## Calculus (3rd Edition)

Total length: $s = 4\pi$ 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: $s = \mathop \smallint \limits_0^\pi \sqrt {{{\left( {4\sin \theta } \right)}^2} + {{\left( {4\cos \theta } \right)}^2}} {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^\pi \sqrt {16{{\sin }^2}\theta + 16{{\cos }^2}\theta } {\rm{d}}\theta$ $s = 4\cdot\mathop \smallint \limits_0^\pi {\rm{d}}\theta = 4\pi$