From zeilberg@math.rutgers.edu Mon Dec 4 19:42:00 2006 Received: (from zeilberg@localhost) by math.rutgers.edu (8.11.7p2+Sun/8.8.8) id kB50fbO18332; Mon, 4 Dec 2006 19:41:37 -0500 (EST) Date: Mon, 4 Dec 2006 19:41:37 -0500 (EST) From: Doron Zeilberger Message-Id: <200612050041.kB50fbO18332@math.rutgers.edu> To: zeilberg@math.rutgers.edu Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu Subject: Today's "quiz" Status: RO Content-Length: 1374 Dear Students, Below is the answer to today's "quiz". About 70% got it perfectly! Stokes' Theorem is one of the most difficult topics, and you should be very proud that you got it. The next (and last) topic, the Divergence Theorem, is not as hard. See you Thurs. Dr. Z. P.S. During Thurs. early morning tutoring, I will continue to review difficult algebra ---------------Solution to "quiz" of Dec. 4, 2006---------- Q. Use Stokes' Theorem to evaluate IntgeralSign IntegralSign curl F. dS where F= xe^(yz) i + (xyz+1) e^(y^3z) j + z e^(4xy) k and S is the hemisphere x^2+y^2+z^2=4, y>=0, oriented to the right. Solution: C is a circle obtained by plugging-in $y=0$ into the equation of the sphere, so we get x^2+0^2+z^2=4 which is the same as x^2+z^2=2^2 This is a circle in the xz-plane of radius 2, so the parametric equation is x=2cos t z=2 sin t (0<=t<=2 pi) and of course y=0 so r= < 2 cos t, 0, 2 sin t > , 0<=t<= 2 pi Next we find dr= < -2 sin t , 0, 2 cos t> dt Not we plug-in into F x=2 cos t y=0 z=2 sin t and get F=<2 cos t e^0, 1 e^0 , 2 sin t e^0>= <2 cos t , 1 , 2 sin t > The integrand is F. dr= (2 cos t)(-2 sin t)+ (1)(0)+ (2 sin t) (2 cos t)= -4 cos t sin t + 0 + 4 sin t cos t =0 and the final answer is IntegralSign_0^{2 pi} 0 dt = 0 Ans.: 0 -----------------------