(a,y)\). Refer to for an illustration of this. Focusing on a single slice allowed us to define partial derivatives and many other tools in .
What happens when we hold a parameter for a parametric surface constant and allow the other to vary? This gives a slice of the parametric surface, but the slice is not necessarily in the direction of one of the coordinates (\(x\), \(y\), or \(z\)). Instead, the slice describes the surface in a constant direction of one parameters, \(s\) or \(t\). The trace given by holding \(s\) constant yields \(\vr(a,t)\), a curve parameterized as a vector-valued function of one variable. This allows the use of the tools from to understand this curve. Specifically, the derivative, \[\begin{aligned}\end{aligned}\] gives a vector that is tangent to the curve given by \(\vr(a,t)\). Similarly, we can look at the trace generated in the direction of a constant value of \(t\). This gives a trace parameterized as \(\vr(s,b)\) with a tangent vector \(\frac{d\vr}{ds}(s,b)\).
shows plot of a 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. Use the sliders at the top of to change the location at which the tangent vectors are shown.
The red and yellow tangent vectors at the highlighted point allow us to describe the tangent plane at that point. A piece of the tangent plane is shown as an orange parallelogram. Note that the orange parallelogram moves and tilts to approximate the curved surface near the highlighted point. Using the properties of the cross product allows us both to compute a vector that is orthogonal to the curved surface at the highlighted point and to find the area of the orange parallelogram. (Refer to and for the original introduction of these ideas.)
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
The Surface Area of Parametrically Defined Surfaces
We are now ready to use the classic calculus approach to compute the surface area of a surface. For step one of the classic calculus approach, we break the parametrized surface into pieces corresponding to a grid of \(s\)- and \(t\)-steps as shown in . On each of these pieces of the \(st\)-grid, we approximate the surface with a parallelogram tangent to the surface at that point. The sum of the areas of these parallelograms gives an approximation of the area of the surface. In step two of the classic calculus approach, we consider smaller step sizes in \(s\) and \(t\) as part of the sum used in the approximation. The third step of the classic calculus approach is a limit of the surface area approximation which leads to a double Riemann sum that corresponds to a double integral.
Our next example shows how to calculate the surface area of sphere using applied to the parameterizations of the sphere from .
Activity
This activity computes the surface area of a right circular cylinder. In particular, consider the cylinder with radius \(a\) and height \(h\) defined parametrically by \[\begin{aligned}\end{aligned}\] for \(0 \leq s \leq 2\pi\) and \(0 \leq t \leq h\), as shown in Figure.
Calculate \(\vr_s, \vr_t, \vecmag{\vr_s \times \vr_t}\) based on the parameterization given above.
Use the calculations from the previous part to set up an iterated integral to determine the surface area of this cylinder.
Evaluate your iterated integral from the previous part.
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 part.
As we noted earlier, we can take any surface \(z = f(x,y)\) and obtain a parameterization for the surface by writing \(\langle s, t, f(s,t) \rangle\). Hence, we can use to find the area of a surface \(z = f(x,y)\) over a given domain.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (4)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Consider the ellipsoid given by the equation \[\begin{aligned}\end{aligned}\]
In Activity, we found that a parameterization of the sphere \(S\) of radius \(R\) centered at the origin is \[\begin{aligned}\end{aligned}\] for \(-\frac{\pi}{2} \leq s \leq \frac{\pi}{2}\) and \(0 \leq t \leq 2\pi\).
Let \((x,y,z)\) be a point on the ellipsoid and let \(X = \frac{x}{4}\), \(Y = \frac{y}{5}\), and \(Z = \frac{z}{3}\). Show that \((X,Y,Z)\) lies on the sphere \(S\). Hence, find a parameterization of \(S\) in terms of \(X\), \(Y\), and \(Z\) as functions of \(s\) and \(t\).
Use the result of part (a) to find a parameterization of the ellipse in terms of \(x\), \(y\), and \(z\) as functions of \(s\) and \(t\). Check your parametrization by substituting \(x\), \(y\), and \(z\) into the equation of the ellipsoid. Then check your work by plotting the surface defined by your parameterization.
Αποκάλυψέ την.
