Calcs look ok, yes. Good work there.
Some additional points:
1) You might want to include a diode across the regulator from output to input. This protects the regulator if the output rises above the input, especially on turn-off when the load might still be connected and it has some capacitance that has not discharged fully. This is definitely needed for any bench supply, but might not be needed in a supply that has a fixed purpose and you control the load.
2) 15 mA might be too bright for I
led, especially if it's a modern LED. You can usually run modern LEDs at 5mA or so. The LED resistor can be the next step up maybe 330 (8 mA) or 470 (6 mA) ohms. It really depends on what you like. I personally don't like bright LEDs
3) consider the power dissipated in the regulators. I always use the max values for this, disregarding the ripple. So you will have about 12V max input with 5V output at 440mA, that's P=VA = (12-5) * 0.440 = 7 * .440 = 7 * .5 = 3.5W
3.5W P
d .. Can it work in free air? (I know it cannot, but let's do the calculation). The datasheet says the TO-220 is 50 degC per watt Junction to Air. So 50 degC/W *3.5W = 175 degC rise above ambient. If ambient is 35-40C, then the junction will be at 40+175 = 215C. The datasheet also says the maximum junction temperature is 125C so clearly 3.5W in free air will not work. You need a heatsink.
So if we say maximum ambient will be 40C and maximum junction temperature allowed is 125C , then the maximum temperature rise that you can tolerate will be 125C - 40C = 85C. So the power dissipated across the thermal resistances cannot allow a rise of more than 85C.
TR = thermal resistance
TR
max = 85C / 3.5W = 24.3 C/W max.
TR
j-c = junction-to-case = 5 deg C/W (from datasheet)
TR
paste = 0.5 deg C/W (varies, but you can use .2 to .5 C/W for paste)
So TR
heatsink = 24.3 - 5 - 0.5 = 18.8 C/W.
So you need a heatsink with a thermal resistance of 18.8 deg C / W.
This one is 20 deg C/W, it will probably work if your ambient is not 40C as I calculated (or in an airflow, where it might be about 11 deg C/W)
However, that's the absolute minimum heatsink you need, because using that one could run the regulator junction at it's max temperature of 125C. We don't like to do that, so let's try to keep the junction at 100C now.
redoing the calculations:
100 - 40C = maximum temperature rise of 60C
60 C / 3.5W = 17 deg C/W
17 - 5 - 0.5 = 11.5 deg C/W for the heatsink. The above heatsink will probably work, in an airflow (fan).
However, if you are in free air, you might want
this one which is 10.5 deg C/W in free air.