Copyright © 1995, 1997, 2003 by James F. Hurley, Department of Mathematics, University of Connecticut, Storrs, CT 06269-3009. All rights reserved

**1. Simple Parametrization**. Think of a parametric surface in ${R}^{3}$ as a quilt consisting of many smooth patches. A *smooth surface patch* is the 2-dimensional analogue in ${R}^{3}$ of a 1-dimensional curve in the plane, namely, a continuous function *f*: *D* -> ${R}^{3}$, where *D* is a region in ${R}^{2}$*. A parametric surface* is the union of a collection of smooth surface patches. **Example 1**. It is simple to parametrize a plane with scalar equation *ax + by + cz = d*, Consider such a plane with *c* != 0. It is then possible to solve for *z* in terms of *x* and *y*: *z = *$\frac{d-\text{}\mathrm{ax}\text{}-\text{}\mathrm{by}}{c}$*.* A natural parametrization of the plane is then

(1) *x = u, y = v, z = f(u, v) = *$\frac{d\text{}\u2013\text{}\mathrm{au}\text{}\u2013\text{}\mathrm{bv}}{c}$*.*

In terms of a parametric vector function **r**: ${R}^{2}$-> ${R}^{3}$this takes the form

(2) **r**(*u, v*) = *u* **i** + *v* **j** + $\frac{d\text{}\u2013\text{}\mathrm{au}\text{}\u2013\text{}\mathrm{bv}}{c}$ **k**.

*Mathematica* can graph a parametric surface by means of its built-in graphics command ParametricPlot3D. The following routine illustrates this for the plane 2*x + *4y - 5*z = *20. To plot other planes, simply change the assignments of the coefficients *a, b, c,* and *d. *(See the notebook *Planes as Surfaces.*) To see *Mathematica*'s default plot, execute the following routine.

In[1]:=

a := 2;

b := 4;

c := -5;

d := 20;

plane = ParametricPlot3D[{u, v, (d - a*u - b*v)/c},

{u, -2, 2}, {v, -2, 2},

AxesLabel -> {x, y, z}]

As before, coordinate axes provide additional orientation. The following routine creates a graphics object called *coordaxes* with red coordinate axes over a portion of ${R}^{3}$appropriate to the graph of 2*x + *4y - 5*z = *20. Activate the routine as usual by hitting the Enter key after placing the cursor at the end.

In[6]:=

coordaxes = Graphics3D[{ RGBColor[1, 0, 0],

Line[{{-2, 0, 0}, {2, 0, 0}}],

Text["x", {2.2, 0, 0}],

Line[{{0, -2, 0}, {0, 2, 0}}],

Text["y", {0, 3, 0}],

Line[{{0, 0, -6}, {0, 0, 2}}],

Text["z", {0, 0, 2.2}] }]

To obtain a plot of the plane with the coordinate axes, ask *Mathematica* to show both plots. The following command carries that out. Try it!

In[8]:=

${\mathrm{Show}\left[\mathrm{plane},\text{}\mathrm{coordaxes}\right]}$

**2. More general surfaces**. A slight extension of the last section's approach provides a parametric representation for the graph of a function *f*:* *${R}^{2}$-> **R**. Just let *x = u,* *y = v,* and *z = f(u, v). *The vector parametrization is then

**r**(*u, v*) =* u* **i **+ *v* **j **+* f(u, v)* **k**,

where *u* and *v* range over an appropriate domain *D* of the *xy-*plane.

Solution. It is natural to use cylindrical coordinates to describe the cylinder. So let

In[9]:=

ParametricPlot3D[{r*Cos[θ], r*Sin[θ], 2*(r*Cos[θ] +

r*Sin[θ])}, {r, 0, 1},

{θ, 0, 2*Pi}]

As before, you may want to include coordinate axes.* *The following routine is just the above *coordaxes*, with some slight editing to display a more appropriate part of the axes.* *Execute it to create the axes.

In[10]:=

coords2 = Graphics3D[{RGBColor[1,0,0],

Line[{{-1, 0, 0}, {1, 0, 0}}],

Text["x", {1.2, 0, 0}],

Line[{{0, -1, 0}, {0, 2, 0}}],

Text["y", {0, 2.2, 0}],

Line[{{0, 0, -3}, {0, 0, 3}}],

Text["z", {0, 0, 3.2}]

}]

As before, display the plot of the part of the plane inside the cylinder with the axes by asking *Mathematica* to show you the second-to-last plot (of the plane) and the object *coords2*.