If \((x,y,z)\) lies on the ellipse and \(X = \frac{x}{4}\), \(Y = \frac{y}{5}\), and \(Z = \frac{z}{3}\), then \[\begin{aligned}X^2+Y^2+Z^2 \amp = \left(\frac{x}{4}\right)^2 + \left(\frac{y}{5}\right)^2 + \left(\frac{z}{3}\right)^2 \\ \amp = \frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} \\ \amp = 1\end{aligned}\]. So \((X,Y,Z)\) lies on the sphere \(S\) and \(S\) has parameterization \[\begin{aligned}\end{aligned}\] for \(-\frac{\pi}{2} \leq s \leq \frac{\pi}{2}\) and \(0 \leq t \leq 2\pi\).
Since \(x = 4X\), \(y = 5Y\), and \(z = 3Z\) we have \[\begin{aligned}\end{aligned}\] for \(-\frac{\pi}{2} \leq s \leq \frac{\pi}{2}\) and \(0 \leq t \leq 2\pi\). Note that \[\begin{aligned}\frac{x^2}{16} + \frac{y^2}{25} + \frac{z^2}{9} \amp = \cos^2(s)\cos^2(t) + \cos^2(s) \sin^2(t) + \sin^2(s) \\ \amp = \cos^2(s)[\cos^2(t) + \sin^2(t)] + \sin^2(s) \\ \amp = \cos^2(s) + \sin^2(s) \\ \amp = 1\end{aligned}\]. So \(x(r,s) = 4\cos(s) \cos(t)\), \(y(s,t) = 5\cos(s) \sin(t)\), and \(z(s,t) = 3sin(s)\) is a parametrization of the ellipse.
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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).
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.
Αποκάλυψέ την.
From Activity we know that a parameterization of a sphere of radius \(R\) centered at the origin is \[\begin{aligned}\end{aligned}\] for \(-\frac{\pi}{2} \leq s \leq \frac{\pi}{2}\) and \(t\) in \([0, 2 \pi]\). By symmetry, we can just calculate the surface area of the sphere in the first octant, then multiply by 8. So we can assume \(0 \leq s, t \leq \frac{\pi}{2}\).
With \(\vr(s,t) = \langle R \cos(s) \cos(t), R\sin(s) \cos(t), R \sin(t) \rangle\) we have \[\begin{aligned}\vr_s(s,t) \amp = \langle -R\sin(s) \cos(t), R\cos(s)\cos(t), 0 \rangle, \\ \vr_t(s,t) \amp = \langle -R\cos(s) \sin(t), -R\sin(s)\sin(t), -R\cos(t) \rangle\end{aligned}\], and so \[\begin{aligned}\lvert \vr_s(s,t) \times \vr_t(s,t) \rvert \amp = \lvert \langle -R\sin(s) \cos(t), R\cos(s)\cos(t), 0 \rangle \times \langle -R\cos(s) \sin(t), -R\sin(s)\sin(t), -R\cos(t) \rangle \rvert \\ \amp = \lvert \langle -R^2\cos(s)\cos^2(t), -R^2\sin(s)\cos^2(t), R^2\cos(t)\sin(t) \rangle \rvert \\ \amp = \sqrt{R^4\cos^2(s)\cos^4(t) + R^4\sin^2(s)\cos^4(t) + R^4\cos^2(t)\sin^2(t)} \\ \amp = R^2\sqrt{\cos^2(t)} \\ \amp = R^2\cos(t)\end{aligned}\]. Therefore, the surface area of the first octant portion of the sphere is \[\begin{aligned}\int_0^{\pi/2} \int_0^{\pi/2} R^2 \cos(t) \, dt \, ds \amp = 8R^2 \int_0^{\pi/2} \sin(t) \restrict{0}{\pi/2} \, ds \\ \amp = R^2 \int_0^{\pi/2} 1 \, ds \\ \amp = \frac{1}{2}\pi R^2\end{aligned}\].
We multiply the previous result by 8 to reproduce the surface area calculation in each of the eight octants. So the surface area of a sphere of radius \(R\) is \(4 \pi R^2\).
