#15, Section 10.4:

r[t_]:=Sqrt[1+Cos[5t]^2];
PolarPlot[r[t],{t,0,2Pi},PlotStyle->{Green,Directive[Dashed,Thick,Orange]}]

t0=0;
t1=2Pi;
L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]

Let’s compare to an appropriate circle, to see if we believe our results:

r=1.2
2Pi*r
Pi*r^2
Clear[r]

Now let’s see what happens if r is not 2pi-periodic:

r[t_]:=Sqrt[1+Cos[t/4]^2];
PolarPlot[r[t],{t,0,2Pi},PlotStyle->{Green,Directive[Dashed,Thick,Orange]}]

t0=0;
t1=2Pi;
L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]

PolarPlot[r[t],{t,0,4Pi},PlotStyle->{Green,Directive[Dashed,Thick,Orange]}]

t0=0;
t1=4Pi;
L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]

Now let’s see what happens if r is pi-periodic:

r[t_]:=Sin[t];
PolarPlot[r[t],{t,0,Pi},PlotStyle->{Green,Directive[Dashed,Thick,Orange]}]

t0=0;
t1=Pi;
L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]

PolarPlot[r[t],{t,0,2Pi},PlotStyle->{Green,Directive[Dashed,Thick,Orange]}]

t0=0;
t1=2Pi;
L=NIntegrate[Sqrt[r[t]^2+r'[t]^2],{t, t0,t1}]
A=NIntegrate[1/2*r[t]^2,{t, t0,t1}]