Parametric Surfaces

Learning Objectives

Find the parametric representations of a cylinder, a cone, and a sphere.
Describe the surface integral of a scalar-valued function over a parametric surface.

A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.

However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve $C$ , we first need to parameterize $C$ . In a similar way, to calculate a surface integral over surface $S$ , we need to parameterize $S$ . That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve.

A parameterized surface is given by a description of the form

$\large{{\bf{r}}(u,v)=\langle{x}(u,v),y(u,v),z(u,v)\rangle}$ .

Notice that this parameterization involves two parameters, $u$ and $v$ , because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. The parameters $u$ and $v$ vary over a region called the parameter domain, or parameter space—the set of points in the $uv$ -plane that can be substituted into ${\bf{r}}$ . Each choice of $u$ and $v$ in the parameter domain gives a point on the surface, just as each choice of a parameter $t$ gives a point on a parameterized curve. The entire surface is created by making all possible choices of $u$ and $v$ over the parameter domain.

definition

Given a parameterization of surface ${\bf{r}}(u,v)=\langle{x}(u,v),y(u,v),z(u,v)\rangle$ , the parameter domain of the parameterization is the set of points in the $uv$ -plane that can be substituted into ${\bf{r}}$ .

Example: parameterizing a cylinder

Describe the surface $S$ parameterized by

[latex]\large{{\bf{r}}(u,v)=\langle\cos{u},\sin{u},v\rangle,-\infty

Show Solution

Solution

To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of $\mathbb{R}^2$ , we can choose any value for $u$ and $v$ and plot the corresponding point. If $u=v=0$ , then ${\bf{r}}(0,0)\langle1,0,0\rangle$ , so point $(1, 0, 0)$ is on $S$ . Similarly, points ${\bf{r}}(\pi,2)=\langle-1,0,2\rangle$ and ${\bf{r}}\left(\frac{\pi}2,4\right)=(0,1,4)$ are on $S$ .

Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize $S$ , we visualize two families of curves that lie on $S$ . In the first family of curves we hold $u$ constant; in the second family of curves we hold $v$ constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.

First, suppose that $u$ is a constant $K$ . Then the curve traced out by the parameterization is $\langle\cos{K},\sin{K},v\rangle$ , which gives a vertical line that goes through point $(\cos{K},\sin{K},v)$ in the $xy$ -plane.

Now suppose that $v$ is a constant $K$ . Then the curve traced out by the parameterization is $\langle\cos{K},\sin{K},v\rangle$ , which gives a circle in plane $z=K$ with radius 1 and center $(0, 0, K)$ .

If $u$ is held constant, then we get vertical lines; if $v$ is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder $x^{2}+y^{2}=1$ (Figure 1).

<img src="/apps/archive/20220422.171947/resources/8222eebba73b30b683ca0265ea289ae54c5f3dd1" data-media-type="image/jpeg" alt="Three diagrams in three dimensions. The first shows vertical lines around the origin. The second shows parallel circles all with center at the origin and radius of 1. The third shows the lines and circle. Together, they form the skeleton of a cylinder." id="4">

Figure 1. (a) Lines $\langle\cos{K},\sin{K},v\rangle$ for $K = 0, \frac{\pi}2, π,$ and $\frac{3\pi}2$ . (b) Circles $\langle\cos{u},\sin{u},K\rangle$ for $K = −2, −1, 1,$ and $2$ . (c) The lines and circles together. As $u$ and $v$ vary, they describe a cylinder.

Notice that if $x=\cos{u}$ and $y=\sin{u}$ , then $x^{2}+y^{2}=1$ , so points from $S$ do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle $\langle\cos{K},\sin{K},v\rangle$ for some $k$ , and therefore each point on the cylinder is contained in the parameterized surface (Figure 2).

<img src="/apps/archive/20220422.171947/resources/7f7bfc8f9889a6e8ebdf1cef0b21bbc3d9f0d8d6" data-media-type="image/jpeg" alt="An image of a vertical cylinder in three dimensions with the center of its circular base located on the z axis." id="5">

Figure 2. Cylinder $x^{2}+y^{2}=r^{2}$ has parameterization ${\bf{r}}(u,v)=\langle{r}\cos{u},r\sin{u},v\rangle$ , $0\leq{u}\leq2\pi,-\infty<v<\infty$ .

Analysis

Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to [latex]0\leq{u}\leq\pi,0

try it

Describe the surface with parameterization [latex]{\bf{r}}(u,v)=\langle2\cos{u},2\sin{u},v\rangle,0\leq{u}\leq2\pi,-\inftyShow Solution

Cylinder $x^{2}+y^{2}=4$ .

