Section 14.5 Parametrized Surfaces and Surface Area
Subsection 14.5.1 Parametrizing surfaces
Definition 14.5.2. Parametrized Surface.
Let ār(u,v)=āØf(u,v),g(u,v),h(u,v)ā© be a vector-valued function that is continuous and one to one on the interior of its domain R in the u-v plane. The set of all terminal points of ār (i.e., the range of ār ) is the surface S, and ār along with its domain R form a parametrization of S.
This parametrization is smooth on R if āru and ārv are continuous and āruĆārv is never ā0 on the interior of R.
Example 14.5.4. Parametrizing a surface over a rectangle.
Parametrize the surface z=x2+2y2 over the rectangular region R defined by ā3ā¤xā¤3, ā1ā¤yā¤1.
There is a straightforward way to parametrize a surface of the form \(z=f(x,y)\) over a rectangular domain. We let \(x=u\) and \(y=v\text{,}\) and let \(\vec r(u,v) = \langle u,v, f(u,v)\rangle\text{.}\) In this instance, we have \(\vec r(u,v) = \langle u,v,u^2+2v^2\rangle\text{,}\) for \(-3\leq u\leq 3\text{,}\) \(-1\leq v\leq 1\text{.}\) This surface is graphed in Figure 14.5.5.
Example 14.5.6. Parametrizing a surface over a circular disk.
Parametrize the surface z=x2+2y2 over the circular region R enclosed by the circle of radius 2 that is centered at the origin.
We can parametrize the circular boundary of \(R\) with the vector-valued function \(\la 2\cos u,2\sin u\ra\text{,}\) where \(0\leq u\leq 2\pi\text{.}\) We can obtain the interior of \(R\) by scaling this function by a variable amount, i.e., by multiplying by \(v\text{:}\) \(\la 2v\cos u,2v\sin u\ra\text{,}\) where \(0\leq v\leq 1\text{.}\)
It is important to understand the role of \(v\) in the above function. When \(v=1\text{,}\) we get the boundary of \(R\text{,}\) a circle of radius 2. When \(v=0\text{,}\) we simply get the point \((0,0)\text{,}\) the center of \(R\) (which can be thought of as a circle with radius of 0). When \(v=1/2\text{,}\) we get the circle of radius \(1\) that is centered at the origin, which is the circle halfway between the boundary and the center. As \(v\) varies from 0 to 1, we create a series of concentric circles that fill out all of \(R\text{.}\)
Thus far, we have determined the \(x\) and \(y\) components of our parametrization of the surface: \(x=2v\cos u\) and \(y=2v\sin u\text{.}\) We find the \(z\) component simply by using \(z = f(x,y) = x^2+2y^2\text{:}\)
Thus \(\vec r(u,v) = \langle 2v\cos u,2v\sin u,4v^2\cos^2u+8v^2\sin^2u\rangle\text{,}\) \(0\leq u\leq 2\pi\text{,}\) \(0\leq v\leq 1\text{,}\) which is graphed in Figure 14.5.7. The way that this graphic was generated highlights how the surface was parametrized. When viewing from above, one can see lines emanating from the origin; they represent different values of \(u\) as \(u\) sweeps from an angle of 0 up to \(2\pi\text{.}\) One can also see concentric circles, each corresponding to a different value of \(v\text{.}\)
Example 14.5.8. Parametrizing a surface over a triangle.
Parametrize the surface z=x2+2y2 over the triangular region R enclosed by the coordinate axes and the line y=2ā2x/3, as shown in Figure 14.5.9.(a).
