My Math Forum Find the average value of the function F(x,y,z) = x^2y^2z^2

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 February 10th, 2014, 08:48 PM #1 Newbie   Joined: Feb 2014 Posts: 5 Thanks: 0 Find the average value of the function F(x,y,z) = x^2y^2z^2 I know it has something to do with arc length and I know how to find arc length for the curve, but I am confused the average value part. Thanks.
 February 11th, 2014, 03:38 AM #2 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Re: Find the average value of the function F(x,y,z) = x^2y^2 In 1D the average value of a function f(x) is defined by $F_A= \frac{1}{b - a}\int_a^bf(x)dx$ Can you see how this is generalized for 3D?
February 11th, 2014, 05:52 AM   #3
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Re: Find the average value of the function F(x,y,z) = x^2y^2

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 Originally Posted by stacigurl12 I know it has something to do with arc length and I know how to find arc length for the curve, but I am confused the average value part. Thanks.
Unless you are asked for the average value on some curve, no, it has nothing to do with finding the arc length. But the problem, as stated, makes no sense since there is no set specified and the values of this function all space are unbounded. To find the average value of a function on a given set, divide the integral of the function over that set by the "measure" of the set (length for a one-dimensional curve, area for a two-dimensional region, volume in three dimensions). For example, if the problem were to find the average of f(x, y, z_ over the parallelpiped, a< x< b, c< y< d, e< z< f, the average value would be $\frac{\int_a^b\int_c^d\int_e^f f(x, y, z) dzdydx}{(b- a)(d- c)(f- e)}$.

 February 11th, 2014, 08:42 AM #4 Newbie   Joined: Feb 2014 Posts: 5 Thanks: 0 Re: Find the average value of the function F(x,y,z) = x^2y^2 Whoops the limits of integration for the curve were 0 to 3qrt2.
 February 12th, 2014, 05:42 PM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Find the average value of the function F(x,y,z) = x^2y^2 Your function has three variables so just saying "0 to $\sqrt{3}$ makes no sense. Which variable is "0 to $\sqrt{3}$"? Or are x, y, and z give by some parametric functions with the parameter from 0 to $\sqrt{3}$?

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