It follows from Example “Parameterizing a Cylinder” that we can parameterize all cylinders of the form $x^{2}+y^{2}=R^{2}$ . If $S$ is a cylinder given by equation $x^{2}+y^{2}=R^{2}$ , then a parameterization of $S$ is

[latex]\large{{\bf{r}}(u,v)=\langle{R}\cos{u},R\sin{u},v\rangle,0\leq{u}\leq2\pi,-\infty

We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

Example: describing a surface

Describe the surface $S$ parameterized by ${\bf{r}}(u,v)=\langle{u}\cos{v},u\sin{v},u^2\rangle,0\le{u}\le\infty,0\le{v}<2\pi$ .

Show Solution

Notice that if $u$ is held constant, then the resulting curve is a circle of radius $u$ in plane $z=u$ . Therefore, as $u$ increases, the radius of the resulting circle increases. If $v$ is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

$\begin{aligned} x^2+y^2&=(u\cos{v})^2+(u\sin{v})^2 \\ &=u^2\cos^2v+u^2\sin^2v \\ &=u^2 \\ &=z \end{aligned}$ .

Therefore, the surface is elliptic paraboloid $x^2+y^2=z$ (Figure 3).

<img src="/apps/archive/20220422.171947/resources/486dd2670ab6d6da4a12308eb513af5b5e439be2" data-media-type="image/jpeg" alt="Two images in three dimensions. The first shows parallel circles on the z axis with radii increasing as z increases. Vertical parabolas opening up frame the circles, forming the skeleton of a paraboloid. The second shows the elliptic paraboloid, which is made of all the possible circles and vertical parabolas in the parameter domain." id="9">

Figure 3. (a) Circles arise from holding $u$ constant; the vertical parabolas arise from holding $v$ constant. (b) An elliptic paraboloid results from all choices of $u$ and $v$ in the parameter domain.

try it

Describe the surface parameterized by [latex]{\bf{r}}(u,v)=\langle{u}\cos{v},u\sin{v},u\rangle,-\inftyShow Solution

Cone $x^{2}+y^{2}=z^{2}$ .

Example: finding a parameterization

Give a parameterization of the cone $x^{2}+y^{2}=z^{2}$ lying on or above the plane $z=-2$ .

Show Solution

The horizontal cross-section of the cone at height $z=u$ is circle $x^{2}+y^{2}=u^{2}$ . Therefore, a point on the cone at height $u$ has coordinates $(u\cos{v},u\sin{v},u)$ for angle $v$ . Hence, a parameterization of the cone is ${\bf{r}}(u,v)=\langle{u}\cos{v},u\sin{v},u\rangle$ . Since we are not interested in the entire cone, only the portion on or above plane $z=-2$ , the parameter domain is given by $-2\leq{u}<\infty,0\leq{v}<2\pi$ (Figure 4).

<img src="/apps/archive/20220422.171947/resources/80b41e00683076a5a361597b58344c406e2d8067" data-media-type="image/jpeg" alt="A three-dimensional diagram of the cone x^2 + y^2 = z^2, which opens up along the z axis for positive z values and opens down along the z axis for negative z values. The center is at the origin." id="12">

Figure 4. Cone $x^{2}+y^{2}=r^{2}$ has parameterization ${\bf{r}}(u,v)=\langle{u}\cos{v},u\sin{v},u\rangle$ , $-\infty<u<\infty, 0\leq{v}\leq2\pi$ .

try it

Give a parameterization for the portion of cone $x^{2}+y^{2}=z^{2}$ lying in the first octant.

Show Solution

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.49” here (opens in new window).

We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. To parameterize a sphere, it is easiest to use spherical coordinates. The sphere of radius $\rho$ centered at the origin is given by the parameterization

$\large{{\bf{r}}(\phi,\theta)=\langle\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi\rangle, \ 0\leq\theta\leq2\pi, \ 0\leq\phi\leq\pi}$ .

The idea of this parameterization is that as $\phi$ sweeps downward from the positive $z$ -axis, a circle of radius $\rho\sin\phi$ is traced out by letting $\theta$ run from 0 to $2\pi$ . To see this, let $\phi$ be fixed. Then

$\begin{aligned} x^2+y^2&=(\rho\cos\theta\sin\phi)^2+(\rho\sin\theta\sin\phi)^2 \\ &=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta) \\ &=\rho^2\sin^2\phi \\ &=(\rho\sin\phi)^2 \\ \end{aligned}$ .