We may begin by letting \(x=u\text{,}\) \(0\leq u\leq 3\text{,}\) and \(y = 2-2u/3\text{.}\) This gives only the line on the āupperā side of the triangle. To get all of the region \(R\text{,}\) we can once again scale \(y\) by a variable factor, \(v\text{.}\)
Still letting \(x = u\text{,}\) \(0\leq u\leq 3\text{,}\) we let \(y = v(2-2u/3)\text{,}\) \(0\leq v\leq 1\text{.}\) When \(v=0\text{,}\) all \(y\)-values are 0, and we get the portion of the \(x\)-axis between \(x=0\) and \(x=3\text{.}\) When \(v=1\text{,}\) we get the upper side of the triangle. When \(v=1/2\text{,}\) we get the line \(y=1/2(2-2u/3) = 1-u/3\text{,}\) which is the line āhalfway upā the triangle, shown in the figure with a dashed line.
Letting \(z = f(x,y) = x^2+2y^2\text{,}\) we have \(\vec r(u,v) = \langle u, v(2-2u/3), u^2+2\big(v(2-2u/3)\big)^2\rangle\text{,}\) \(0\leq u\leq 3\text{,}\) \(0\leq v\leq 1\text{.}\) This surface is graphed in Figure 14.5.9.(b). Again, when one looks from above, we can see the scaling effects of \(v\text{:}\) the series of lines that run to the point \((3,0)\) each represent a different value of \(v\text{.}\)
Another common way to parametrize the surface is to begin with \(y=u\text{,}\) \(0\leq u\leq 2\text{.}\) Solving the equation of the line \(y=2-2x/3\) for \(x\text{,}\) we have \(x = 3-3y/2\text{,}\) leading to using \(x=v(3-3u/2)\text{,}\) \(0\leq v\leq 1\text{.}\) With \(z=x^2+2y^2\text{,}\) we have \(\vec r(u,v) = \langle v(3-3u/2),u, \big(v(3-3u/2)\big)^2+2v^2\rangle\text{,}\) \(0\leq u\leq 2\text{,}\) \(0\leq v\leq 1\text{.}\)
Example 14.5.10. Parametrizing a surface over a triangle.
Parametrize the surface z=x2+2y2 over the triangular region R enclosed by the lines y=3ā2x/3, y=1 and x=0 as shown in Figure 14.5.11.(a).
While the region \(R\) in this example is very similar to the region \(R\) in the previous example, and our method of parametrizing the surface is fundamentally the same, it will feel as though our answer is much different than before.
We begin with letting \(x=u\text{,}\) \(0\leq u\leq 3\text{.}\) We may be tempted to let \(y = v(3-2u/3)\text{,}\) \(0\leq v\leq 1\text{,}\) but this is incorrect. When \(v = 1\text{,}\) we obtain the upper line of the triangle as desired. However, when \(v=0\text{,}\) the \(y\)-value is 0, which does not lie in the region \(R\text{.}\)
We will describe the general method of proceeding following this example. For now, consider \(y = 1+v(2-2u/3)\text{,}\) \(0\leq v\leq 1\text{.}\) Note that when \(v=1\text{,}\) we have \(y=3-2u/3\text{,}\) the upper line of the boundary of \(R\text{.}\) Also, when \(v=0\text{,}\) we have \(y=1\text{,}\) which is the lower boundary of \(R\text{.}\) With \(z=x^2+2y^2\text{,}\) we determine \(\vec r(u,v) = \langle u, 1+v(2-2u/3), u^2+2\big(1+v(2-2u/3)\big)^2\rangle\text{,}\) \(0\leq u\leq 3\text{,}\) \(0\leq v\leq 1\text{.}\)
The surface is graphed in Figure 14.5.11.(b).
Key Idea 14.5.12. Parametrizing Surfaces.
Let a surface S be the graph of a function f(x,y), where the domain of f is a closed, bounded region R in the xy-plane. Let R be bounded by aā¤xā¤b, g1(x)ā¤yā¤g2(x), i.e., the area of R can be found using the iterated integral ā«baā«g2(x)g1(x)dydx, and let h(u,v)=g1(u)+v(g2(u)āg1(u)).
S can be parametrized as
Example 14.5.13. Parametrizing a cylindrical surface.
Find a parametrization of the cylinder x2+z2/4=1, where ā1ā¤yā¤2, as shown in Figure 14.5.14.
