My Math Forum changing order of integration

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 November 30th, 2012, 12:37 PM #1 Newbie   Joined: Nov 2012 Posts: 1 Thanks: 0 changing order of integration Hi! I'm solving this problem and I'm not sure how to solve it I have definite triple integral of function f(x,y,z). It's domain is set $M = \{0 I would like to get function g(z) = \int\int f(x,y,z) dxdy but I'm quite confused with borders of integration (and od course domain of the function g) -> Could you give me some help? Thanks in advance! --- -> and here goes my idea (but I'm not sure about it): - for each$z$and$y$I'm able to get an interval for$x$:$\max \left(0, \frac{z-2y^2+3ay}{y+a} \right) < x < \min \left( a , \frac{a^2 - 2y^2+2ay + z}{y} \right)$- therefore I can get function \int_max(...)^min(...) f(x,y,z)dxdydz = [h]_{\max(...)}^{\min(...)}$ - for each $z$ I'm able to say which values functions min/max takes - so the function g(z) could be sum of integrals of k(y,z) (the number of integrals and their domains depends on values of min/max function..) -> so, is it good idea or really wrong way of thinking about this problem? -> and what about domain of g(z)? (could it be [min_(x,y)f(x,y,z);max_(x,y)f(x,y,z)] ?) --- Once again thanks in advance! (and sorry for my English.. ). Have a nice day! Doxxik
 December 14th, 2012, 05:40 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: changing order of integration All you are told about x and y is that 0< x< a and 0< y< a so that the integrals, with respect to x and y, are from 0 to a. The result will be a function of z and its domain will be the possible values of z. What are the maximum and minimum values of the two functions bounding z on the square 0< x< a and 0< y< a?

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