The result of part (c) is the standard formula for the surface area of a sphere of radius \(R\).
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Consider the plane generated by \(z = f(x,y) = 24 - 2x - 3y\) over the region \(D = [0,2]\times[0,3]\).
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 \(D\).
Use Equation to determine the surface area generated by \(f\) over the domain \(D\).
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 \(D\). Find the vector \(\vv\) such that \(\vu\) and \(\vv\) together span the parallelogram that represents the surface defined by \(f\) over \(D\), and hence compute \(| \vu \times \vv |\). What do you observe about the value you find?
Αποκάλυψέ την.
The solid looks like a box with a slanted top.
If we let \(x(s,t) = s\) and \(y(s,t) = t\), then \(z(s,t) = 24-2s-3t\) provides a parameterization of the plane. So we can let \(\vr(s,t) = s \vi + t \vj + (24-2s-3t) \vk\).
In our situation we have \(\vr_s(s,t) = \vi - 2 \vk\) and \(\vr_t(s,t) = \vj - 3 \vk\). So \(\vr_s(s,t) \times \vr_t(s,t) = 2 \vi + 3\vj + \vk\). So the surface area generated by \(f\) over the domain \(D\) is found by \[\begin{aligned}\int_0^{3} \int_0^{2} | \langle 2,3,1 \rangle | \, dt \, ds \amp = \sqrt{14} \int_0^{3} \int_0^{2} \, dt \, ds \\ \amp = 6 \sqrt{14}\end{aligned}\].
Take \(\vv = \langle 0, 2, -9 \rangle\) to be the vector that points from \((0,0,24)\) to \((0,3,15)\) along one side of the surface generated by the plane \(f\) over \(D\). Then \(\vu\) and \(\vv\) together span the parallelogram that represents the surface defined by \(f\) over \(D\). Here we have \[\begin{aligned}\end{aligned}\]. Since \(| \vu \times \vv |\) is the area of the parallelogram determined by \(\vu\) and \(\vv\), we should expect this area to be the same as the surface area of the surface defined by the plane \(f\) over \(D\).
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A cone with base radius \(a\) and height \(h\) can be realized as the surface defined by \(z = \frac{h}{a} \sqrt{x^2+y^2}\), where \(a\) and \(h\) are positive.
Find a parameterization of the cone described by \(z = \frac{h}{a} \sqrt{x^2+y^2}\). (Hint: Compare to the parameterization of a cylinder as seen in Activity.)
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 \(a\).
Αποκάλυψέ την.
Let \(t\) run along the positive \(z\)-axis from 0 to \(h\). By similar triangles, the cross section of the cone at height \(t\) parallel to the \(x\)-\(y\) plane has radius \(\frac{h}{a}t\). We can think of the cone as made of circles with these varying radii, so a parameterization of the cone is \[\begin{aligned}\end{aligned}\], for \(0 \leq s \leq 2 \pi\) and \(0 \leq t \leq a\).
We have \[\begin{aligned}\end{aligned}\], and \[\begin{aligned}\end{aligned}\]. By symmetry, we calculate the area of the cone in the first octant and multiply by 4 to obtain the area of the surface of the cone as \[\begin{aligned}\end{aligned}\].
Evaluating the iterated integral yields \[\begin{aligned}4\int_0^{\pi/2} \int_0^a \frac{a}{h^2} \sqrt{h^2+a^2} t \, dt \, ds \amp = 4\frac{a}{h^2} \sqrt{h^2+a^2} \int_0^{\pi/2} \frac{1}{2}t^2 \restrict{0}{h} \, ds \\ \amp = 2 a \sqrt{h^2+a^2} \int_{0}^{\pi/2} \, ds \\ \amp = \pi a \sqrt{h^2+a^2}\end{aligned}\].
Symbols used here
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
Ratios of sides in a right triangle; coordinates on the unit circle.
Instantaneous rate of change; slope of the graph.
Inequalities that allow equality; < and > exclude it.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Parameterizations of Surfaces and Surface Area
- What does it mean to parameterize a surface?
- How can parameterizations of curved surfaces be used to compute surface area?
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Δοκίμασε μόνος σου.
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.
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