The equation \(x^2+z^2/4=1\) can be envisioned to describe an ellipse in the \(xz\)-plane; as the equation lacks a \(y\)-term, the equation describes a cylinder (recall Definition 10.1.13) that extends without bound parallel to the \(y\)-axis. This ellipse has a vertical major axis of length 4, a horizontal minor axis of length 2, and is centered at the origin. We can parametrize this ellipse using sines and cosines; our parametrization can begin with
where we still need to determine the \(y\) component.
While the cylinder \(x^2+z^2/4=1\) is satisfied by any \(y\) value, the problem states that all \(y\) values are to be between \(y=-1\) and \(y=2\text{.}\) Since the value of \(y\) does not depend at all on the values of \(x\) or \(z\text{,}\) we can use another variable, \(v\text{,}\) to describe \(y\text{.}\) Our final answer is
Example 14.5.15. Parametrizing an elliptic cone.
Find a parametrization of the elliptic cone z2=x24+y29, where ā2ā¤zā¤3, as shown in Figure 14.5.16.
One way to parametrize this cone is to recognize that given a \(z\) value, the cross section of the cone at that \(z\) value is an ellipse with equation \(\frac{x^2}{(2z)^2} + \frac{y^2}{(3z)^2}=1\text{.}\) We can let \(z=v\text{,}\) for \(-2\leq v\leq 3\) and then parametrize the above ellipses using sines, cosines and \(v\text{.}\)
We can parametrize the \(x\) component of our surface with \(x=2z\cos u\) and the \(y\) component with \(y=3z\sin u\text{,}\) where \(0\leq u\leq 2\pi\text{.}\) Putting all components together, we have
When \(v\) takes on negative values, the radii of the cross-sectional ellipses become ānegative,ā which can lead to some surprising results. Consider Figure 14.5.17, where the cone is graphed for \(0\leq u\leq \pi\text{.}\) Because \(v\) is negative below the \(xy\)-plane, the radii of the cross-sectional ellipses are negative, and the opposite side of the cone is sketched below the \(xy\)-plane.
Example 14.5.18. Parametrizing an ellipsoid.
Find a parametrization of the ellipsoid x225+y2+z24=1 as shown in Figure 14.5.19.(a).
Recall Key Idea 10.2.20 from Section 10.2, which states that all unit vectors in space have the form \(\langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\rangle\) for some angles \(\theta\) and \(\varphi\text{.}\) If we choose our angles appropriately, this allows us to draw the unit sphere. To get an ellipsoid, we need only scale each component of the sphere appropriately.
The \(x\)-radius of the given ellipsoid is 5, the \(y\)-radius is 1 and the \(z\)-radius is 2. Substituting \(u\) for \(\theta\) and \(v\) for \(\varphi\text{,}\) we have
where we still need to determine the ranges of \(u\) and \(v\text{.}\)
Note how the \(x\) and \(y\) components of \(\vec r\) have \(\cos v\) and \(\sin v\) terms, respectively. This hints at the fact that ellipses are drawn parallel to the \(xy\)-plane as \(v\) varies, which implies we should have \(v\) range from \(0\) to \(2\pi\text{.}\)
One may be tempted to let \(0\leq u\leq 2\pi\) as well, but note how the \(z\) component is \(2\cos u\text{.}\) We only need \(\cos u\) to take on values between \(-1\) and \(1\) once, therefore we can restrict \(u\) to \(0\leq u\leq \pi\text{.}\)
The final parametrization is thus
In Figure 14.5.19.(b), the ellipsoid is graphed on \(\frac{\pi}{4}\leq u\leq \frac{2\pi}{3}\text{,}\) \(\frac{\pi}4\leq v\leq \frac{3\pi}2\) to demonstrate how each variable affects the surface.
Subsection 14.5.2 Surface Area
Theorem 14.5.22. Surface Area of Parametrically Defined Surfaces.
