Finding the Value of a Line Integral

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:

1. \(x(t) =t\) and \(y(t) =2t,\quad 0\leq t\leq 1\)
2. \(x(t) =\sin t\) and \(y(t) =2\sin t,\quad 0\leq t\leq \dfrac{\pi }{2}\)

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).