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<9>   Problem set 26, Problem 74, Schaum's p. 134
     This problem was assigned by mistake. It requires material in section
     5.2 (Green's formulas, especially the 3rd) which will not be assumed
     known until after Lecture 27.
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<8>   Stokes: line integrals instead of surface integrals
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Subject : Re: HW #21 4.9.10
----- Message Text -----
Here the trick is to use Stokes' thm. so you end up computing a surface
integral instead of a line integral. The boundary is an ellipse,
as it is the intersection of a cylinder and a plane. You may find the
problem easier in cylindrical coordinates. The ellipse is at a constant   
distance from the z-axis, since it lies on the cylinder.
On the other hand, the surface is the plane y=z, with easy normal and
easy limits. And curl F is simple...
 Hope this helps

On Wed, 9 Apr 2003 
> I am very lost doing this problem...  I started out with finding curlF
> to use Stoke's theorem integral of curlF*dS... however finding dS is 
> alluding me, should I cross partial r wrt phi partial r wrt theta and
> then dot with curlF ( I found to be i-j+k), however, in order to    
> integrate, this doesn't work bc r is varying over the ellipse.

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<7>   Converting between coordinate systems: an example 
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      Consider the conversion between cartesian and cylindrical coordinates
    in 2D:
  (I will write r for rho and t for theta; d will mean partial derivative)
   x = r*cos(t) ; y = r*sin(t)

   We can compute dr/dx, dt/dx etc by differentiating the above expressions 
  wrt x (and similarly for y):
   
   1 = (dr/dx)*cos(t) - r*sin(t)*(dt/dx)
   0 = (dr/dx)*sin(t) + r*cos(t)*(dt/dx)
from which you can solve for (dr/dx) and dt/dx):
 (dr/dx) = cos(t)
 (dt/dx) = (1/r)*sin(t)

You can use similar method to compute dr/dy, dt/dy, and the same thing
can be also employed to compute the terms necessary for the conversion to
spherical coordinates
(now differentiate wrt x,y, and z the relationships:
 x = r*cos(t)*cos(p); y = r*sin(t)*cos(p); z = r*sin(p) 
and solve for dr/dx, dt/dx, dp/dx and similarly for y and z). 
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<6>      Homework 12-tips
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     Problem 4.5.3(b)
     Gradient properties
 (2) gives direction of gradient
 (3) gives magnitude
   For field phi = -1/R, you know direction of increase (this gives you
   unit vector) and you know magnitude (max increase: what is it?).
   Combine to get gradient -- the function phi depends on R alone, so it stays 
   constant when R is constant
     Problem 4.5.8(b)
   Here assume the field is a gradient (i.e. it's curl is zero) and proceed
   with the systematic approach we followed in class:
   partial phi/partial x =.... --> x = ... + C(y,z)  etc.
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<5>     SIMPLY CONNECTED DOMAINS
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%****"Simply connected" means that 
ANY LOOP CAN BE SHRUNK DOWN TO A POINT WIHOUT LEAVING THE DOMAIN.
This means that the interior of a donut is not simply
connected, while the region between two concentric spheres IS simply
connected. I'll say more about this in class.

> In case you don't intenend to cover this in class, I need some help
> understanding simply connected domains. The book describes them as
> domains without any holes, but it also says that the region between two
> spheres is a simply connected domain, even though it has a hole in the 
> middle of it. Why is the region in (x^2 + y^2 > 0) not a domain while
> the region in (x^2 + y^2 + z^2 > 4) is a simply connected domain?
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<4>           Subject: Re: Problem 4.1.5
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> This problem seems very straight forward however, I am trying to solve
> it and coming up with a much different answer than the book, even when
> I use the F*dR= f(Q)-f(P) (from section 4.3, suggested by the solutions
> appendix)  I come up with an answer of 31, and for the line integral
> method I keep getting 103/6.  The book says the answer is 40.  I'm not
> sure what I'm doing wrong.
>
> For example, when I solve the integral 20t2-15t+3, I get ( 20/3t3-
> 15/2t2+3t) from 0 to -1 and so on...

