Hello, graeme87!
Welcome aboard!
Here's the first one . . .
Formula: . ^2 + \left(\frac{dy}{dt}\right)^2} \,dt)
We have: . ![\begin{array}{ccc}\dfrac{dx}{dt} &=& 1 -\cos t \\ \\[-3mm] \dfrac{dy}{dt} &=& \sin t \end{array}](http://latex.codecogs.com/png.latex?\begin{array}{ccc}\dfrac{dx}{dt} &=& 1 -\cos t \\ \\[-3mm] \dfrac{dy}{dt} &=& \sin t \end{array})
^2 + \left(\frac{dy}{dt}\right)^2 \;=\;(1\;-\;\cos t)^2 + \;\sin^2\!t \quad\Rightarrow\quad 1 \;-\; 2\cos t \;+\; \underbrace{\cos^2\!t + \sin^2\!t }_{\text{This is 1}})
. . . . . .  \;=\;4\left(\frac{1-\cos t}{2}\right) \;=\;4\sin^2\tfrac{t}{2} )
Hence: . ^2 + \left(\frac{dy}{dt}\right)^2} \;=\;2\sin\tfrac{t}{2})
We have: . ![L \;=\;\int^{2\pi}_0 \!\!2\sin\tfrac{t}{2}\,dt \;=\;-4\cos\tfrac{t}{2}\,\bigg]^{2\pi}_0 \;=\;\bigg(-4\cos\pi\bigg) - \bigg(-4\cos 0\bigg)](http://latex.codecogs.com/png.latex?L \;=\;\int^{2\pi}_0 \!\!2\sin\tfrac{t}{2}\,dt \;=\;-4\cos\tfrac{t}{2}\,\bigg]^{2\pi}_0 \;=\;\bigg(-4\cos\pi\bigg) - \bigg(-4\cos 0\bigg))
. . . . . . (\text{-}1) - (\text{-}4) \;=\; 4 + 4 \;=\;\boxed{8})
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
That is a surprising answer . . .
We have a circle . . . and we are rolling it.
We would expect to see 
in the arc length.
Instead, the answer is an integer.
Fact: The arc length of one arch of a cycloid is equal to the perimeter
. . . . of the square circumscribed about the generating circle.
Fact: The area under one arch of a cycloid
. . . . is 3 times the area of the generating circle. Code:
* * *...
* * *::*..
* * *:::::*
* * *::::::::*
* :::::::::::*
* * πrē *::::::::::::*
πrē ::::: πrē::::::
* * *:::::::::::::::*
* *::::::::::::::::::
* *::::::::::::::::::::
* - - - - - - - - - - - * * * - - - - - - - - - - - *
That shark-fin-shaped region is equal to the generating circle.
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