This results in the desired circle (Figure 5).

<img src="/apps/archive/20220422.171947/resources/7b37adfac687f3ede4c12b1624213f7c408c0a68" data-media-type="image/jpeg" alt="A three-dimensional diagram of the sphere of radius rho." id="14">

Figure 5. The sphere of radius $\rho$ has parameterization ${\bf{r}}(\phi,\theta)=\langle\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi\rangle$ , $0\leq\theta\leq2\pi, 0\leq\phi\leq\pi$ .

Finally, to parameterize the graph of a two-variable function, we first let $z=f(x, y)$ be a function of two variables. The simplest parameterization of the graph of $f$ is ${\bf{r}}(x,y)=\langle{x},y,f(x,y)\rangle$ , where $x$ and $y$ vary over the domain of $f$ (Figure 6). For example, the graph of $f(x, y)=x^{2}y$ can be parameterized by ${\bf{r}}(x,y)=\langle{x},y,x^2y\rangle$ , where the parameters $x$ and $y$ vary over the domain of $f$ . If we only care about a piece of the graph of $f$ —say, the piece of the graph over rectangle $[1,3]\times[2,5]$ —then we can restrict the parameter domain to give this piece of the surface:

$\large{{\bf{r}}(x,y)=\langle{x},y,x^2y\rangle, \ 1\leq{x}\leq3, \ 2\leq{y}\leq5}$ .

Similarly, if $S$ is a surface given by equation $x=g(y, z)$ or equation $y=h(x, z)$ , then a parameterization of $S$ is

${\bf{r}}(y,z)=\langle{g}(y,z),y,z\rangle$ or ${\bf{r}}(x,z)=\langle{x},h(x,z),z\rangle$ , respectively. For example, the graph of paraboloid $2y=x^{2}+z^{2}$ can be parameterized by ${\bf{r}}(x,z)=\langle{x},\frac{x^2+z^2}2,z\rangle$ , $0\leq{x}\leq\infty, \ 0\leq{z}\leq\infty$ . Notice that we do not need to vary over the entire domain of $y$ because $x$ and $z$ are squared.

<img src="/apps/archive/20220422.171947/resources/0dc3336b5ac8a3f543f7960e6dffd31b165532bd" data-media-type="image/jpeg" alt="A three-dimensional diagram of a surface z = f(x,y) above its mapping in the two-dimensional x,y plane. The point (x,y) in the plane corresponds to the point z = f(x,y) on the surface." id="15">

Figure 6. The simplest parameterization of the graph of a function is ${\bf{r}}(x,y)=\langle{x},y,f(x,y)\rangle$ .

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization ${\bf{r}}(t)$ , $a\leq{t}\leq{b}$ is regular if ${\bf{r}}^\prime(t)\ne0$ for all $t$ in $[a,b]$ . For a curve, this condition ensures that the image of ${\bf{r}}$ really is a curve, and not just a point. For example, consider curve parameterization ${\bf{r}}(t)=\langle1,2\rangle$ , $0\leq{t}\leq5$ . The image of this parameterization is simply point $(1, 2)$ , which is not a curve. Notice also that ${\bf{r}}^\prime(t)=0$ . The fact that the de
rivative is zero indicates we are not actually looking at a curve.

Analogously, we would like a notion of regularity for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider parameterization

$\large{{\bf{r}}(u,v)=\langle0,\cos{v},1\rangle, \ 0\leq{u}\leq1, \ 0\leq{v}\leq\pi}$ .

Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure 7). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that ${\bf{r}}_u=\langle0,0,0\rangle$ and ${\bf{r}}_v=\langle0,-\sin{v},0\rangle$ , and the corresponding cross product is zero. The analog of the condition ${\bf{r}}^\prime(t)=0$ is that ${\bf{r}}_u\times{\bf{r}}_v$ is not zero for point $(u, v)$ in the parameter domain, which is a regular parameterization.

<img src="/apps/archive/20220422.171947/resources/745227e56c04a5df74ee49af2b05fce09d079046" data-media-type="image/jpeg" alt="A three-dimensional diagram of a line on the x,z plane where the z component is 1, the x component is 1, and the y component exists between -1 and 1." id="16">

Figure 7. The image of parameterization ${{\bf{r}}(u,v)=\langle0,\cos{v},1\rangle, \ 0\leq{u}\leq1, \ 0\leq{v}\leq\pi}$ is a line.

definition

Parameterization ${\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle$ is a regular parameterization if ${\bf{r}}_u\times{\bf{r}}_v$ is not zero for point $(u, v)$ in the parameter domain.

