HOMEWORK #7   WD SMith  math127(3)
5 Exercises in book  (perf score 50/50)   Due THURS 27 Mar
**THE ANSWERS**
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17.6 p1158 ex 8   and also find the surface area of this surface

Simplest is just to
use x and y as the parameters.         2   2
              x=x, y=y, z = x+2,      x + y  < 1.

SurfaceArea = integral integral of SquareRoot(1+1) dx dy  over the unit circle.
(The second 1 is dz/dx squared.)
That is:   SquareRoot(2) pi.

Harder (but also valid) is to use  
x=r cos(theta),   y=r sin(theta),   z=r sin(theta)+2
for 0<=r<=1, 0<theta<2pi.

                 1       2pi
SurfaceArea = integral integral of SquareRoot(1+1) r dtheta dr =
                 0        0
    1
=integral 2pi r SquareRoot(2) dr = pi SquareRoot(2).
    0


17.6 p1158 ex22. Surface is  z = SquareRoot(2xy)  with 0<x<a  and  0<y<b.

Surface area =                 y     x
   integral of SquareRoot(1 + --- + ---) dx dy  over 0<x<a and 0<y<b
                              2x    2y
where               2                         2
    y/(2x) = (dz/dx)     and  x/(2y) = (dz/dy).

TRICK(spotted by two students - but I missed it myself!) rewrite as
Surface area =            2   2
    b   a                x + y  + 2xy
   int int of SquareRoot(------------) dx dy 
    0   0                      2xy

and recognize the numerator as the square of (x+y)...
Surface area =      
    b   a         x + y
   int int of --------------- dx dy
    0   0     SquareRoot(2xy)
now it is easy to do the integrals of SquareRoot(x/(2y)) and SquareRoot(y/(2x)) 
and we get...
Surface area =  (a+b) SquareRoot(8ab) / 3.

17.8 p1175 ex 10:

                x
For F(x,y,z) = e  (x,y,z):

                 x      x
curl F =  (0,  -e  z,  e  y)

div F  =   (x+3) exp(x)


17.9 p1181-2 ex 2:
If a vector field F is  F(x,y,z) = [(1-x),  (2-y),  (3-z)],  then
the flux out of the unit ball  |(x,y,z)|<1  is gotten as follows.
 (1,2,3)  leads to 0 net flux (inward and outward fluxes exactly cancel on 
                               the other side of the sphere)
          or (second way):  div(1,2,3) = 0   (since constant so all derivatives 0)
                            so divergence thm says flux is 0 = teriple integral of 0 dxdydz.
 (x,y,z)  leads to net flux  4 pi
    because either:  this vector (on the sphere surface) always has unit length 
                     and always is outward-pointing, so we get the surface area
                     as the flux-integral (and the surface area is   4 pi   since radius=1)
    or (2nd way):    use the divergence theorem:
                        div(x,y,z)=1+1+1=3  so  solid triple integral of divF  dxdydz =
                                                = 3*volume=3*(4 pi/3)=4 pi.
    So COMBINED ANSWER is -4 pi.

17.9 p1181-2 ex 22
Use the fact that  div(x,y,z)=3=1+1+1  everywhere.
Use the divergence theorem.
CONCLUDE:
  volume(R) = triple integral      1  dxdydz
                for (x,y,z) in R

     =                 surface integral        (x,y,z)/3 . dAreaVector 
   by divergence thm   (x,y,z) on surface of R

     =            (1/3) int int        r(u,v) . (dr/du X dr/dv) du dv.

where r = (x,y,z)  is a vector-valued function of u and v.

Also we could use the fact div(x,0,0)=1 instead, then would get

     =                 surface integral        (x,0,0) . dAreaVector 
   by divergence thm   (x,y,z) on surface of R

     =                  int int        x(u,v) . (dr/du X dr/dv)  du dv.
                                                               1

This offers a way (in fact 2 ways, since I gave a 2nd solution) to 
compute volume as a surface (2D)
integral - instead of a triple integral
as would normally be expected.

Similarly (but in 1 dimension less)
we can compute AREA as a curve (1D) integral instead of a 
double integral as would normally be expected, via

  area(R) = double integral      1  dxdy
            for (x,y) in R

     =    (1/2) curve integral        (x,y) . (unit outward normal vector) dArcLength 
   by          (x,y) on boundary of R
divergence thm  


     =    (1/2) integral     (x(s),y(s)) . (unit outward normal vector(s)) ds
       
where x(s), y(s)  is a parameterization of the curve that bounds R.

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