Find the mass $M$ of a thin piece of wire in the shape of a semicircle $x=2\cos t$, $y=2\sin t$, $0\leq t\leq \pi$, if the variable density of the wire is $\rho (x,y)=y+2$.

Solution Using (2), we find that the mass $M$ of the wire is $$\begin{eqnarray*} M&=&\int_{C}\rho (x,y)\,ds=\int_{C}(y+2)\,ds=\int_{0}^{\pi }(y+2) \sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}dt\\[4pt] &=& \int_{0}^{\pi }(2\sin t+2)\sqrt{(-2\sin t)^{2}+(2\cos t)^{2}}\,dt \hspace{1.5pc}\color{#0066A7}{\dfrac{dx}{dt} =-2\sin t; \dfrac{dy}{dt} =2 \cos t}\\ &=&\int_{0}^{\pi }4(\sin t+1) \,dt=4\big[-\cos t+t\big]_{0}^{\pi }=4[\pi +2] \end{eqnarray*}$$