Answer
Total length: $s = 4\pi $
Work Step by Step
We have
$r = f\left( \theta \right) = 4\sin \theta $, ${\ \ }$ $f'\left( \theta \right) = 4\cos \theta $
Since $x = r\cos \theta $ and $y = r\sin \theta $, we get the rectangular coordinates:
$\left( {x,y} \right) = 4\sin \theta \left( {\cos \theta ,\sin \theta } \right)$
We compute several points in the interval $0 \le \theta \le \pi $ and list them in the following table:
$\begin{array}{*{20}{c}}
\theta &{\left( {x,y} \right)}\\
0&{\left( {0,0} \right)}\\
{\frac{\pi }{4}}&{\left( {2,2} \right)}\\
{\frac{\pi }{2}}&{\left( {0,4} \right)}\\
{\frac{{3\pi }}{4}}&{\left( { - 2,2} \right)}\\
\pi &{\left( {0,0} \right)}
\end{array}$
Then we plot the points and sketch the circle by joining these points.
From the graph we see that the entire circle is traced out for the interval $0 \le \theta \le \pi $. Using Eq.(7), we get the total length:
$s = \mathop \smallint \limits_0^\pi \sqrt {{{\left( {4\sin \theta } \right)}^2} + {{\left( {4\cos \theta } \right)}^2}} {\rm{d}}\theta $
$s = \mathop \smallint \limits_0^\pi \sqrt {16{{\sin }^2}\theta + 16{{\cos }^2}\theta } {\rm{d}}\theta $
$s = 4\cdot\mathop \smallint \limits_0^\pi {\rm{d}}\theta = 4\pi $
