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These are replies submitted by gulliet

Thank you for the clarification between unevaluated expression and inert representation. (I was not aware of the latter.) I have just read the relevant help pages and learned quite a few things.
Hi, My understanding is that multiple assignments require to use sequences of arguments on both sides of the assignment operator. The LHS must be a sequence of names, i.e. names enclosed within parentheses, whereas the RHS must be a sequence of values of same length that the LHS. Moreover, since the constant NULL evaluates immediately to tne null sequence, we must use the inert form of the expression on the RHS. (I hope that I am shedding more light than obscurity!) > a, b := 1, 2; 1, 2 > a, b := 'NULL', 'NULL'; NULL, NULL > a; > b; > HTH, -- Jean-Marc
To solve a system of linear equations, you can use the command "solve" as in solve({y = 10*x, y = 5*x-4}, {x, y}); / -4 \ { x = --, y = -8 } \ 5 / Regards, -- Jean-Marc
Scott, Many thanks for the short delay to answer my question (I had started wondering what was wrong with my account :-) Regards, -- Jean-Marc
Whenever I try to listen to the recording (several times today, at different hours, and being logged into my account), an error web page is displayed with the following message: "Content Not Available: Your memberhip account does not have sufficient security rights to access this content." How can I access the podcast? (Note that it is not available on the page -- Jean-Marc
Thanks a lot for the detailed and swift answer. Having access to the ACM Portal web site via my own institution, I shall choose this path. Best regards, Jean-Marc
That sounds very interesting. Could you, please, post the bibliographic reference for this paper (or a link to access it online, if available)? Regards, Jean-Marc
You are right. Indeed, having thought more carefully about the initial problem (and done some hand drawings), I released the volume that is common to both cylinders is not a sphere, but a more complicated shape. Right now, I can think about nothing else but integration to solve this problem. Regards, Jean-Marc


Your answer to the first question is almost right. You made a mistake in calculating the volume of the sphere that is common to both cylindrical holes: b is the diameter of the hole, so the radius is b/2 and this is this value that you must use to compute the volume of the sphere (this what you did to compute the volume of one cylinder).

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