copyright © 2000 by Paul Green and Jonathan Rosenberg

In this notebook, we will study integrals over parametrized
surfaces. Recall that a surface is an object in 3-dimensional space that *locally*
looks like a plane. In other words, the
surface is given by a vector-valued function ** P** (encoding the

To see how this works, let us compute the surface area of
the ellipsoid whose equation is _{}. We may parametrize this ellipsoid as we have done in the past,
using modified spherical coordinates:

ellipsoid=[2*sin(p)*cos(t),3*sin(p)*sin(t),cos(p)]

[ 2*sin(p)*cos(t), 3*sin(p)*sin(t), cos(p)]

To check that this really is a parametrization, we verify the original equation:

simplify(subs((x^2/4)+(y^2/9)+z^2,[x,y,z],ellipsoid))

1

And we can also draw a picture with **ezsurf**:

ezsurf(ellipsoid(1),ellipsoid(2),ellipsoid(3),[0,pi,0,2*pi])

This is as it should be. Now we compute the surface area factor:

surffactor=simple(veclength(cross(diff(ellipsoid,t),diff(ellipsoid,p))))

(-(5*cos(t)^2*cos(p)^2-32*cos(p)^2-5*cos(t)^2-4)*sin(p)^2)^(1/2)

This is going to be too complicated to integrate symbolically, so we do a numerical integration, using the numerical integration toolbox.

addpath H:\class\math241\common\nit

area=newnumint2(surffactor,p,0,pi,t,0,2*pi)

48.8822

To integrate a function, such as _{}, over the surface, we must express it in terms of the parameters
and insert the result as a factor in the integrand.

func=subs(x^2+2*z^2,[x,y,z],ellipsoid)

4*sin(p)^2*cos(t)^2+2*cos(p)^2

integral=newnumint2(surffactor*func,p,0,pi,t,0,2*pi)

100.5002

Here's another example: suppose we want the surface area of
the portion of the cone z^{2} = x^{2} + y^{2 }between z = 0 and z = 4. The equation of the cone in cylindrical
coordinates is just z = r, so we can take as our parameters r and t
(representing q).

syms r; cone=[r*cos(t),r*sin(t),r]

[ r*cos(t), r*sin(t), r]

The picture is:

ezsurf(cone(1),cone(2),cone(3),[0,4,0,2*pi])

conesurffactor=simple(veclength(cross(diff(cone,r),diff(cone,t))
))

2^(1/2)*csgn(r)*r

Since r is positive, we can ignore the sign csgn(r), and the
surface area is

conesurf=symint2(2^(1/2)*r,r,0,4,t,0,2*pi)

16*2^(1/2)*pi

This is right since we can also "unwrap" the cone
to a sector of a circular disk, with radius 4_{} and outer circumference 8p (compared to 8p_{}for the whole circle), so the surface area is p(4_{})^{2} / _{}, or indeed 16p_{}.

(b) Evaluate _{}, where _{} is the surface whose
area you found in part (a).

The formula _{} also allows us to
compute flux integrals over parametrized surfaces.

Let us compute _{} where the integral is
taken over the ellipsoid of Example 1, **F
** is the vector field defined by the
following input line, and **n** is the
outward normal to the ellipsoid.

[ x, y, z]

ndS=cross(diff(ellipsoid,p),diff(ellipsoid,t))

[ 3*sin(p)^2*cos(t), 2*sin(p)^2*sin(t), 6*sin(p)*cos(t)^2*cos(p)+6*sin(p)*sin(t)^2*cos(p)]

We can check that we
have the outward normal by setting p=_{} and t=0, giving us the point of the ellipsoid that is on the
positive x-axis. The outward normal should point in the positive x direction.

3.0000 0 0.0000

Before we can evaluate the integral, we must evaluate F in terms of the parameters.

Fpar=subs(F,[x,y,z],ellipsoid)

[ 2*sin(p)*cos(t), 3*sin(p)*sin(t), cos(p)]

flux=symint2(realdot(Fpar,ndS),p,0,pi,t,0,2*pi)

24*pi

[ 2*x, y, z-2]

3

We must now parametrize the solid ellipsoid. We can do this
most easily by adding a radial factor r that takes values between 0 and 1. Note that since **ellipsoid** is a
vector, all we have to do is "scalar multiply" it by r:

solid=r*ellipsoid

[ 2*r*sin(p)*cos(t), 3*r*sin(p)*sin(t), r*cos(p)]

Recall that to integrate in the coordinates r, p, and t, we are going to have to insert a scale factor, the absolute value of the determinant of the Jacobian matrix of the change of variables. As we indicated is last week's session, it's usually easiest to leave off the absolute value at first, and then change the sign if the scale factor comes out negative.

scale=simple(det(jacobian(solid,[r,p,t])))

6*sin(p)*r^2

That's clearly positive, and it's simple enough to make a symbolic integration feasible.

answer=int(symint2(3*scale,r,0,1,p,0,pi),t,0,2*pi)

24*pi

This agrees with our calculation of the flux. The numerical value is:

75.3982

1. Check the accuracy of the computation in Example 1 above by repeating the integration over the ellipsoid, using x and y as the parameters and solving for z as a function of x and y. (Hint: since there are two solutions for z, but the surface is symmetric, just integrate over the top half of the ellipsoid and then double the result.)

2. Let *f* = 2x^{2}
+ 3y^{2} +6z^{2 }.
Compute the flux of the gradient of *f* through the ellipsoid

x^{2} + 4y^{2} + 9z^{2} = 36, both
directly and by using the Divergence Theorem.

3. Let *T* be the torus with equation z^{2 }+
(r-2)^{2}
= 1 in cylindrical coordinates.
Parametrize the torus and use the answer to compute the surface
area.

4. Let *T* be the same torus as in Additional Problem 3
just above. Compute the volume enclosed
by the torus two ways: by triple integration, and by computing the flux of the
vector field **F** = (*x*, *y*, *z*) through *T* and by
using the Divergence Theorem. Check
that the results agree, and also that they agree with the prediction of *Pappus's
Theorem* (Ellis and Gulick, p. 531): the volume of a solid obtained by
rotating a region *R* in the x-z plane around the z-axis is the area of *R*
times the circumference of the circle traced out by the center of mass of *R*
as it rotates around the z-axis.