CHAPTER 16. VECTOR CALCULUS 239

16.6 Parametric Surfaces and Their Areas

Comments. On the next page we summarize some ways of looking at graphs in CalcI, Calc II, and Calc III, and pose a question about what comes next.

Definition. Let r : R2 → R3 and let D ⊆ R2. A parametric surface S is the set of allpoints 〈x, y, z〉 such that 〈x, y, z〉 = r(u, v) for some 〈u, v〉 ∈ D. In other words, it’sthe set of points you get using

x = some function of u, v

y = some function of u, v

z = some function of u, v

Example 1. (a) Describe the surface z = −x2− y2 + 2 over the region D : −1 ≤ x ≤1, −1 ≤ y ≤ 1 as a parametric surface. Graph it in MATLAB.

(b) Describe the surface z = −x2 − y2 + 2 over the region D : x2 + y2 ≤ 1 as aparametric surface. Graph it in MATLAB.

Solution: (a) In this case we “cheat”: we already have z as a function of x and y, soanything we do to get rid of x and y will work

x = u

y = v

z = −u2 − v2 + 2

−1 ≤ u ≤ 1, −1 ≤ v ≤ 1

[u,v]= meshgrid(linspace (-1,1,35));

x=u;

y=v;

z=2-u.^2-v.^2;

surf(x,y,z)

CHAPTER 16. VECTOR CALCULUS 240

How graphs look in different contexts:

Pre-Calc, and Calc I graphs:

y = f(x)

• 1 number gets plugged in (usually x)

• 1 number comes out (usually y)

• Graph is essentially one dimensional (if you zoomin enough it looks like a line, plus you only needone number, x, to specify any location on thegraph)

• We need to picture the graph in a two dimen-sional world (to contain 1# in + 1 out)

Calc II parametric graphs:

x = f(t), y = g(t)

• 1 number gets plugged in (usually t)

• 2 numbers come out (x and y)

• graph is essentially one dimensional (but wedon’t picture t at all)

• We need to picture the graph in two dimensionalworld.

Calc III, chapter 14 and 15 graphs:

• 2 numbers get plugged in (usually x and y)

• 1 number comes out (usually z)

• graph is essentially two dimensional (if you zoomin enough it looks like a plane, plus you need twonumbers, x and y, to specify a location on thegraph).

• We need to picture the graph inside a three di-mensional world (to contain 2 #s in + 1 out).

What goes here?

(b) In this case, we have to work a little bit more, not because of the equation,but because of the domain. Since the domain is circular, it’s convenient to use polarcoordinates:

x = r cos(θ)

y = r sin(θ)

z = 2− r2

0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1

[r,theta ]= meshgrid(linspace (-1,1,30),linspace (0,2*pi ,35));

x=r.*cos(theta);

y=r.*sin(theta);

z=2-r.^2;

surf(x,y,z)

Example 2. Use MATLAB to graph the following paramatric surface:

x = (3 + sin(v)) cos(u)

y = (3 + sin(v)) sin(u)

z = cos(v)

0 ≤ u ≤ 2π, 0 ≤ v ≤ 2π

Describe what it looks like. Analyze what u and v represent in the shape.

Solution:

CHAPTER 16. VECTOR CALCULUS 242

[u,v]= meshgrid(linspace (0,2*pi ,50));

x=(3+ sin(v)).*cos(u);

y =(3+ sin(v)).*sin(u);

z=cos(v);

surf(x,y,z);

axis equal

It looks like a donut. The parameter u corresponds to how far around the large circle apoint is: as u increases you get a path that travel counter clockwise around the shape.The curves that look “flat”, like lines of lattitude, or like the main equator, are generatedby holding v equal to a constant and allowing u to increase. In fact, for a constant valueof v we have x = ∗ cos(u) and y = ∗ sin(u) where “∗” just represents some number, butthe point is that this generates a circle in x and y.

