Evaluate the line integral $\int_C 2+x^2y\ ds$, where $C$ is the upper half of the unit circle $x^2+y^2=1$.

Solution

Since we evaluate a line integral in terms of $t$, we need to parametrize $C$. We can do it as follows:

$\begin{aligned} x &= \cos t\ y &= \sqrt{1-x^2} = \sqrt{1-\cos^2 t} = \sin t \end{aligned}$

Since this is the upper half of the circle, $0≤t≤π$.

Then $\vec{r}\ '(t) = \langle -\sin t, \cos t \rangle$, so $|\vec{r}\ '(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1$. Therefore, the line integral becomes

$\begin{aligned} \int_0^\pi &2+\cos^2 t\ \sin t \ dt\ &= 2t-\dfrac{1}{3}\cos^3 t\bigg|_0^\pi \ \ \ \ \textrm{ by u-substitution}\ &= \ans{2\pi + \dfrac{2}{3}} \end{aligned}$

Alternate Solution

What if we parameterized the curve differently (recall that parameterization isn't unique)? It would be simpler to parameterize it as follows, but we'll find that the integral is tricker:

$\begin{aligned} x &= t\ y &= \sqrt{1-t^2} \end{aligned}$

In this case, $−1≤t≤1$:

Then $\vec{r}\ '(t) = \langle 1,\dfrac{-t}{\sqrt{1-t^2}} \rangle$, so $|\vec{r}\ '(t)| = \sqrt{1+\left(\dfrac{-t}{\sqrt{1-t^2}}\right)^2} = \dfrac{1}{\sqrt{1-t^2}}.$

To integrate this, we need to use trigonometric substitution: if we define $t=\sin θ$, then $dt=\cos θ\ dθ$.

$\begin{aligned} \int \dfrac{2}{\sqrt{1-t^2}}\ dt &= \int \dfrac{2}{\cos \theta} \cdot \cos \theta\ d\theta\ &= 2\theta = 2\sin^{-1} t \end{aligned}$

Going back to the line integral:

$\begin{aligned} \int_{-1}^1 &\dfrac{2}{\sqrt{1-t^2}} + t^2 \ dt\ &= 2\sin^{-1} t+\dfrac{1}{3}t^3\bigg|_{-1}^1\ &= \ans{2\pi + \dfrac{2}{3}} \end{aligned}$

Notice, of course, that we get the same answer in both cases; one had a simpler parameterization and an easier integral, and vice versa.

Evaluate the line integral $\displaystyle\int_C xy+2z\ ds$, where $C$ is the line segment from $(1,0,0)$ to $(0,1,1)$.

Solution

Start by parameterizing the line segment $C$: $r(t) = \langle 1-t,t,t \rangle, \ \ \ \ 0 \leq t \leq 1$

Then $r⃗ '(t)=⟨−1,1,1⟩$, so $|r⃗ '(t)|=\sqrt{3}$.

Rewriting the line integral in terms of $t$: $\begin{aligned} \int_C xy+2z\ ds &= \int_0^1 \big((1-t)t+2t\big)\sqrt{3}\ dt\ &= \sqrt{3} \int_0^1 3t-t^2\ dt\ &= \sqrt{3} \bigg[\dfrac{3}{2}t^2-\dfrac{1}{3}t^3 \bigg]_0^1\ &= \ans{\dfrac{7\sqrt{3}}{6}} \end{aligned}$

The flight of an eagle is given by $\vec{r}(t) = \left\langle 2400 \cos \dfrac{t}{2}, 2400 \sin \dfrac{t}{2}, 500t \right\rangle$

in ft. How far does the eagle travel over $0≤t≤10$?

Solution

$\begin{aligned} \vec{r}\ '(t) &= \langle -1200 \sin \dfrac{t}{2}, 1200 \cos \dfrac{t}{2}, 500 \rangle\ |\vec{r}\ '(t)| &= \sqrt{1200^2\left(\sin^2 \dfrac{t}{2} + \cos^2 \dfrac{t}{2}\right) + 500^2} = 1300\ \implies L &= \int_C |\vec{r}\ '(t)| \ dt\ &= \int_0^{10} 1300 dt = \ans{13,000\ ft} \end{aligned}$

Note that the magnitude of the velocity vector is $1300$, meaning that the eagle traveled at $1300$ ft/min for $10$ minutes, for a total of $13,000$ ft.

Electric Force Field

An electric force field is described by $\vec{F} = \dfrac{k \langle x,y,z \rangle}{(x^2+y^2+z^2)^{3/2}},$ where $k$ is a constant. Suppose that $k=1$ and find the work done by this force field in moving a particle from $(1,1,1)$ to $(3,3,3)$.

Solution

Start by parameterizing the line segment $C$: $r(t) = \langle 1+t,1+t,1+t \rangle, \ \ \ \ 0 \leq t \leq 2$

Then $r⃗ '(t)=⟨1,1,1⟩$ and the force is written in parametric form as $\vec{F} = \dfrac{1}{\big(3(1+t^2)\big)^{3/2}} \langle 1+t,1+t,1+t \rangle,$ so $\vec{F} \cdot \vec{r}\ '(t) = \dfrac{3(1+t)}{\sqrt{27}(1+t)^3} = \dfrac{1}{\sqrt{3}(1+t)^2}.$

The work, therefore, is $\begin{aligned} W &= \int_a^b \vec{F} \cdot \vec{r}\ '(t)\ dt\ &= \int_0^2 \dfrac{1}{\sqrt{3}(1+t)^2}\ dt\ &= -\dfrac{1}{\sqrt{3}(1+t)} \bigg|_0^2 = \ans{\dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}} \end{aligned}$