If parameterization ${\bf{r}}$ is regular, then the image of ${\bf{r}}$ is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations ${\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle$ are assumed to be regular.

Recall that curve parameterization ${\bf{r}}(t)$ , $a\leq{t}\leq{b}$ is smooth if ${\bf{r}}^\prime(t)$ is continuous and ${\bf{r}}^\prime(t)\ne0$ for all $t$ in $[a,b]$ . Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.

definition

A surface parameterization ${\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle$ is smooth if vector ${\bf{r}}_u\times{\bf{r}}_v$ is not zero for any choice of $u$ and $v$ in the parameter domain.

A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist.

Example: identifying smooth and nonsmooth surfaces

Which of the figures in Figure 8 is smooth?

<img src="/apps/archive/20220422.171947/resources/0c3ca1bca3bc9aceb9cd677eaa0899d9c3a8afa3" data-media-type="image/jpeg" alt="Two three-dimensional figures. The first surface is smooth. It looks like a tire with a large hole in the middle. The second is piecewise smooth. It is a pyramid with a rectangular base and four sides." id="20">

Figure 8. (a) This surface is smooth. (b) This surface is piecewise smooth.

Show Solution

The surface in Figure 8(a) can be parameterized by

${\bf{r}}(u,v)=\langle(2+\cos{v})\cos{u},(2+\cos{v})\sin{u},\sin{v}\rangle \ 0\leq{u}\leq2\pi, \ 0\leq{v}\leq2\pi$

(we can use technology to verify). Notice that vectors

${\bf{r}}_u=\langle-(2+\cos{v})\sin{u},(2+\cos{v})\cos{u},0\rangle\text{ and }{\bf{r}}_v=\langle-\sin{v}\cos{u},-\sin{v}\sin{u},\cos{v}\rangle$

exist for any choice of $u$ and $v$ in the parameter domain, and

$\begin{aligned} {\bf{r}}_v&=\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ -(2+\cos{v})\sin{u}&(2+\cos{v})\cos{u}&0 \\ -\sin{v}\cos{u}&-\sin{v}\sin{u}&\cos{v} \end{vmatrix} \\ &=[(2+\cos{v})\cos{u}\cos{v}]{\bf{i}}+[(2+\cos{v})\sin{u}\cos{v}]{\bf{j}}+[(2+\cos{v})\sin{v}\sin^2u+(2+\cos{v})\sin{v}\cos^2u]{\bf{k}} \\ &=[(2+\cos{v})\cos{u}\cos{v}]{\bf{i}}+[(2+\cos{v})\sin{u}\cos{v}]{\bf{j}}+[(2+\cos{v})\sin{v}]{\bf{k}} \end{aligned}$

The $\bf{k}$ component of this vector is zero only if $v=0$ or $v=\pi$ . If $v=0$ or $v=\pi$ , then the only choices for $u$ that make the $\bf{j}$ component zero are $u=0$ or $u=\pi$ . But, these choices of $u$ do not make the $\bf{i}$ component zero. Therefore, ${\bf{r}}_u\times{\bf{r}}_v$ is not zero for any choice of $u$ and $v$ in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.

In the pyramid in Figure 8(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of five smooth faces, and thus this surface is piecewise smooth.

Key Takeaways

Is the surface parameterization ${\bf{r}}(u,v)=\langle{u}^{2v},v+1,\sin{u}\rangle$ , $0\leq{u}\leq2, \ 0\leq{v}\leq3$ smooth?

Show Solution

Try It

Surface Area of a Parametric Surface

Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.

Let $S$ be a surface with parameterization ${\bf{r}}(u,v)=\langle{x}(u,v),y(u,v),z(u,v)\rangle$ over some parameter domain $D$ . We assume here and throughout that the surface parameterization ${\bf{r}}(u,v)=\langle{x}(u,v),{y}(u,v),{z}(u,v)\rangle$ is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that $D$ is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle $D$ into subrectangles $D_{ij}$ with horizontal width $\nabla{u}$ and vertical length $\nabla{v}$ . Suppose that $i$ ranges from $1$ to $m$ and $j$ ranges from $1$ to $n$ so that $D$ is subdivided into $mn$ rectangles. This division of $D$ into subrectangles gives a corresponding division of surface $S$ into pieces $S_{ij}$ . Choose point $P_{ij}$ in each piece $S_{ij}$ . Point $P_{ij}$ corresponds to point $(u_i,v_j)$ in the parameter domain.

