How can you calculate (or convert) the $\rm Wh$ of a capacitor whose energy is given in Farads? When trying to compare the energy in a battery to the energy in a capacitor, the units don't match up. How can one compare a battery whose Ah are 10 and Voltage is 3 (for a total of 30 Wh) to a capacitor whose Farads is X and voltage is Y?
 A: To begin, let's call things with their names (no offense). What is measured in Farads is capacitance $C$. What is measured in $Ah$ is the charge that can be stored in a battery or a capacitor.
From the definition of capacitance, the charge on the walls of a capacitor with capacitance $C$ and potential difference $V$ is 
$$ q = CV $$
so you obtain the value you're interested in. However, if $C$ is in Farad and $V$ is in volts, $q$ will be measured in Coulombs.
$$ \text{Ampere} = \frac{\text{Coulomb}}{\text{Second}} $$
so
$$ 1\text{ Ampere}\cdot\text{hour} = \text{1 Coulomb}\cdot\frac{\text{hour}}{\text{second}} = 3600 \text{ Coulombs}$$
So, in your notation, the stored charge is $XY/3600$.
A: Energy in a capacitor is $CV^2/2 = QV/2$ because its voltage starts at 0 when uncharged (unlike a battery, where the voltage is more or less constant).  See note 8 in the Maxwell doc you referenced, which uses this formula and then converts from joules to watt-hours in the same fashion correctly described by Bzazz's answer.
A: V=J/C
30Wh x 60min x 60sec = 10800Ws(or J) / 3v = 3600C
A: In my real-world example I had to estimate the same for Maxwell Nesscap UltraCapacitors 2.7V 3000F Cells 18 Pack ESHSR-3000C0-002R7B5
(For a trailer whose 10000W inverter occasionally needs a bigger punch to get the Air Conditioning unit rolling, and also healthier for deep-cycle batteries to smooth out their solar charging)
So applying **J = C x V^2 /2 **
For one 2.7V capacitor:
2.7V^2 x 3000F /2 = 7.29 x 1500 = 10935‬ Watt-seconds
 / 3600 seconds per hour
 = 3.0375‬ Wh


x 18 for the whole pack = 54.675 Wh

Of course, I'll need to get some high-current buck boost DC-DC converter to be able to access most of that stored energy, and feed its ouput to the controller that charges lead-acid batteries.  Otherwise the capacitors wouldn't discharge to near 0V and trap the leftover charge as unusable.
