From zeilberg@math.rutgers.edu  Thu Dec  7 16:06:05 2006
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Date: Thu, 7 Dec 2006 16:05:46 -0500 (EST)
From: Doron Zeilberger <zeilberg@math.rutgers.edu>
Message-Id: <200612072105.kB7L5kN23230@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: 1313


Dear Students,

  Below is the answer to today's "quiz". About 97% got
it perfectly! 


                    See you Monday,

                            Dr. Z.

P.S. Some reminders:

1. You can take the "deal eligibility" test 
on Either  7:40-8:30am Or 9:10-10:00 am, Monday Dec. 11, 2006,
at Hill 705.

2. This coming Monday (Dec. 11, 2006) is the last-day-of-classes
and will dedicated to review. It is very important, so even if you
are very sick (unless you are contagious) try to come.

3. Optional (additional) review is Thurs., Dec. 14, 2006, 8:40am-1:20pm,
ARC 103.

4. There is no "practice final" since all the Homework and
the handouts and the previous exams and the "attendance quizes"
(soon to be posted) should give you plenty of practice.

---------------Solution to "quiz" of Dec. 4, 2006----------

Q. By using the divergence theorem, or otherwise,
compute 

IntgeralSign IntegralSign curl F.  dS 

where

S is the sphere (x-1)^2+(y-2)^2+(z+4)^2=25, and
F(x,y,z)= (2x+3yz+z^3)i +(x^2-4y+xz) j+ (3x^2+7y+2z)k

Solution: div(F)=2+(-4)+(2)=0, hence

IntgeralSign IntegralSign  IntegralSign_E (div F)=

IntgeralSign IntegralSign  IntegralSign_E 0 =  0.


Ans.: 0

Comment: since we lucked-out and the divergence happened to
be 0, we don't even have to think about E.

-----------------------