Let ār(u,v) be a smooth parametrization of a surface S over a closed, bounded region R of the u-v plane.
The surface area differential dS is: dS=||āruĆārv||dA.
-
The surface area S of S is
S=ā¬SdS=ā¬R||āruĆārv||dA.
Example 14.5.24. Finding the surface area of a parametrized surface.
Using the parametrization found in Example 14.5.6, find the surface area of z=x2+2y2 over the circular disk of radius 2, centered at the origin.
In Example 14.5.6, we parametrized the surface as \(\vec r(u,v) = \la 2v\cos u, 2v\sin u, 4v^2\cos^2u+8v^2\sin^2u\ra\text{,}\) for \(0\leq u\leq 2\pi\text{,}\) \(0\leq v\leq 1\text{.}\) To find the surface area using Theorem 14.5.22, we need \(\snorm{\vec r_u\times\vec r_v}\text{.}\) We find:
Thus the surface area is
There is a lot of tedious work in the above calculations and the final integral is nontrivial. The use of a computer-algebra system is highly recommended.
Exercises 14.5.3 Exercises
Terms and Concepts
In the following exercises, parametrize the surface defined by the function z=f(x,y) over each of the given regions R of the xy-plane.
3.
z=3x2y;
R is the rectangle bounded by ā1ā¤xā¤1 and 0ā¤yā¤2.
R is the circle of radius 3, centered at (1,2).
R is the triangle with vertices (0,0), (1,0) and (0,2).
R is the region bounded by the x-axis and the graph of y=1āx2.
4.
z=4x+2y2;
R is the rectangle bounded by 1ā¤xā¤4 and 5ā¤yā¤7.
R is the ellipse with major axis of length 8 parallel to the x-axis, and minor axis of length 6 parallel to the y-axis, centered at the origin.
R is the triangle with vertices (0,0), (2,2) and (0,4).
R is the annulus bounded between the circles, centered at the origin, with radius 2 and radius 5.
In the following exercises, a surface S in space is described that cannot be defined as the graph of a function f(x,y). Give a parametrization of S.
In the following exercises, a domain D in space is given. Parametrize each of the bounding surfaces of D.
9.
D is the domain bounded by the planes z=12(3āx), x=1, y=0, y=2 and z=0.
10.
D is the domain bounded by the planes z=2x+4yā4, x=2, y=1 and z=0.
11.
D is the domain bounded by z=2y, y=4āx2 and z=0.
12.
D is the domain bounded by y=1āz2, y=1āx2, x=0, y=0 and z=0.
13.
D is the domain bounded by the cylinder x+y2/9=1 and the planes z=1 and z=3.
14.
D is the domain bounded by the cone x2+y2=(zā1)2 and the plane z=0.
15.
D is the domain bounded by the cylinder z=1āx2 and the planes y=ā1, y=2 and z=0.
16.
D is the domain bounded by the paraboloid z=4āx2ā4y2 and the plane z=0.
In the following exercises, find the surface area S of the given surface S. (The associated integrals are computable without the assistance of technology.)
17.
S is the plane z=2x+3y over the rectangle ā1ā¤xā¤1, 2ā¤vā¤3.
18.
S is the plane z=x+2y over the triangle with vertices at (0,0), (1,0) and (0,1).
19.
S is the plane z=x+y over the circular disk, centered at the origin, with radius 2.
20.
S is the plane z=x+y over the annulus bounded by the circles, centered at the origin, with radius 1 and radius 2.
In the following exercises, set up the double integral that finds the surface area S of the given surface S, then use technology to approximate its value.
21.
S is the paraboloid z=x2+y2 over the circular disk of radius 3 centered at the origin.
22.
S is the paraboloid z=x2+y2 over the triangle with vertices at (0,0), (0,1) and (1,1).
23.
S is the plane z=5xāy over the region enclosed by the parabola y=1āx2 and the x-axis.
24.
S is the hyperbolic paraboloid z=x2āy2 over the circular disk of radius 1 centered at the origin.