Finding the Value of a Line Integral Along a Smooth Curve

Find \(\int_{C}y\,ds\) if \(C\) is the curve defined by the parametric equations \(x(t)=t\) and \(y(t)=\sqrt{t}\), \(2\leq t\leq 6\)

Solution The curve \(C\) is smooth and is part of the graph of \(y=\sqrt{x}\), from \((2,\sqrt{2})\) to \((6,\sqrt{6})\), as shown in Figure 13.

The differential \(ds\) of arc length along \(C\) is given by \[ ds=\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}\,dt \]

Since \(\dfrac{dx}{dt}=1\) and \(\dfrac{dy}{dt}=\dfrac{1}{2\sqrt{t}},\) we have \[ ds=\sqrt{1^{2}+\left( \dfrac{1}{2\sqrt{t}}\right) ^{2}}\,dt=\sqrt{1+\dfrac{1}{4t}}\,dt=\sqrt{\dfrac{4t+1}{4t}}\,dt=\dfrac{\sqrt{4t+1}}{2\sqrt{t}}\,dt \]

Now we use formula (1). \[ \int_{C}y\,ds\underset{\underset{\color{#0066A7}{\hbox{\(y=\sqrt{t}\)}}}{\color{#0066A7}{\uparrow}}}{=}\int_{2}^{6}\sqrt{t}\dfrac{\sqrt{4t+1}}{2\sqrt{t}}\,dt=\dfrac{1}{2}\int_{2}^{6}\sqrt{4t+1}\,dt=\dfrac{1}{2}\left[ \dfrac{(4t+1)^{3/2}}{4\,{\cdot}\, \dfrac{3}{2}}\right] _{2}^{6}=\dfrac{49}{6} \]