************************* seems like two different errors:
on the line integral, probably the path parametrization
(the equation I get looks different).
  The path is a straight line from A(1,0,2) to B(3,4,1)
with
 R(t) = A + t(B-A) and dR/dt = (B-A) = v
 /
 | F*v dt = v1(2xy)+v2(x^2+z)+v3(y) then plug in x(t), y(t), z(t)
/
   and integrate with t from 0 to 1. I also got 40.
For the potential method, check  curl_F = 0 (it is), and one can find
scalar func. phi(x,y,z) such that grad_phi = F then
phi(B) - phi(A) gives you also 40 (be sure to check that grad_phi really
does equal F for the phi you are using!)
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<3>               Sebject: math website access problems         2/11/03;13:30
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We are currently experiencing a problem with access to the math.unm.edu
website from the following locations:

   ESC pod (confirmed)
   Zimmerman (reported)
   CSEL (reported)

CIRT Networking is working on the problem.
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<2>               Subject : Re: Math 311 Question               2/10/03;18:25
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> I have one brief question concerning part (b)in problem 13.1.1 on page
> 112.
> If R = x + y + z and f = 1/|R|,
> then shouldn't grad(f) = (1/|R|)(2x + 2y + 2z) = 2R/|R|?
> The book says that the answer is -R/|R|^3
******** |R|=(x^2+y^2+z^2)^(1/2)  
   |R|^(-1) = (x^2+y^2+z^2)^(-1/2)
  (|R|^(-1))' = (-1/2)(x^2+y^2+z^2)^(-1/2-1)*(x^2+y^2+z^2)'=... 
Another way:           
  f(r) = r^-1  
  df/dx = df/dr*dr/dx (d = partial here)
        = -r^(-2) * dr/dx
  r = (x^2+y^2+z^2)^(1/2) => dr/dx = (1/2)(x^2+y^2+z^2)^(-1/2)*2x =
                                   = x/r etc.

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<1>  Homework questions Monday 2_3_03  (random questions I got+answers)  
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>
> In problem # 11, it asks us to simplify this expression:
>
> [Ax(AxB)]xA*C
>
>I think this last part is asking us to cross a vector [Ax(AxB)] and a
>scalar A*C...  What I am coming up with is something like this:
>(A*A)(A*C)(A-B)
>Am I on the right track here?
************* Careful, you cannot cross a vector and a scalar!
An expression like:   AxB*C must be evaluated as (AxB)*C
i.e. a proper parenthesis is implied when
an expression makes sense only one way and not the other.
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> Questions 2,3 and 5 on pg. 85 of the text.
> To find the arc length
> s = ^[(dx'dt)^2 + (dy/dt)^2 + (dz/dt)^2 ]^(1/2) dt
> where s is the arc length and the integral is from (t=lower bound) to
> (t=upper bound)
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> Question 2:
> I found that R' is:
> x' = cost +tsint + cost************--->x' = cost +tsint - cost
> y' = -sint +tcost +sint
> z' = 2t
> so the |R'|, after squaring and canceling the terms, is
> (4 + 5t^2)^(1/2)********----> Different formula, see above
****************               (cancellations)!
> I am not sure how to take the integral of this to find s. Do I need to
> use integration by parts?
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> Question 3:
> Using the formula the text specifies:
> dx^2 + dy^2 + dz^2 = (dx^2)(1/4^^2) + (cost)^2 + (sint)^2
> The integral of that would be [(1/4^^2) + 1]^(1/2)
> Does this seem right? The book does not agree with me. It places the
> 4^^2 in the numerator, but I do not see how to do that.
***********  use algebra: (1/a)^2+1 = (1+a^2)/a^2
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> Question 5:
> For R':
> x' = -(e^t)sint + (e^t)cost
> y' = (e^t)cost + (e^t)sint
> z' = 0
> When I take the magnitude of R' I get a nasty number:
> |R'| =[ 2(e^2t) - 4(e^4t)sintcost ]^(1/2)
> This does not seem right at all and I have no idea how to find the
> integral for this.
*******if you do the algebra carefuly, the nasties cancel out; try again!
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