Some polar calculus....
I like that curve we looked at in class: r(t)=ln(t). And since I noticed that little closed loop in the middle, I wanted to know how long its perimeter was, and what the enclosed area was. Here’s the original think, that got me excited:

t0=0
t1=100
r[t_]:=Log[t]
Manipulate[
PolarPlot[r[t],{t,t0,tstop}],
{tstop,t0,t1}
]

Now I need to find where the two thetas are that lead to the intersection that closes the curve. They’ll be two values of theta, separated by Pi, that have r values of opposite sign but equal size: that is, they’ll satisfy the equation 
ln(t)=-ln(t-Pi)

Solve[Log[t]==-Log[t-Pi],t]

If we simplify it ourselves, using logs, we introduce an additional solution:

Solve[t^2-t*Pi-1==0,t]

t0=N[(1/2) (π+(Sqrt[4+(π^2)]))]-Pi
t1=N[(1/2) (π+(Sqrt[4+(π^2)]))]

L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]

r[t_]:=Log[t]
Manipulate[
PolarPlot[r[t],{t,t0,tstop}],
{tstop,t0,t1}
]