From zeilberg@math.rutgers.edu  Thu Oct 26 15:40:38 2006
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Date: Thu, 26 Oct 2006 15:40:02 -0400 (EDT)
From: Doron Zeilberger <zeilberg@math.rutgers.edu>
Message-Id: <200610261940.k9QJe2925188@math.rutgers.edu>
To: zeilberg@math.rutgers.edu
Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu
Subject: MathIsFun. Today's "quiz"
Status: RO
Content-Length: 3001


Dear Students,

   Most people got the first question right. Many people got the second question right, and
most people were on the right track.


--------Solution to the "quiz" of Oct. 26, 2006---------------------

1. Find the Jacobian of the transformation

x=u^2+v^2

y=u^2-v^2


Sol. to 1:

J= | x_u x_v  |  =  | 2u    2v |
   |          |     |          |
   | y_u y_v  |     | 2u   -2v |
   

=(2u)(-2v)-(2u)(2v)= -4uv-4uv= -8uv


Ans. to 1.:  -8uv



2. Evaluate the following iterated integral by converting it into polar coordinates

IntegralSign_{-1}^{1} IntgeralSign_{-sqrt(1-y^2)}^{sqrt(1-y^2)} (x^2+y^2) dx dy


First Step; Write down the region in Type II form

D={(x,y)| -1<=y<=1, -sqrt(1-y^2)<= x <=sqrt(1-y^2) }

Second Step: Draw this!

The projection on the y-axis of our region is from y=-1 to y=1. A horizontal cross-section
(at y) ranges from  x=-sqrt(1-y^2) to x=sqrt(1-y^2) } . These are both parts of
the circle x^2+y^2=1. So it turns out that our region is the WHOLE circle
center origin radius 1.

(Note: if instead of -1<=y<=1 it would have been 0<=y<=1 then it would be the UPPER semi-circle
if instead of -1<=y<=1 it would have been -1<=y<=0 then it would be the LOWER semi-circle

if instead of -sqrt(1-y^2)<= x <=sqrt(1-y^2) }it would have been 0 <= x <=sqrt(1-y^2)  then it would have been the RIGHT semi-circle

if instead of -sqrt(1-y^2)<= x <=sqrt(1-y^2) }it would have been -sqrt(1-y^2)<= x <= 0  then it would have been the LEFT semi-circle )

Third Step; Express the region in POLAR form:

D={ 0<=theta<=2Pi, 0<=1<=r}

Fourth Step: Set up the integral in POLAR replacing x^2+y^2 by r^2 (and if necessary[not in this problem]
x by rcos(theta) and y by rsin(theta)) and dxdy by r dr dtheta


IntegralSign_{0}^{2Pi} IntgeralSign_{0}^{1} r^2 r dr dtheta=
IntegralSign_{0}^{2Pi} IntgeralSign_{0}^{1} r^3 dr dtheta

Fifth Step: Evaluate it.


Ans.= IntegralSign_{0}^{2Pi} [IntgeralSign_{0}^{1} r^3 dr ]dtheta

We first do the Inner integral


IntgeralSign_{0}^{1} r^3 dr =(r^4/4) |_{0}^{1}=1/4

Outer integral

IntegralSign_{0}^{2Pi} [IntgeralSign_{0}^{1} r^3 dr ]dtheta

=IntegralSign_{0}^{2Pi} [1/4] dtheta= theta/4 |_{0}^{2Pi} =Pi/2


Ans.: Pi/2.


COMMON MISTAKES (watch out!):

1. People replaced x^2+y^2 by r INSTEAD of by r^2

2. People replaced dx dy by dr dtheta INSTEAD of  r dr dtheta

3. In the Polar description of D, they had 0<=theta<=Pi INSTEAD of 0<=theta<=2Pi

4. In the Polar description of D, they had -1 <=r<= 1 INSTEAD of 0<= r <=1

 Warning: THIS IS A HUGE BOOBOO (enoght to lose credit for the whole problem!)

r is "distance from the origin" DISTANCE IS NEVER NEGATIVE, the following is nonsense

NegativeNumber <=r<= ...

5. In the Converted integral in polar they had vestiges of x and/or y.
That's another big BOOBOO.

 It is either RECTANGULAR (ONLY featuring x and/or y)

or POLAR (ONLY featuring r and/or theta)

If it has both mixed up it is MEANINGLESS!!!! (and you would get 0 points for this problem).

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