The parameter v corresponds to how far around the small circle a point is: as vincreases you get a path that travels “up” over the shape, down into the hole, throughthe hole, and back around back up to where it started. The curves that go up and down,like longitude, are generated by holding u equal to a constant and allowing v to increase.In fact, for a constant value of u, and say when y = 0, then we have x = 3 + sin(v) andz = cos(v). This generates a circle in the (x, z) plane where the center is z = 0 andx = 3.

This is where we ended on Wednesday, April 24

Example 3. Based on the previous example, what will happen to the graph if we changethe parametric equations so that z = u+ cos(v) and 0 ≤ u ≤ 6π?

Solution: Everything is the same, except now z will increase because of the “u+” term.Thus, we will have a donut, but it won’t close, it will go around and around. If we hadenough batter we could make an endless donut.

CHAPTER 16. VECTOR CALCULUS 243

[u,v]= meshgrid(linspace (0,6*pi ,75),linspace (0,2*pi ,15));

x=(3+ sin(v)).*cos(u);

y =(3+ sin(v)).*sin(u);

z=u+cos(v);

surf(x,y,z);

axis equal

16.6.1 Tangent planes

Definition. Let r(u, v) be a parametric surface, and write the component functions asx(u, v), y(u, v) and z(u, v). The tangent plane at the point (u0, v0) is the plane thatcontains the point r(u0, v0) and which is normal to rv(u0, v0)× ru(u0, v0) where

rv(u0, v0) =

⟨∂x

∂v(u0, v0),

∂y

∂v(u0, v0),

∂z

∂v(u0, v0)

⟩ru(u0, v0) =

⟨∂x

∂u(u0, v0),

∂y

∂u(u0, v0),

∂z

∂u(u0, v0)

⟩

Recall that the equation of a plane is easily found by specifying what a point that itshould contain, and a normal vector (see Section 12.5 Definition 12.5.2.)

Example 4. Find the equation of the tangent plane of the donut from Example 2 atpoint u = π, v = π/4. Graph the result.

Solution:

r(π, π/4) = 〈(3 + sin(π/4)) cos(π), (3 + sin(π/4)) sin(π), cos(π/4)〉

=⟨−(3 + 1/

√2), 0, 1/

√2⟩

ru = 〈−(3 + sin(v)) sin(u), (3 + sin(v)) cos(u), 0〉∣∣∣π,π/4

= 〈−(3 + sin(π/4)) sin(π), (3 + sin(π/4)) cos(π), 0〉

CHAPTER 16. VECTOR CALCULUS 244

=⟨

0, −(3 + 1/√

2), 0⟩

rv = 〈(cos(v)) cos(u), (cos(v)) sin(u), − sin(v)〉∣∣∣π,π/4

= 〈cos(π/4) cos(π), cos(π/4) sin(π), − sin(π/4)〉

=⟨−1/√

2, 0, −1/√

2⟩

ru × rv =

∣∣∣∣∣∣i j k

0 −(3 + 1/√

2) 0

−1/√

2 0 −1/√

2

∣∣∣∣∣∣= i((3 + 1/

√2)(1/

√2))− j(0))− k

((3 + 1/

√2)(1/

√2))

=

⟨3√2

+1

2, 0, − 3√

2− 1

2

⟩Plane:

(3√2

+1

2

)(x+ 3 + 1/

√2)

+ 0−(

3√2

+1

2

)(z − 1/

√2)

= 0(x+ 3 + 1/

√2)−(z − 1/

√2)

= 0

z = x+ 3 +√

2

[u,v]= meshgrid(linspace (0,2*pi ,50));

x=(3+ sin(v)).*cos(u);

y =(3+ sin(v)).*sin(u);

z=cos(v);

surf(x,y,z,'FaceAlpha ' ,0.5,'EdgeAlpha ' ,0);axis equal

hold on

u= pi;

v=pi/4;

x=(3+ sin(v)).*cos(u);

y =(3+ sin(v)).*sin(u);

z=cos(v);

