16.6 Parametric Surfaces and their Areas

These notes are intended as a supplement to the material covered in section 16.6 of the book.  Please read the material in the book before proceeding.

Recall parametric curves, C: (OverVector[r])(t)=⟨x(t),y(t),z(t)⟩.  This is a vector function with one independent variable and produces a curve.  For example:

ParametricPlot3D[{Cos[t],Sin[t],t/(6Pi)},{t,0,6Pi}]

This is a helix.

A little piece of arc-length is given by Δ s=|(OverVector[r])'(t)| Δ t.

Now look at a vector function of two independent variables, S: (OverVector[r])(u,v)=⟨x(u,v),y(u,v),z(u,v)⟩.  This is a surface.  For example:

ParametricPlot3D[{Sin[u]Cos[v],Sin[u]Sin[v],Cos[u]},{u,0,Pi/2},{v,0,2Pi}]

This is the hemisphere or upper half of a sphere of radius 1.  Recall that the change of variables between rectangular and spherical was x=ρ Sin[ϕ]Cos[θ], y=ρ Sin[ϕ]Sin[θ], and z=ρ Cos[ϕ], where ρ is distance from the origin, ϕ is measured down from the positive z-axis and θ is the same angle as in polar (or cylindrical) coordinates.

Recall from section section 15.6 that when a surface is given by z=f(x,y), then the surface area was

Δ S=|(OverVector[a])×(OverVector[b])|=(Sqrt[(((Subscript[f, x])(x,y))^2)+(((Subscript[f, y])(x,y))^2)+1])Δ x Δ y

where (OverVector[a])=⟨Δ x,0,(Subscript[f, x])(x,y)⟩ and (OverVector[b])=⟨0,Δ y,(Subscript[f, y])(x,y)⟩.  In a similar fashion, the area of a parametric surface is given by

Δ S=|(OverVector[(Subscript[r, u])])×(OverVector[(Subscript[r, v])])|Δ u Δ v.

Then the total surface area would be the double integral

(Underscript[∫∫, D])|(OverVector[(Subscript[r, u])])(u,v)×(OverVector[(Subscript[r, v])])(u,v)|dudv,

where D is the (u,v) region that maps the part of the surface of interest.

Example (i)

Find a parameterization for the surface 2x+y+z=6 and use the parameterization to find the area of the part of the surface that is in the first octant.

First: because we can write the equation as z as a function of x and y, we can use x and y as the parameters.  Thus, one parameterization of the surface is

(OverVector[r])(x,y)=⟨x,y,6-2x-y⟩.

Next, look at a graph

ParametricPlot3D[{{x,y,6-2x-y},{x,y,0}},{x,0,3},{y,0,6},PlotRange->{{0,3},{0,6},{-1,6}},AxesLabel->{x,y,z},BoxRatios->{1,1,1}]

The part of the surface that is in the first octant has parameter values in a triangular region:

Plot[6-2x, {x,0,3}]

The “increment” of surface area is

d S=|(OverVector[(Subscript[r, x])])×(OverVector[(Subscript[r, y])])| d x  d y=|⟨1,0,-2⟩×⟨0,1,-1⟩| d x  d y=|⟨2,1,1⟩| d x  d y.

Note that in this case, (OverVector[(Subscript[r, x])])×(OverVector[(Subscript[r, y])])=⟨2,1,1⟩, the normal vector given by the coefficients of the equation of the plane in standard form.  So the area of the surface is the double integral:

(∫_0^3)(∫_0^(6-2x))|⟨2,1,1⟩|dydx=(Sqrt[(2^2)+(1^2)+(1^2)])((1/2)(3)(6))=9(Sqrt[6]).

Note that we do not need to actually do the integral (of a constant), because the value of the integral is the area of the region times the (constant) integrand.

Example (ii)

Find a parameterization of the top half of (x^2)+(y^2)-(z^2)=1.

Note that we can rewrite the equation as (x^2)+(y^2)=(z^2)+1.  So, this is a hyperboloid in one sheet.  For fixed z, we have a circle of radius (Sqrt[(z^2)+1]).  Thus, one parameterization is

(OverVector[r])(θ,z)=⟨(Sqrt[(z^2)+1])Cos[θ],(Sqrt[(z^2)+1])Sin[θ],z⟩,  0≤θ≤2π,  z≥0.

The increment of surface area for this one is a bit more complicated:

(OverVector[(Subscript[r, θ])])=⟨-(Sqrt[(z^2)+1])Sin[θ],(Sqrt[(z^2)+1])Cos[θ],0⟩,

(OverVector[(Subscript[r, z])])=⟨(z/(Sqrt[(z^2)+1]))Cos[θ],(z/(Sqrt[(z^2)+1]))Sin[θ],1⟩,

so we have

|(OverVector[(Subscript[r, θ])])×(OverVector[(Subscript[r, z])])|=|⟨(Sqrt[(z^2)+1])Cos[θ],(Sqrt[(z^2)+1])Sin[θ],-z⟩|=(Sqrt[2(z^2)+1]).