Math 127(3)  WD Smith,  HW#8  Due Tues 8 Apr 2003 ANSWERS
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6 exercises total.

sec 17.1 ex# 8
Line-Integrate 
 (exp(x-y), exp(x+y))  
over the line segment from
 (-1,1) to (1,2).
ANS:
1. Parameterize the line seg:  x=2t-1, y=1+t  over 0<t<1.
2. integrate  (exp(x-y), exp(x+y)) . dsVECTOR  from t=0 to t=1.
That is:
integrate  (exp(t-2), exp(3t)) . (dx/dt, dy/dt) dt  from t=0 to t=1.
That is:
  integrate  (exp(t-2), exp(3t)) . (2, 1) dt  from t=0 to t=1.
= integrate  2*exp(t-2) + 1*exp(3t) dt  from t=0 to t=1.
                              1
=  2 exp(t-2) + exp(3t)/3    |
                              0
=  2 exp(-1) + 1/3 exp(3) - 2 exp(-2) - 1/3

=  6.826933956.


sec 17.2 ex# 2
Curve-Integrate 
 (x+y, y)
over the ellipse surve   x=a cos(u),  y = b sin(u),   0<u<2pi.
ANS:
  integrate  (x+y, y) . dsVECTOR  from u=0 to u=2pi
= integrate  (a cos(u) + b sin(u),  b sin(u)) . (dx/du, dy/du) du  from u=0 to u=2pi
= integrate  (a cos(u) + b sin(u),  b sin(u)) . (-a sin(u), b cos(u)) du  from u=0 to u=2pi

                2                        2    2
= integrate [ -a cos(u) sin(u)- ab sin(u)  + b sin(u) ] du  from u=0 to u=2pi

                2  2                          2 
= integrate [ (b -a ) cos(u) sin(u)- ab sin(u)  ] du  from u=0 to u=2pi

easy integral since over full period.  The cos(u) sin(u) becomes 0 
and the sin(u) squared (always non-neg) becomes 2pi/2=pi. 
(Note: cos^2 and sin^2 add up to 1, hence their integrals over
a full period are same as integrating 1/2 over full period.)
So get

= - a b pi.


sec 17.4 ex# 32 
for the vector field
                   2    2               2
F(x,y,z) = (2xy + z ,  x - 2yz,  2xz - y ).
(a) show F is a gradient.
ANS:
Two ways to do it. First we could evaluate 
curlF = (0,0,0)
proving it is a gradient.  
This curl is easy because of the symmetry of the expression for F.
Thus the 1st component of the curl is 
dF2/dz - dF3/dy = -2y + 2y = 0.

Second, we could explictly find Q(x,y,z) so gradQ = F.
We have
                    2           2     2
dQ/dx = F1 = 2xy + z    so Q = x y + z x + C1(y,z).

              2                  2     2
dQ/dy = F2 = x  - 2yz    so Q = x y - y z + C2(x,z)

                    2            2     2
dQ/dz = F3 = 2xz - y     so Q = z x - y z + C3(x,y)

                                             2     2     2
so, combining the information... Q(x,y,z) = x y + z x - y z + C.

(b) Evaluate  integral F(x,y,z) dsVECTOR
along ANY curve from (1,0,1) to (3,2,-1).
ANS:
Two ways to do it. 
method 1: Since answer is same for ANY curve, easiest is a line segment.
Parameterize line seg:  x=1+2t, y=2t,  z=1-2t  for 0<t<1.
dsVECTOR = (dx/dt, dy/dt, dz/dt) dt = (2,2,-2) dt.
  integrate  F(x,y,z) . (2,2,-2) dt  from t=0 to t=1

                          2      2                      2
= integrate   [4 x y + 2 z  + 2 x  - 4 y z - 4 x z + 2 y ] dt  for 0<t<1

              
= integrate   from 0 to 1
                                2              2                                              2
   [ 8 (1 + 2 t) t + 2 (1 - 2 t)  + 2 (1 + 2 t)  - 8 t (1 - 2 t) - 4 (1 + 2 t) (1 - 2 t) + 8 t ] dt

                  2
= integrate   72 t    from 0 to 1  (big cancellation when you simplify integrand!)

= 72/3 = 24.

method 2:
Use Newton's theorem of conservation of energy (see sec 17.3 p1130-1131 in my book).
The integral is of   force.distance = work.
Work done (by the force field) has to be the same as the  potential energy decrease
which is  (since -Q is the potential energy)
  -Q(1,0,1) + Q(3,2,-1) = -(0+1-0) + (3*3*2 + 1*1*3 + 2*2*1) = 24.
Note we got same answer... Newton was right.


sec 17.5 ex# 4:
            2      2                                                        2
integrate (y dx + x dy)   round the closed curve bounded by the parabola y=x and the line x=y.
[Note: the points where these 2 curves hit are (0,0) and (1,1).]
In my notation this is
                     2   2
  curve-integrate  (y , x )  dsVECTOR.
Do two ways: plain and using Green's theorem (Which is the 2D case of Stokes's theorem).

                                       2            2
The "curl" (z-component only) is    d(x ) / dx - d(y )/dy = 2x-2y.
Green theorem says double-integrating this curl inside the area yields same result
as our "circulation" single-integral round the boundary curve.

OK. To do the curve integral:        2
parameterize the parabola:   x=t, y=t ,   0<t<1.  
 dsVECTOR=(dx/dt, dy/dt)dt = (1,2t)dt.
Get           4    2
   integral (t  , t ) . (1, 2t) dt  from t=0 to 1
               4    3
 = integral [ t + 2t  ] dt  from t=0 to 1

 = 1/5 + 2/4 = 7/10.