CHAPTER 16. VECTOR CALCULUS 245

plot3(x,y,z,'or','MarkerSize ',10,'MarkerFaceColor ','r');[x,y]= meshgrid(linspace (-4.5,-3,10),linspace ( -0.5 ,0.5 ,10));

z=x+3+ sqrt (2);

surf(x,y,z,'FaceColor ','green ','FaceAlpha ' ,0.5);hold off

16.6.2 Surface Area

Definition. If S is the surface parameterized by r(u, v) for (u, v) ∈ D, then the surfacearea is

A(S) =

∫∫D

|ru × rv| dA

where

rv =

⟨∂x

∂v,∂y

∂v,∂z

∂v

⟩ru =

⟨∂x

∂u,∂y

∂u,∂z

∂u

⟩|ru × rv| = length of the vector ru × rv

Example 5. Find the surface area of the donut from Example 2

Solution:

ru = 〈−(3 + sin(v)) sin(u), (3 + sin(v)) cos(u), 0〉rv = 〈cos(v) cos(u), cos(v) sin(u), − sin(v)〉

ru × rv =⟨− (3 + sin(v)) cos(u) sin(v), −(3 + sin(v)) sin(u) sin(v),

−(3 + sin(v)) sin2(u) cos(v)− cos(v) cos2(u)(3 + sin(v))−(3 + sin(v)) sin2(u) cos(v)− cos(v) cos2(u)(3 + sin(v))⟩

For the last component, we factor out−(3+sin(v)) cos(v) and this leaves behind sin2(u)+cos2(u) = 1.

ru × rv = 〈−(3 + sin(v)) cos(u) sin(v), −(3 + sin(v)) sin(u) sin(v), −(3 + sin(v)) cos(v)〉

|ru × rv| =√

(3 + sin(v))2 cos2(u) sin2(v) + (3 + sin(v))2 sin2(u) sin2(v)(3 + sin(v))2 cos2(u) sin2(v) + (3 + sin(v))2 sin2(u) sin2(v) + (3 + sin(v))2 cos2(v)

From the first two terms we factor out (3 + sin(v))2 sin2(v) and this leaves behindcos2(u) + sin2(u) = 1:

|ru × rv| =√

(3 + sin(v))2 sin2(v) + (3 + sin(v))2 cos2(v)

=

√(3 + sin(v))2

(sin2(v) + cos2(v)

)=√

(sin(v) + 3)2

= sin(v) + 3

A(S) =

∫∫D

sin(v) + 3 dA

CHAPTER 16. VECTOR CALCULUS 246

=

2π∫0

2π∫0

sin(v) + 3 dv du

=

2π∫0

6π du

= 12π2

By the way, the general formula for the surface area of a torus with radii r and R isπ2Rr (the radii in this case are 1 and 3).

Comments. To me, the examples in this section lead one to think about DifferentialGeometry. In this subject, you start with a surface, like the torus, and you ask basicgeometric questions about life on that surface. If I were an ant living on a torus, whatwould a straight line look like? Once I knew that, I could say what a triangle lookedlike. What would be the area formula for a triangle? How would that relate to thecurvature of the surface? Would there be theorems about similarity, etc?

“Straight lines” are those curves that minimize distance between points. Curveslike this are usually renamed “geodesics”. It turns out, even on the torus (which is arelatively simple surface), it’s hard to describe the geodesics. Here’s some follow up info:

• To get a simple path on a torus, you can use our equations given above, but withan easy relation between u and v, like v = 10u. This will create path that wrapsitself around small circly of the torus 10 times as it goes around the big circle.Maybe, you think, a geodesic will be a spiral like this?

• How to find “straight lines” Mark Iron’s Site

• How to construct an architectural torus

• An application of a torus to a festival

• A mind bending realization of a flat torus (It is flat in the sense that distancesalong this torus are the same as they would be if you cut it open and flattened itout: you can flatten it without any stretching!)

• What number of colors suffice to fill in any map on a torus?

• Do the spirals on the torus correspond to the motion of particles on vortex rings?

This is where we ended on Thursday, April 25