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From: JEMebius on 8 Aug 2010 12:46 Michael Robinson wrote: > "Michael Robinson" <nospam(a)billburg.com> wrote in message > news:zyy7o.50983$f_3.12262(a)newsfe17.iad... >> I have this surface integral I can't get to come out right. >> I posted it in a place where I can use math characters. Here's a link: >> http://burlington.craigslist.org/forums/?forumID=21 >> if you have time to look at it. Otherwise I will ask my math teacher >> tomorrow. >> It is a problem from a book, but it's not homework. I'm practicing. > That link only takes you to the forum. The post is titled > "surface integral," posted early morning 08/08. > > A quick and dirty reply... I calculated the book's result pi/60.(391.sqrt(17) + 1) = 84.463523... and your result pi/240.(391.sqrt(17) + 1) = 21.115880... One shows easily that the result 84.463523 cannot be correct: consider a circular cylinder of radius 2. Its perpendicular cross-section is 4.pi = 12.57...; its cross-section with the maximal slope of the part of the paraboloid z = x^2 + y^2 that lies inside the cylinder, is 4.pi.sqrt(17) = 51.78... . The area of the paraboloid within the cylinder is certainly smaller than this amount. In conclusion I am afraid that your result is correct. I did some sketchy calculus to confirm your result, but I am not yet ready. Good luck: JOhan E. Mebius |