


\section{Line Integrals}

\begin{enumerate}[1.]





\item {\it Evaluate the integral
$\int_{C} \sqrt{x^2+y^2}\,ds$, where $C$ is the parametrized curve $\bfr(t)=(t\cos t, t\sin t)$, $0\leq t\leq 2\pi$.}

\medskip

We have $\sqrt{x(t)^2+y(t)^2}=\sqrt{t^2\cos^2t+t^2\sin^2t}=t$ and 
\begin{align*}
ds&=|\bfr'(t)|=\sqrt{(-t\sin t+\cos t)^2+(t\cos t+\sin t)^2}dt\\
&=\big(t^2\sin^2t-2t\sin t\cos t+\cos^2 t+t^2\cos^2 t+2t\cos t\sin t+\sin^2 t\big)^{1/2}dt\\
&=\sqrt{t^2+1}dt.\\
\end{align*}
Hence the integral is
$$
\int_0^{2\pi}t\sqrt{t^2+1}dt=\frac{1}{2}\frac{(t^2+1)^{3/2}}{3/2}\Big|_0^{2\pi}
=\frac{1}{3}\Big((4\pi^2+1)^{3/2}-1\Big).
$$





\item 
{\it Evaluate the line integral
$\int_C f\,ds$, where $f(x,y,z)=z-x$ and ${\bf r}(t)=e^{-t}{\bf i}+\sqrt{2}
t {\bf j}+e^t{\bf k}$, $0\leq t\leq 1$.}


\medskip

We have ${\bf r}'(t)=-e^{-t}{\bf i}+\sqrt{2}{\bf j}+e^t {\bf k}$, so that
$$
|{\bf r}'|=\sqrt{e^{-2t}+2+e^{2t}}=e^t+e^{-t},
$$
$$
\int_C f\,ds=\int_0^1(e^t-e^{-t})(e^t + e^{-t})dt=\int_0^1 (e^{2t}-e^{-2t})dt
$$
$$=\frac{e^{2t}+e^{-2t}}{2}\Big|_{t=0}^1 =\frac{e^2+e^{-2}}{2}-1.
$$



\medskip


\item {\it 
Evaluate the line integral ${\int_C 2yz\,ds}$, if $C$ is parametrized by ${\bf r}(t)=t\ii + 2e^t\jj + e^{2t}\kk$ for $0\leq t\leq 1$. }


\medskip
We have ${\bf r}'(t)=\ii + 2e^t\jj +2 e^{2t}\kk$, so that 
$$ds=|{\bf r}'(t)|dt=\sqrt{1+4e^{2t}+4e^{4t}}dt=(1+2e^{2t})dt,$$\begin{align*}
\int_C 2yz\,ds =& \int_0^1 4e^{3t}(1+2e^{2t})dt
 \\
& = \int_0^{1} (4e^{3t}+8e^{5t})dt
\\
&= \frac{4}{3}e^{3t}+\frac{8}{5}e^{5t} \Big|_0^{1}= \frac{4}{3}(e^3-1)+\frac{8}{5}(e^5-1).
\end{align*}







\medskip

\item 
{\it Evaluate the line integral
$\int_C xydx-xy^2dy$, where $C$ is the curve with equation $x^2=y^3$, from $(-1,1)$ to $(1,1)$.}

\medskip

Parametrize the curve as $x=t^3,y=t^2$, $-1\leq t\leq 1$. Then
$$
\int_C xydx-xy^2dy=\int_{-1}^1 ( t^3t^2\cdot 3t^2 - t^3t^4\cdot 2t ) dt
=\int_{-1}^1 (3t^7-2t^8)dt
$$
$$=\frac{3t^8}{8}-\frac{2t^9}{9}\Big|_{t=-1}^1=-\frac{4}{9}.
$$




\item 
{\it Evaluate the line integral
$\int_{C} xdy-ydz$, where $C$ is the oriented line segment from $(1,0,0)$
to $(3,1,6)$.}
\medskip

We parametrize the line segment as
$$
\bfr(t)=(1-t)(1,0,0)+t(3,1,6)=(1+2t,t,6t),\ 0\leq t\leq 1.$$
Hence the integral is
$$
\int_0^1 (1+2t)dt-t\cdot 6dt=\int_0^1(1-4t)dt=t-2t^2\big|_0^1=1-2=-1.
$$

\item
{\it Evaluate the integral 
${\int_C ydx+xdy}$, where $C$ is the arc of the circle $(x-1)^2+y^2=1$
from $(0,0)$ to $(1,1)$ in the first quadrant of the $xy$-plane.}

\medskip

Parametrize the arc by $x=1-\cos\theta$, $y=\sin\theta$, $0\leq\theta\leq \pi/2$. Then
\begin{align*}
{\int_C ydx + xdy} & = \int_0^{\pi/2} (\sin\theta)(\sin\theta) + (1-\cos\theta)(\cos\theta)d\theta
\\
&= \int_0^{\pi/2} (\sin^2\theta +\cos\theta - \cos^2\theta) d\theta 
= \int_0^{\pi/2} (\cos\theta - \cos(2\theta)) d\theta 
\\
&= \sin\theta - \frac{\sin(2\theta)}{2} \Big|_0^{\pi/2} = 1 - 0 =1.
\end{align*}



\bigskip

\item {\it
Evaluate the integral
$\int_C {\bf F}\cdot d{\bf r}$, where ${\bf F}=y\ii -z\jj$ and $C$ is the part of the curve of
intersection of the surfaces $x+y+z=2$ and $y=z^2$ from $(0,1,1)$ to $(2,0,0)$.
}