Note that we can form a grid with lines that are parallel to the $u$ -axis and the $v$ -axis in the $uv$ -plane. These grid lines correspond to a set of on surface $S$ that is parameterized by ${\bf{r}}(u,v)$ . Without loss of generality, we assume that $P_{ij}$ is located at the corner of two grid curves, as in Figure 9. If we think of ${\bf{r}}$ as a mapping from the $uv$ -plane to $\mathbb{R}^3$ , the grid curves are the image of the grid lines under ${\bf{r}}$ . To be precise, consider the grid lines that go through point $(u_i,v_j)$ . One line is given by $x=u_i$ , $y=v$ ; the other is given by $x=u$ , $y=v_j$ . In the first grid line, the horizontal component is held constant, yielding a vertical line through $(u_i,v_j)$ . In the second grid line, the vertical component is held constant, yielding a horizontal line through $(u_i,v_j)$ . The corresponding grid curves are ${\bf{r}}(u_i,v)$ and ${\bf{r}}(u,v_j)$ , and these curves intersect at point $P_{ij}$ .

<img src="/apps/archive/20220422.171947/resources/554bdb200e243c7a7fe7695f959bcc99d6c61c86" data-media-type="image/jpeg" alt="Two diagrams, showing that grid lines on a parameter domain correspond to grid curves on a surface. The first shows a two-dimensional rectangle in the u,v plane. The horizontal rectangle is in quadrant 1 and broken into 9x5 rectangles in a grid format. One rectangle Dij has side lengths delta u and delta v. The coordinates of the lower left corner are (u_i *, v_j *). In three dimensions, the surface curves above the x,y plane. The D_ij portion has become S_ij on the surface with lower left corner P_ij." id="22">

Figure 9. Grid lines on a parameter domain correspond to grid curves on a surface.

Now consider the vectors that are tangent to these grid curves. For grid curve ${\bf{r}}(u_i,v)$ the tangent vector at $P_{ij}$ is

${\bf{t}}_v(P_{ij})={\bf{r}}_v(u_i,v_j)=\langle{x}_v(u_i,v_j),y_v(u_i,v_j),z_v(u_i,v_j)\rangle$ .

For grid curve ${\bf{r}}(u,v_j)$ , the tangent vector at $P_{ij}$ is

${\bf{t}}_u(P_{ij})={\bf{r}}_y(u_i,v_j)=\langle{x}_v(u_i,v_j),y_u(u_i,v_j),z_u(u_i,v_j)\rangle$ .

If vector ${\bf{N}}={\bf{t}}_u(P_{ij})\times{\bf{t}}_u(P_{ij}$ exists and is not zero, then the tangent plane at $P_{ij}$ exists (Figure 10). If piece $S_{ij}$ is small enough, then the tangent plane at point $P_{ij}$ is a good approximation of piece $S_{ij}$ .

<img src="/apps/archive/20220422.171947/resources/7a8f20e54fbe300cbc4ec7c1c79f355a7c0f670a" data-media-type="image/jpeg" alt="Two diagrams. The one on the left is two dimensional and in the first quadrant of the u,v coordinate plane. A point u_0 is marked on the horizontal u axis, and a point v_0 is marked on the vertical v axis. The point (u_0, v_0) is shown in the plane. The diagram on the right shows the grid curve version. Now, the u_0 is marked as r(u_0, v) and the v_0 is marked as r(u, v_0). The (u_0, v_0) point is labeled P. Coming out of P are three arrows: one is a vertical N arrow, and the other two are t_u and t_v for the tangent plane." id="23">

Figure 10. If the cross product of vectors ${\bf{t}}_u$ and ${\bf{t}}_v$ exists, then there is a tangent plane.

The tangent plane at $P_{ij}$ contains vectors ${\bf{t}}_u(P_{ij})$ and ${\bf{t}}_v(P_{ij})$ , and therefore the parallelogram spanned by ${\bf{t}}_u(P_{ij})$ and ${\bf{t}}_v(P_{ij})$ is in the tangent plane. Since the original rectangle in the $uv$ -plane corresponding to $S_{ij}$ has width $\Delta{u}$ and length $\Delta{v}$ , the parallelogram that we use to approximate $S_{ij}$ is the parallelogram spanned by $\Delta{u}{\bf{t}}_u(P_{ij})$ and $\Delta{v}{\bf{t}}_v(P_{ij})$ . In other words, we scale the tangent vectors by the constants $\Delta{u}$ and $\Delta{v}$ to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of $S_{ij}$ is

$\Delta{S_{ij}}\approx||(\Delta{u}{\bf{t}}_u(P_{ij}))\times(\Delta{v}{\bf{t}}_v(P_{ij}))=||{\bf{t}}_u(P_{ij})\times{\bf{t}}_v(P_{ij})||\Delta{u}\Delta{v}$ .