In[11]:=

Show[%%, coords2]

**3. Surface Area**. *Mathematica* is a convenient tool for checking calculations of ||${r}_{u}$× ${r}_{v}$||, whose integral gives the surface area of a paraametric surface patch *S* with parametrizing vector function **r**: *D *->* *${R}^{3}$. The following examples illustrate that.**Example 3**. Find the area of the part of the plane *z = *3 *x* - 2 *y* that lies inside the cylinder ${x}^{2}$* + *${y}^{2}$* = *4.**Solution**. As in Exercise 4 above, it is natural to parametrize the surface as

**X**(*r, *θ) = * r *cos* *θ* ***i** +*r *sin * *θ* ***j** +* *(3* r *cos* *θ* - *2* r *sin * *θ*) ***k**, *r *∈* *[0, 2], θ* * ∈ [0, 2π].* *

* ** *The following *Mathematica* routine calculates ${X}_{r}$= $\frac{\partial X}{\partial r}\text{\hspace{0.17em}}$and ${r}_{\theta}$=$\text{\hspace{0.17em}}\frac{\partial X}{\partial \theta}$*.* Try it!

In[12]:=

X[r_, θ_] := {r*Cos[θ], r*Sin[θ],

3*r*Cos[θ] - 2*r*Sin[θ]}

Xr = D[X[r, θ], r]

Xtheta = D[X[r, θ], θ]

Out[13]=

$\left\{\mathrm{Cos}\left[\theta \right],\mathrm{Sin}\left[\theta \right],3\mathrm{Cos}\left[\theta \right]-2\mathrm{Sin}\left[\theta \right]\right\}$

Out[14]=

$\left\{-r\mathrm{Sin}\left[\theta \right],r\mathrm{Cos}\left[\theta \right],-2r\mathrm{Cos}\left[\theta \right]-3r\mathrm{Sin}\left[\theta \right]\right\}$

To calculate the cross product of the two partial derivatives of **X** with respect to *r *and θ, previous versions of *Mathematica* required use of a special package, but now the built-in command Cross is available without further work. The next routine carries out the calculation. Execute it and compare to your hand calculation.

In[15]:=

dS := Cross[Xr, Xtheta]

dS

Out[16]=

$\left\{-3r{\mathrm{Cos}\left[\theta \right]}^{2}-3r{\mathrm{Sin}\left[\theta \right]}^{2},2r{\mathrm{Cos}\left[\theta \right]}^{2}+2r{\mathrm{Sin}\left[\theta \right]}^{2},r{\mathrm{Cos}\left[\theta \right]}^{2}+r{\mathrm{Sin}\left[\theta \right]}^{2}\right\}$

The output is not in simplest form, so ask *Mathematica* to simplify it as much as it can by executing the following command:

In[17]:=

Simplify[dS]

Out[17]=

$\left\{-3r,2r,r\right\}$

Finally, the area of the given surface is the double integral of the length of that cross product over the rectangular domain *D * = [0, 2] × [0, 2π]. The following little routine uses the fact that the length of the vector **dS **is just $\sqrt{\mathrm{dS}\xb7\mathrm{dS}}$. Try it!

In[18]:=

dA := Sqrt[Dot[dS, dS]]

dA

Out[19]=

$\sqrt{{\left(-3r{\mathrm{Cos}\left[\theta \right]}^{2}-3r{\mathrm{Sin}\left[\theta \right]}^{2}\right)}^{2}+{\left(r{\mathrm{Cos}\left[\theta \right]}^{2}+r{\mathrm{Sin}\left[\theta \right]}^{2}\right)}^{2}+{\left(2r{\mathrm{Cos}\left[\theta \right]}^{2}+2r{\mathrm{Sin}\left[\theta \right]}^{2}\right)}^{2}}$

Again, the complexity of this expression prompts a request for simplication:

In[20]:=

Simplify[dA]

Out[20]=

$\sqrt{14}\sqrt{{r}^{2}}$

Impressive programming, isn't it? Since we have not specified an allowable range of values for the variable *r*, *Mathematica* can't express the square root of ${r}^{2}$ in a simpler form. Once we specify the region of integration over which to evaluate the double integral of *dA*, *Mathematica* can and does provide a simple answer. Try the following!

In[21]:=

Integrate[dA, {r, 0, 2}, {θ, 0, 2 Pi}]

Out[21]=

$4\sqrt{14}\pi $

Converted by *Mathematica*
(June 24, 2003)