Erickson refers to this as indirect power. See page 5 of ECEN5807 Lecture 3.
Given on that slide: $$P_{indirect}=(DV_g)(D'I_L)$$
$$I_L*D'=I_g$$
$$P_{indirect}=D*V_g*I_g$$
$$P_{direct}=P_{in}=P_{out}=V_g*I_g$$
$$P_{direct}/P_{indirect}=1/D$$
This is close to what Mohan found, but I'm getting the inverse. Doing a similar process for a boost I get the same inverted result (1/D').
I hope this helps.
EDIT: It makes more sense for the switch utilization to be \$P_T/P_O\$ because the power "through" the transistor must always be less then the output power (keeping the ratio below 1.0). I think this is a typo in the book.
If this is true the equations for Buck:
$$P_T/P_O = D $$
For Boost:
$$P_T/P_O = D'= (1-D)$$
For Buck Boost:
$$ P_T/P_O = D*D'$$
The values are different in Table 6.1 of Erickson because (I believe) he's using the RMS values of the switched Voltage/Current so these values can be related to switch loss.
This means, for example, that conduction losses of the transistor in a buck converter are related to \$\sqrt{D}\$.
Galvanic isolation means that no (significant) current can flow between two parts of the circuit (here, the input side and the output side). It doesn't mean the output voltage of a circuit has to be lower than the input voltage.
Galvanic isolation is very important in mains power supplies, because a failure to galvanically isolate the two sides may put the low-voltage output side of the power supply (and whatever it's powering) at mains potential.
This galvanic isolation is the subject of safety regulations, and can be a real concern with low-quality power supplies. See for example this teardown of a fake Apple charger, or this teardown of a USB charger (that is actually putting mains on the USB ports!)
Galvanic isolation in step-up converters
In a step-up converter (and indeed, anywhere else), galvanic isolation still means the same thing: no (significant) current can flow between two parts of a circuit.
One way this can be relevant is if the input side of your circuit is (or can become) earth-referenced. With galvanic isolation, there is only a voltage across the two output terminals, but the output terminals are floating with respect to earth ground. A shock risk should only exist when someone touches both terminals.
However, if the galvanic isolation fails, one of the output terminals may become earth-referenced, putting the other terminal at a high voltage relative to earth ground, creating a shock risk when just the one terminal is touched!
Best Answer
If you look at literature, you will find that switching converters operate by charging and discharging an inductor. The main building block is composed by an inductor and two switches, connected in such a way that one switch charges the inductor and the other discharges it.
Now, you have to give a signal to those switches, so one of them must be a controllable switch (aka transistor; usually MOSFETs are used).
The other switch is usually substituted by a diode, because its usual behavior (allowing current to flow in just one direction) is enough to guarantee the proper functioning of the device.
Anyway of course you can remove the diode and substitute it with another MOSFET. Your control logic then becomes more complicated (it needs to control two switches), bit the performances increase (the voltage drop on the diode, usualli 0.3V, is now reduced to almost 0). These converters are called synchronous converters (see here for more infos).
The maximum reverse voltage depends on the device. For buck converters the reverse voltage on the diodes is roughly Vin, while on boost ones it is Vout.