Varying point $P_{ij}$ over all pieces $S_{ij}$ and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 11).

<img src="/apps/archive/20220422.171947/resources/5446eff034279beb6c8d1c545bd42983566a72fb" data-media-type="image/jpeg" alt="A surface S_ij that looks like a curved parallelogram. Point P_ij is at the bottom left corner, and two blue arrows stretch from this point to the upper left and lower right corners of the surface. Two red arrows also stretch out from this point, and they are labeled t_v delta v and t_u delta u. These form two sides of a parallelogram that approximates the piece of surface of S_ij. The other two sides are drawn as dotted lines." id="24">

Figure 11. The parallelogram spanned by ${\bf{t}}_u$ and ${\bf{t}}_v$ approximates the piece of surface $S_{ij}$ .

definition

Let ${\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle$ with parameter domain $D$ be a smooth parameterization of surface $S$ . Furthermore, assume that $S$ is traced out only once as $(u, v)$ varies over $D$ . The surface area of $S$ is

$\large{\displaystyle\iint_D||{\bf{t}}_u\times{\bf{t}}_v||dA}$ ,

where ${\bf{t}}_u=\left\langle\frac{\partial{x}}{\partial{u}},\frac{\partial{y}}{\partial{u}},\frac{\partial{z}}{\partial{u}}\right\rangle\text{ and }{\bf{t}}_v=\left\langle\frac{\partial{x}}{\partial{v}},\frac{\partial{y}}{\partial{v}},\frac{\partial{z}}{\partial{v}}\right\rangle$ .

Example: calculating surface area

Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height $h$ and radius $r$ .

Show Solution

Solution

Before calculating the surface area of this cone using The Surface Area Equation, we need a parameterization. We assume this cone is in $\mathbb{R}^3$ with its vertex at the origin (Figure 12). To obtain a parameterization, let $\alpha$ be the angle that is swept out by starting at the positive $z$ -axis and ending at the cone, and let $k=\tan\alpha$ . For a height value $v$ with $0\leq{v}\leq{h}$ , the radius of the circle formed by intersecting the cone with plane $z=v$ is $kv$ . Therefore, a parameterization of this cone is

${\bf{s}}(u,v)=\langle{k}v\cos{u},kv\sin{u},v\rangle, \ 0\leq{u}\leq2\pi, \ 0\leq{v}\leq{h}$ .

The idea behind this parameterization is that for a fixed $v$ value, the circle swept out by letting $u$ vary is the circle at height $v$ and radius $kv$ . As $v$ increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.

<img src="/apps/archive/20220422.171947/resources/b15442ba83e41b09bb6363dbd68df733254ba936" data-media-type="image/jpeg" alt="A right circular cone in three dimensions, opening upwards on the z axis. It has radius r = kh and height h with the given parameterization. Alpha is the angle that is swept out by starting at the positive z-axis and ending at the cone. It is noted that k is equal to the tangent of alpha." id="27">

Figure 12. The right circular cone with radius $r = kh$ and height $h$ has parameterization ${\bf{s}}(u,v)=\langle{k}v\cos{u},kv\sin{u},v\rangle$ , $0\leq{u}\leq2\pi, 0\leq{v}\leq{h}$ .

With a parameterization in hand, we can calculate the surface area of the cone using The Surface Area Equation. The tangent vectors are ${\bf{t}}_u=\langle-kv\sin{u},kv\cos{u},0\rangle$ and ${\bf{t}}_v=\langle{k}\cos{u},k\sin{u},1\rangle$ . Therefore,

$\begin{aligned} {\bf{t}}_u\times{\bf{t}}_v&=\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ -kv\sin{u}&kv\cos{u}&0 \\ {k}\cos{u}&k\sin{u}&1 \end{vmatrix} \\ &=\langle{k}v\cos{u},kv\sin{u},-k^2v\sin^2y-k^2v\cos^2u\rangle \\ &=\langle{k}v\cos{u},kv\sin{u},-k^2v\rangle \end{aligned}$ .

