5v regulator getting too hot

When I use a Xbee shield on top of a diecimila board the 5 v voltage regulator is getting really hot.
is this a problem?
what can I do to improve this situation that I do not like?

It's getting hot because you are requiring it to dissipate too much power. This can be either because of too much current or the regulator having to drop too much voltage.

What voltage are you feeding the system with. Anything over 7V is generating more heat than needed.

See:- Power & Heat

I've powered my Arduino + Xbee shield from a 12volt battery for a long period of time and it doesn't get hot with me.
But for I'm only using the Xbee 1mw and not the 50mw pro, I guess you are drawing more power than me.

Power Consumption:
XBee - 55mA
XBee Pro - 215mA

I had the same problem with 12V power supply and Ethernet Shield...
I have solved the problem using a Traco power supply on a usb connector just near the board.

It can be great to have arduino boards with a heavier power supply and optionally an heat sink

Best regards

Theres 12v power supplies and "12v" power supplies, One I used with an ethernet shield made the regulator get hot, when I actually measured it rather than believing what it said on the label I found it was kicking out almost 16 volts. It only got mildly warm when actually fed with 12 volts.

thanks a lot for all these answers.
i am using a 12v battery to power the board.
on the Xbee shield there is the possibility to switch the power supply from the icsp to an external connection (5v i suppose?). so, i am going just to do that: not to draw extra power from arduino.

Another trick to reduce the power dissipated by the regulator is to insert a resistor in between the battery and regulator. Adjust the value so you see maybe 7 or 8 volts ahead of the 5V regulator. of course the resistor will now get hot, make sure it is big enough to handle the wattage. Start with 50 ohms ( like 2 100 ohm 1/2 W resistors in parallel ) If the total draw is .1A then the voltage drop across the resistor will be IxR or 5 volts and the power shifted to the resistor is V squared over R or 1/2 W so there is plenty of margin with 2- 1/2W resistors. If the current differs from 100 ma then you have to adjust the resistor values so this 'trick' is application specific. More current, add a 3rd 100 ohm resistor in parallel, etc.