Re: Double integral problem

Quote:

Originally Posted by

**richardsim10** I'm trying to prove that J=>0. J is defined as the double integral over region R of f(x,y)dA with R = [-2,2]x[2,4] and f = (x^5+y^5)^(1/5). I know I have to set f smaller and bigger than my region R, but then how can I translate that to the integral?

The Function will have its absolute minimum on the recangle

$\displaystyle [-2, 2] \times [2,4]$

at the point $\displaystyle (-2,2)$

and it will have its absolute max at the point

$\displaystyle (2,4)$.

This gives.

$\displaystyle 0 \le f(x,y) \le 2\sqrt[5]{33}$

But more importantly why does it have these extreem values

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Re: Double integral problem

Function as a surface:

$\displaystyle f(\text{x},\text{y})\text{=}\left(x^5+y^5\right)^{ \frac{1}{5}}$

http://mathhelpforum.com/attachment....1&d=1348807314