The magnitude of this vector is

$\begin{aligned} ||\langle{k}v\cos{u},kv\sin{u},-k^2v\rangle||&=\sqrt{k^2v^2\cos^2u+k^2v^2\sin^2u+k^4v^2} \\ &=\sqrt{k^2v^2+k^4v^2} \\ &=kv\sqrt{1+k^2} \end{aligned}$ .

By The Surface Area Equation, the surface area of the cone is

$\begin{aligned} \displaystyle\iint_D||{\bf{t}}_u\times{\bf{t}}_v||dA&=\displaystyle\int_0^h\displaystyle\int_0^{2\pi}kv\sqrt{1+k^2}dudv \\ &=2\pi{k}\sqrt{1+k^2}\displaystyle\int_0^hvdv \\ &=2\pi{k}\sqrt{1+k^2}\left[\frac{v^2}2\right]_0^h \\ &=\pi{k}h^2\sqrt{1+k^2} \end{aligned}$ .

Since $k=\tan\alpha=r/h$ ,

$\begin{aligned} \pi{k}h^2\sqrt{1+k^2}&=\pi\frac{r}hh^2\sqrt{1+\frac{r^2}{h^2}} \\ &=\pi{r}h\sqrt{1+\frac{r^2}{h^2}} \\ &=\pi{r}\sqrt{h^2+h^2\left(\frac{r^2}{h^2}\right)} \\ &=\pi{r}\sqrt{h^2+r^2} \end{aligned}$ .

Therefore, the lateral surface area of the cone is $\pi{r}\sqrt{h^2+r^2}$ .

Analysis

The surface area of a right circular cone with radius $r$ and height $h$ is usually given as $\pi{r}^2+\pi{r}\sqrt{h^2+r^2}$ . The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $\pi{r}^2$ is added to the lateral surface area $\pi{r}\sqrt{h^2+r^2}$ that we found.

try it

Find the surface area of the surface with parameterization ${\bf{r}}(u,v)=\langle{u}+v,u^2,2v\rangle$ , $0\leq{u}\leq3$ , $0\leq{v}\leq2$ .

Show Solution

$\approx43.02$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.51” here (opens in new window).

Example: calculating surface area

Show that the surface area of the sphere $x^{2}+y^{2}+z^{2}=r^{2}$ is $4\pi{r}^2$ .

Show Solution

The sphere has parameterization

$\langle{r}\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi\rangle, \ 0\leq\theta\leq2\pi, \ 0\leq\phi\leq\pi$ .

The tangent vectors are

${\bf{t}}_\theta=\langle-r\sin\theta\sin\phi,r\cos\theta\sin\phi,0\rangle\text{ and }{\bf{t}}_\phi=\langle{r}\cos\theta\cos\phi,r\sin\theta\cos\phi,-r\sin\phi\rangle$ .

Therefore,

$\begin{aligned} {\bf{t}}_\phi\times{\bf{t}}_\theta&=\langle{r}^2\cos\theta\sin^2\phi,r^2\sin\theta\sin^2\phi,r^2\sin^2\theta\sin\phi\cos\phi+r^2\cos^2\theta\sin\phi\cos\phi\rangle \\ &=\langle{r}^2\cos\theta\sin^2\phi,r^2\sin\theta\sin^2\phi,r^2\sin\phi\cos\phi\rangle \end{aligned}$ .

Now,

$\begin{aligned} ||{\bf{t}}_\phi\times{\bf{t}}_\theta||&=\sqrt{r^4\sin^4\phi\cos^2\theta+r^4\sin^4\phi\sin^2\theta+r^4\sin^2\phi\cos^2\phi} \\ &=\sqrt{r^4\sin^4\phi+r^4\sin^2\phi\cos^2\phi} \\ &=r^2\sqrt{\sin^2\phi} \\ &=r^2\sin\phi \end{aligned}$ .

Notice that $\phi\geq0$ on the parameter domain because $0\leq\phi\leq\pi$ , and this justifies equation $\sqrt{\sin^2\phi}=\sin\phi$ . The surface area of the sphere is

$\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}r^2\sin\phi{d}\phi{d}\theta=r^2\displaystyle\int_0^{2\pi}2d\theta=4\pi{r}^2$ .

We have derived the familiar formula for the surface area of a sphere using surface integrals.

try it

Show that the surface area of cylinder $x^2+y^2=r^2$ , $0\leq{z}\leq{h}$ is $2\pi{r}h$ . Notice that this cylinder does not include the top and bottom circles.

Show Solution

With the standard parameterization of a cylinder, The Surface Area Equation shows that the surface area is $2\pi{r}h$ .

