Find $$\int_{C}(x^{2}+y)\,ds$$

if $C$ is the line segment from $(0,0)$ to $(1,2)$ and $C$ is expressed using the parametric equations:

Figure 14 $y = 2x\, \hbox{from}\, (0,0)\, \hbox{to}\,(1,2)$

Solution The two sets of parametric equations in (a) and (b) are just different ways of representing a part of the line $y=2x$ from $(0,0)$ to $(1,2)$, as shown in Figure 14.

(a) For $x(t) =t$ and $y(t) =2t,$ we have $\dfrac{dx}{dt}=1$ and $\dfrac{dy}{dt}=2$. The differential $ds$ of the arc length $s$ is $ds=\sqrt{1^{2}+2^{2}}dt=\sqrt{5}\,dt$. Then $$\int_{C}(x^{2}+y)\,ds \underset{ \underset{ \underset{\color{#0066A7} {\hbox{\(y=2t\)}}} {\color{#0066A7}{\hbox{\(x=t\)}}}} {\color{#0066A7}{\uparrow }}} {=} \int_{0}^{1}(t^{2}+2t)\sqrt{5}\,dt=\left. \sqrt{5}\left( \dfrac{t^{3}}{3} +t^{2}\right) \right] _{0}^{1}=\dfrac{4\sqrt{5}}{3}$$

980

(b) For $x(t) =\sin t$ and $y(t) =2\sin t,$ we have $\dfrac{dx}{dt}=\cos t$ and $\dfrac{dy}{dt}=2\cos t$. Since $\cos t\geq 0$ on $0\leq t\leq \dfrac{\pi }{2},$ the differential $ds$ of the arc length $s$ is $$ds=\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}} \,dt=\sqrt{\cos ^{2}t+4\cos ^{2}t}\,dt=\sqrt{5\cos ^{2}t}\,dt=\sqrt{5}\cos t\,dt$$

Then $$\begin{eqnarray*} \int_{C}(x^{2}+y)\,ds \underset{\underset{\underset{\color{#0066A7}{\hbox{\(y=2\sin t\)}}}{\color{#0066A7}{\hbox{\(x=\sin t\)}}}} {\color{#0066A7}{\uparrow }}}{=}\int_{0}^{\pi /2}(\sin ^{2}t+2\sin t)\sqrt{5}\cos t\,dt= \sqrt{5}\int_{0}^{\pi /2}(\sin ^{2}t+2\sin t)\cos t\,dt \\[-10pt] =\sqrt{5}\left[ \dfrac{\sin ^{3}t}{3}+\sin ^{2}t\right] _{0}^{\pi /2}=\dfrac{4\sqrt{5}}{3} \end{eqnarray*}$$

the same value found in (a).