Now parameterize the line (counterclockwise!)  y=1-t, x=1-t,  from t=0 to t=1.
 dsVECTOR=(dx/dt, dy/dt)dt = (-1, -1) dt
Get               2       2
   integral -(1-t)  -(1-t)    dt from t=0 to 1
                   2
=  integral -2(1-t)  dt   from t=0 to 1
               2
=  integral 2 m  dm   from m=1 to 0   (using m=1-t, dm = -dt)

=  -2/3.

FINAL RESULT = 7/10 - 2/3 = 1/30.


Now, to do the area double-integral:
    x=1         y=x
  integrate  integrate      (2x-2y) dy dx 
    x=0         y=xx

    x=1            2   y=x
= integrate  (2xy-y ) |              dx 
    x=0                y=xx

    x=1        2    3   4
= integrate  (x - 2x  +x ) dx  = 1/3 - 2/4 + 1/5 = 1/30.
    x=0       
Green's theorem worked!

COORDINATE SYSTEMS - WHERE TO FIND IN BOOK:
Cylindrical coords sec 16.9
Spherical coords   sec 16.10


sec 17.6 ex# 32
I will use my own notation, not the book's here.
An ellipsoid
       2        2        2
  (x/a)  + (y/b)  + (z/c)  = 1                              (*)
can be parameterized by "stretched spherical coordinates"
(stretched by a,b,c in the x,y,z directions respectively)
  x = a cos(theta) sin(phi)
  y = b sin(theta) sin(phi)
  z = c cos(phi)
                        0<phi<pi   0<theta<2pi.
Since an ellipsoid is just a stretched sphere.               2     2
You can verify the sum-of-squares identity as usual using cos + sin = 1,
same thing happens as in spherical coords since the stretching factors
cancel out thanks to the a,b,c in the denominators of (*).

They ask for the surface area in the (surface of revolution case) a=b.
Well, integrate dphi frpm 0 to pi, and inside that integrate 
(round the revolution) dtheta from 0 to 2pi.  
We need to compute the area elenment in this coordinate system:

  dSurfaceArea = |(dx/dphi, dy/dphi, dz/dphi) X (dx/dtheta, dy/dtheta, dz/dtheta)| dtheta dphi

     = |(a cos(theta) cos(phi),  b sin(theta) cos(phi),  -c sin(phi)) X
        (-a sin(theta) sin(phi),  b cos(theta) sin(phi),  0)|   dtheta dphi

do X:
                     2                           2
     = |(b c sin(phi)  cos(theta),   a c sin(phi)  sin(theta),  

                            2                                    2
               a b cos(theta)  cos(phi) sin(phi) + a b sin(theta)  cos(phi) sin(phi)  )| dtheta dphi

       2     2
use sin + cos= 1:
                     2                           2
     = |(b c sin(phi)  cos(theta),   a c sin(phi)  sin(theta),  a b cos(phi) sin(phi)  )| dtheta dphi
use a=b:
                     2                           2               2
     = |(a c sin(phi)  cos(theta),   a c sin(phi)  sin(theta),  a  cos(phi) sin(phi)  )| dtheta dphi
do length:
                   2 2        4           2      2 2        4           2    4         2         2
     = SquareRoot(a c sin(phi)  cos(theta)  +   a c sin(phi)  sin(theta) +  a  cos(phi)  sin(phi)  ) dtheta dphi
       2     2
use sin + cos= 1:
                   2 2        4     4         2         2
     = SquareRoot(a c sin(phi)  +  a  cos(phi)  sin(phi)  ) dtheta dphi

factor out a sin(phi):

                              2        2     2         2 
     = a sin(phi) SquareRoot(c sin(phi)  +  a  cos(phi)  ) dtheta dphi
so...

               phi=pi    theta=2pi                          2        2     2        2
SurfaceArea = integral  integral     a sin(phi) SquareRoot(c sin(phi)  +  a cos(phi) ) dtheta dphi
               phi=0     theta=0

The inner integral over theta is easy since integral does not depend on theta
so this integration just multiplies by 2pi:

                      phi=pi                          2        2     2        2
SurfaceArea = 2 pi a integral    sin(phi) SquareRoot(c sin(phi)  +  a cos(phi) ) dphi
                       phi=0  

which is the answer the book wanted (my phi is their v; my c is their b).


sec 17.7 ex# 32
Find the flux of the gravity-force function
                                        2   2   2 3
   F(x,y,z) = GmM (x,y,z) / SquareRoot(x + y + z )
out of the sphere
                      2   2   2    2
                     x + y + z  = a.

ANS: 
Peanuts. The force F is always normal to the sphere (radial)
so that the F.(unit normal) is just |F|    [since a.b = |a| |b| cos(angle)  when angle=0].
That makes it easy.
So
     flux = int int   Fvector . dAreavector
             sphere
reduces to
     flux = int int   |F| dAreascalar
             sphere
and now it is even easier since
              2          2
 |F| = GmM / r  = GmM / a   always on the sphere (i.e. is CONSTANT), so answer is
                   2                                      2         2
    flux = (GmM / a  )  (Surface Area of Sphere) = GmM / a    4 pi a  

         = 4 pi G m M.

The trick here was to see the spherical symmetry makes everything the same no matter
which direction you go.

WARNING: trying to use divergence theorem does not quite work.   The problem
is that the divergence of F is 0 everywhere EXCEPT at (x,y,z)=(0,0,0) where it is 
infinite!  Ouch!   So when you try to integrate the divergence you get 0, PLUS,
over an infinitesimally small region, you have to
integrate something infinite... result is nonzero...!    Physicists
call this a "point charge".  Moral:  those bad places where you divide by 
zero - they actually can matter.

Moral: the divergence and Stokes and Newton-energy theorems only work
for DIFFERENTIABLE F... "corners" and jumps in F can screw them up.

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