In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let $y=f(x)\geq0$ be a positive single-variable function on the domain $a\leq{x}\leq{b}$ and let $S$ be the surface obtained by rotating $f$ about the $x$ -axis (Figure 13). Let $\theta$ be the angle of rotation. Then, $S$ can be parameterized with parameters $x$ and $\theta$ by

${\bf{r}}(x,\theta)=\langle{x},f(x)\cos\theta,f(x)\sin\theta\rangle, \ a\leq{x}\leq{b}, \ 0\leq{x}\leq2\pi$ .

<img src="/apps/archive/20220422.171947/resources/62e003469f226d24fb2fc7ce6d152985d58f6522" data-media-type="image/jpeg" alt="Two diagrams, a and b, showing the surface of revolution. The first shows three dimensions. In the (x,y) plane, a curve labeled y = f(x) is drawn in quadrant 1. A line is drawn from the endpoint of the curve down to the x axis, and it is labeled f(x). The second shows the same three dimensional view. However, the curve from the first diagram has been rotated to form a three dimensional shape about the x axis. The boundary is still labeled y = f(x), as the curve in the first plane was. The opening of the three dimensional shape is circular with the radius f(x), just as the line from the curve to the x axis in the plane of the first diagram was labeled. A point on the opening’s boundary is labeled (x,y,z), the distance from the x axis to this point is drawn and labeled f(x), and the height is drawn and labeled z. The height is perpendicular to the x,y plane and, as such, the original f(x) line drawn from the first diagram. The angle between this line and the line from the x axis to (x,y,z) is labeled theta." id="32">

Figure 13. We can parameterize a surface of revolution by ${\bf{r}}(x,\theta)=\langle{x},f(x)\cos\theta,f(x)\sin\theta\rangle$ , $a\leq{x}\leq{b}, 0\leq{x}\leq2\pi$ .

Example: calculating surface area

Find the area of the surface of revolution obtained by rotating $y=x^2$ , $0\leq{x}\leq{b}$ about the $x$ -axis (Figure 14).

<img src="/apps/archive/20220422.171947/resources/7d61bcda55b02e38101709cea6667b3baa333904" data-media-type="image/jpeg" alt="A solid of revolution drawn in two dimensions. The solid is formed by rotating the function y = x^2 about the x axis. A point C is marked on the x axis between 0 and x’, which marks the opening of the solid." id="34">

Figure 14. A surface integral can be used to calculate the surface area of this solid of revolution.

Show Solution

This surface has parameterization

${\bf{r}}(x,\theta)\langle{x},x^2\cos\theta,x^2\sin\theta\rangle, \ 0\leq{x}\leq{b}, \ 0\leq{x}\leq2\pi$ .

The tangent vectors are

${\bf{t}}_x=\langle1,2x\cos\theta,2x\sin\theta\rangle\text{ and }{\bf{t}}_\theta=\langle0,-x^2\sin\theta,-x^2\cos\theta\rangle$ .

Therefore,

$\begin{aligned} {\bf{t}}_x\times{\bf{t}}_\theta&=\langle2x^3\cos^2\theta+2x^3\sin^2\theta,-x^2\cos\theta,-x^2\sin\theta\rangle \\ &=\langle2x^3,-x^2\cos\theta,-x^2\sin\theta\rangle \end{aligned}$

and

$\begin{aligned} {\bf{t}}_x\times{\bf{t}}_\theta&=\sqrt{4x^6+x^4\cos^2\theta+x^4\sin^2\theta} \\ &=\sqrt{4x^6+x^4} \\ &=x^2\sqrt{4x^2+1} \end{aligned}$ .

The area of the surface of revolution is

$\begin{aligned} \displaystyle\int_0^b\int_0^\pi{x}^2\sqrt{4x^2+1}d\theta{d}x&=2\pi\int_0^bx^2\sqrt{4x^2+1}dx \\ &=2\pi\left[\frac1{64}(2\sqrt{4x^2+1}(8x^3+x)\sin{h}^{-1}(2x))\right]_0^b \\ &=2\pi\left[\frac1{64}(2\sqrt{4b^2+1}(8b^3+b)\sin{h}^{-1}(2b))\right] \end{aligned}$ .

try it

Use The Surface Area Equation to find the area of the surface of revolution obtained by rotating curve $y=\sin{x}, \ 0\leq{x}\leq\pi$ about the $x$ -axis.

Show Solution

$2\pi(\sqrt2+\sin{h}^{-1}(1))$

Module 6: Vector Calculus