\medskip


If we use the parametrization $z=t$, $y=t^2$, $x=2-t-t^2$, $0\leq t\leq 1$, 
then this parametrizes the
curve $C$, but in the opposite direction (from $z=0$ to $z=1$, instead of the other way around).
We are going to use this parametrization anyway and just add a minus sign.
We have ${\bf r}'(t)=(-2t-1)\ii +2t\jj +\kk$, so that
$$
\int_C {\bf F}\cdot d{\bf r}=- \int_0^1 \Big[ t^2(-2t-1) -t (2t)\Big] dt
=\int_0^1 (2t^3 +3t^2)dt
$$
$$=\frac{t^4}{2}+t^3\Big|_{t=0}^1 =\frac{1}{2}+1=\frac{3}{2}.
$$


\bigskip

\item {\it 
Evaluate the integral $\int_C {\bf F}\cdot d{\bf r}$, where ${\bf F}=y\ii +z\jj+x^2\kk$ and $C$ is the 
intersection curve of the surfaces $z=y^2$ and $x^2+y^2=4$, oriented counterclockwise when viewed from the direction of the positive $z$-axis.} 


\medskip

We use the parametrization $x=2\cos t$, $y=2\sin t$, $z=y^2=4\sin^2 t$, $0\leq t\leq 2\pi$.
Then ${\bf r}'(t)=-2\sin t\, \ii +2\cos t \,\jj +8\sin t\cos t \,\kk$, so that
\begin{align*}
\int_C {\bf F}\cdot d{\bf r}&= \int_0^{2\pi} \Big[ 
(2\sin t)(-2\sin t) + (4\sin^2 t)(2\cos t) + (4\cos^2 t)(8\sin t\cos t)\Big] dt
\\
&= \int_0^{2\pi} \Big[ 
-4\sin^2 t + 8\sin^2 t\cos t + 32\cos^3 t\sin t\Big] dt
\end{align*}
We have $\cos(2t)=1-2\sin^2 t$, so that
$$
\int_0^{2\pi} (-4\sin^2t)dt = \int_0^{2\pi}(2\cos(2t)-2)dt
=\left[\sin(2t)-2t\right]_{0}^{2\pi}=-4\pi,
$$
and also
\begin{align*}
& \int_0^{2\pi} 8\sin^2 t\cos t\,dt=\left[\frac{8}{3}\sin^3 t\right]_0^{2\pi}=0,
\\
& \int_0^{2\pi}32\cos^3 t\sin t \,dt=\left[-8\cos^4 t\right]_0^{2\pi}=0.
\end{align*}
Therefore 
$$\int_C {\bf F}\cdot d{\bf r}=-4\pi+0+0=-4\pi.$$



\item {\it 
Consider the integral $\int_C f\, ds$, where $f(x,y,z)=e^{-(x^2+y^2+z^2)}$ and $C$ is the 
line where the planes $x+y-z=0$ and $x+2y+z=0$ intersect. Prove that this integral is finite.

\smallskip\noindent
(Hint: if you set up this integral using the ``usual'' parametrization of a line where the velocity is constant, you will get an improper integral. 
If you do not know how to evaluate it, 
it will suffice if you use inequalities to prove that the integral is finite.}



\medskip
\noindent{\bf Solution 1.} We first parametrize the line. From the first plane equation we have $x=z-y$, which we plug into the second equation:
$$
0=x+2y+z=(z-y)+2y+z=y+2z,\ \ y=-2z,\ \ x=z-y=z-(-2z)=3z.
$$
We get a parametrization $\bfr(t)=3t\ii -2t\jj +t\kk$, with $|\bfr'|=\sqrt{14}$. The line integral is
$$
\int_C f\,ds=\int_{-\infty}^\infty e^{-14t^2} \sqrt{14}\, dt.
$$
One way to complete the proof is to use the fact that $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$. (This can be done using polar coordinates in the plane -- see if you can find it in your textbook!) Then by a change of variables $u=t\sqrt{14}$, $du=dt\sqrt{14}$,
$$
\int_C f\,ds=\int_{-\infty}^\infty e^{-u^2} \, du=\sqrt{\pi}.
$$

\medskip
\noindent{\bf Solution 2.} Start as above, but if
 you do not know or do not remember the last integral, there are alternative solutions using inequalities. For example, we can use that $e^{-14t^2}\leq 1$ for $|t|\leq 1$ and $e^{-14t^2}\leq e^{-|t|}$ for $|t|\geq 1$. Then
\begin{align*}
\int_C f\,ds& =2 \int_0^\infty e^{-14t^2} \sqrt{14}\, dt
\\
&\leq 2 \int_0^1  \sqrt{14}\, dt +2 \int_1^\infty e^{-t} \sqrt{14}\, dt
\\
&=2\sqrt{14}+2 \sqrt{14}\left[ -e^{-t}\right]_{1}^\infty
=4\sqrt{14},
\end{align*}
which again is finite.

\medskip
\noindent{\bf Solution 3.} Since $(0,0,0)$ lies on both planes, the line $C$ passes through it. Now, notice that $f(x,y,z)=e^{-(x^2+y^2+z^2)}$ is radially symmetric! Therefore the integral of $f$ on any line passing through the origin is the same. So, instead of integrating on the line indicated, we can make our lives easier and integrate on the $x$-axis instead, with ${\bf r}=t\ii$ and $f(x,0,0)=e^{-x^2}$:
$$
\int_C f\,ds=\int_{-\infty}^\infty e^{-t^2} \, dt=\sqrt{\pi}.
$$













\end{enumerate}

