Math 225: Calculus III   Solution 10

Initialization

<<Calculus`Math225`

i = {1, 0, 0}; j = {0, 1, 0}; k = {0, 0, 1};

Problem 1

Compute Div[F], Curl[F], and Div[Curl[F]] for the following vector fields.
a) F = (x^2 y + z) i + (y^2 z + x) j + (z^2 x + y) k
b) F = Cos[x+y] i + Sin[x-y] j
c) F = y E^z i + z E^x j + x E^y k
d) "Prove" using Mathematica that for an arbitrary vector field,
    F = m[x,y,z]i + n[x,y,z]j + p[x,y,z]k
     we always get Div[Curl[F]] == 0.

Solution

a)

F = (x^2 y + z) i + (y^2 z + x) j + (z^2 x + y) k

Div[F]

Curl[F]

b)

F = Cos[x+y] i + Sin[x-y] j

Div[F]

Curl[F]

c)

F = y E^z i + z E^x j + x E^y k

Div[F]

Curl[F]

d)

F = m[x,y,z]i + n[x,y,z]j + p[x,y,z]k

Div[F]

Curl[F]

Div[Curl[F]]

Problem 2

Let F = (y-z) i + (x-z) j - (x+y) k. Compute the flow integral "F.dr" over each of the following curves by using the appropriate formula with Integrate[]. Compare your answer to FlowIntegrate[]:
a) r[t_] = (2t^3-t) i + (t^2+t)/2 j + t k, 0<= t <=1
b) r[t_] = (1-Cos[t])/2 i + Sin[t/2]j + t/Pi k, 0<= t <= Pi
c) Explain why the answers to a) and b) have to be the same, even though the curves are quite different.

Solution

F = (y-z) i + (x-z) j - (x+y) k

a)

r[t_] = (2t^3-t) i + (t^2+t)/2 j + t k

F /. {x->r[t][[1]], y->r[t][[2]], z->r[t][[3]]}

Integrate[%.r'[t], {t,0,1}]

FlowIntegrate[F,r[t],{t,0,1}]

b)

r[t_] = (1-Cos[t])/2 i + Sin[t/2]j + t/Pi k

F /. {x->r[t][[1]], y->r[t][[2]], z->r[t][[3]]}

Integrate[%.r'[t], {t,0,Pi}]

FlowIntegrate[F,r[t],{t,0,Pi}]

c) The integrals in a) and b) give the same answer because the vector field is conservative, i.e.,
F = Grad[f[x,y,z]] and the paths have the same end points.

f[x_,y_,z_] = x y - x z - y z;

Grad[f[x,y,z]]

Grad[f[x,y,z]] == F

By the Fundamental Theorem of Line Integrals, the flow integral of F over wither pathe equals
f evaluated at the end points.

f[1,1,1] - f[0,0,0]

Problem 3

a) Integrate
(y^3 - y + z)dx + (3x y^2 - x + 2y z)dy + (y^2 + x + 2z)dz
over the curve C defined by t^2 i + t j + (t^3-1)k, 0 <= t <= 1,
(using Integrate[], or FlowIntegrate[]).
b) Find a function f[x,y,z] such that
Grad[f[x,y,z]] == 
    (y^3 - y + z)i + (3x y^2 - x + 2y z)j + (y^2 + x + 2z)k
c) Use the Fundamental Theorem of Line Integrals to evaluate the the integral in a) using the result of b).

Solution

a)

FlowIntegrate[
    {y^3 - y + z, 3x y^2 - x + 2y z, y^2 + x + 2z},
    {t^2, t, t^3-1}, {t,0,1}]

b) We must find a function f[x,y,z] that satisfies the equations
    D[f[x,y,z],x] == y^3 - y + z
    D[f[x,y,z],y] == 3x y^2 - x + 2y z
    D[f[x,y,z],z] == y^2 + x + 2z
Integrate the first equations iwth respect to x, adding a "constant" of integration that could depend on y and z:

Clear[f,g,h]

f[x_,y_,z_] = Integrate[y^3 - y + z,x] + g[y,z]

Now use the second equation to learn more about g[y,z]:

D[f[x,y,z],y] == 3x y^2 - x + 2y z

Therefore, D[g[y,z],y] == 2y z and so

g[y_,z_] = Integrate[2y z, y] + h[z]

The third equation determines h[z]:

D[f[x,y,z],z] == y^2 + x + 2z

So

h[z_] = Integrate[2z,z]

and

f[x,y,z]

c) By the Fundamental Theorem of Line Integrals, the flow integral of Grad[f[x,y,z]] equals f[x,y,z] evaluated at the end points of the curve:

r[t_] = t^2 i + t j + (t^3-1)k

r[0]

r[1]

f[1,1,0] - f[0,0,-1]

This is the same answer as in part a).

Problem 4

Let C be the boundary of the "triangular" region R in the first quadrant enclosed by the x-axis, the line x==1, and the curve x==y^3. Use Green's Theorem to transform the integral of
2x y^2dx + x^2 y dy over C into a double integral over R and then evaluate this double integral.

Solution

Here's a plot of the region (not required).

ParametricPlot[{{x,x^(1/3)},{1,x}},{x,0,1},AspectRatio->1]

By Green's Theorem, the integral of x y^2dx + x^2 y dy over C equals the double integral over R of

g = D[x^2 y,x] - D[2x y^2,y]

The region R is described horizontally by
    y^3 <= x <= 1
      0 <= y <= 1
so the value of the original line integral equals

Integrate[g,{y,0,1},{x,y^3,1}]

Problem 5

Produce a plot of a surface that you think has interesting features. The surface does not have to be the graph of a function and can be done in any coordinate system you wish. Explain what features of the surface are interesting and why.

Solution

The Moebius Band. The Moebius band is an example of a surface for which a unit normal vector cannot be assigned to every point on the surface in a continuous way since the surface has only one "side". If you start with a unit vector pointing up at one point, say, and then slide this vector around the surface and come back to the same point, the vector will be pointing down.
The surface is built starting with a horizontal circle and then adding a band that twists 180 degrees as we go around the complete circle. We'll use a horizontal circle of radius three:

u = {Cos[t], Sin[t], 0}

circ = 3 u

The band is built out of multiples of a vector pointing away from the circle and spinning around on its own. This vector can be written as a combination of u and k. The factor of 1/2 in the trig functions assures that as we move around the horizontal circle the band gets only a half twist.

v = u Cos[t/2] + k Sin[t/2]

We paramaterize the surface using the angle t for the horizontal circle and the spinning vector v, and we use another parameter s to thicken the circle into a band in the direction of v as we go around the circle. The result is a band with a "half-twist" in it.

band = ParametricPlot3D[circ + s v,{s,-1,1},{t,0,2Pi}]

Incrreasing the range for the parameter s creates a surface that intersects itself and appears to have tubular holes in it.

surf = ParametricPlot3D[circ + s v,{s,-8,8},
            {t,0,2Pi}]

Show[%,ViewPoint->{1.093, 1.562, 2.796}]