%
\section{Line Integrals}\label{sec:lineIntegrals}
Let $C$ be a smooth curve on $\mathbb{R}^n$ (where $n = 2$ or $n = 3$) with endpoints, and let
$\mathbf{r}: [a,b] \rightarrow \mathbb{R}^n$ be a smooth parametrization of $C$.


\subsection{Scalar Line Integrals as Limits of Riemann Sums} Let $f : D \rightarrow \mathbb{R}$ be a function on some domain $D \subset \mathbb{R}^n$ which contains the curve $C$.  We define the \textbf{line integral} of $f$ over $C$
\begin{equation}\label{23eq-lineintegral}
\int_C f(x,y,z) ds, \textrm{ where } ds \textrm{ is arc-length},
\end{equation}
as the limit of the Riemann sums
$$
S : = \sum\limits_{i=1}^{n} f\big(\mathbf{r}(t_i^*)\big) \vert \Delta \mathbf{r}_i \vert = \sum\limits_{i=1}^{n} f (x_i^*, y_i^*, z_i^*) \vert \Delta \mathbf{r}_i \vert.
$$
Let us unpack this definition.  The points $t_i$ are coming from some partition
$$
a = t_0 < t_1 < \cdots < t_n = b
$$
of the interval $I = [a,b]$.  Having fixed this partition, we then define $\Delta \mathbf{r}_i = \mathbf{r}(t_i) - \mathbf{r}(t_{i-1})$, and let
$$
\mathbf{r}(t_i^*) : = (x_i^*, y_i^*, z_i^*), \textrm{ for some } t_i^* \in [t_{i-1}, t_i].
$$
We call these points $\mathbf{r}(t_i^*)$ {\it evaluation points} or {\it sampling points}.  The situation is depicted below. We then take the limit over all such partitions as $\max_i |t_i-t_{i-1}|\to 0$. For the integral (\ref{23eq-lineintegral}) to exist, we need the limit to exist and to be independent of the choice of the partitions and sampling points.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .4\linewidth]{figures/23fig1.png}
\caption{The curve $C$ (in orange) is contained in the domain $D \subset \mathbb{R}^n$ (the green oval).  The ``cross-marks'' are the $\mathbf{r}(t_i)$.}
\end{minipage}
\end{figure}

To derive a formula for evaluating line integrals, 
we first make the approximation
\begin{equation}\label{23eq-approx1}
\Delta \mathbf{r}_i = \frac{\mathbf{r}(t_i) - \mathbf{r}(t_{i-1})}{t_i - t_{i-1}} (t_i - t_{i-1}) \approx \mathbf{r}' (t_i) \Delta t_i,
\end{equation}
where $\Delta t_i = t_i - t_{i-1}$. The approximation will have less error as $\Delta t_i \rightarrow 0$. Using \eqref{23eq-approx1}, we have
\begin{equation}\label{23eq-approx2}
S_n \approx \sum\limits_{i=1}^{n} f\big(\mathbf{r}(t_i) \big) \vert \mathbf{r}' (t_i) \vert \Delta t_i = S_n^*.
\end{equation}
But, $S_n^*$ is a Riemann sum associated to the following (one-dimensional) definite integral
$$
\int_a^b f\big(\mathbf{r}(t) \big) \vert \mathbf{r}' (t) \vert dt.
$$
Recalling that arc-length parametrization gives $ds =  \vert \mathbf{r}' (t) \vert dt$, we arrive at the following formula.

\medskip

\noindent
\fbox{
\begin{minipage}{\textwidth}
\begin{center}

\medskip

 \textbf{Evaluating Scalar Line Integrals}
\end{center}

\vspace{.2cm}

Suppose that $C \subset D$ is a smooth arc and $f : D \rightarrow \mathbb{R}$ is continuous on $C$.  Then, the line integral of $f$ on $C$ satisfies the equality
\begin{equation}\label{23-lineintegralcalc}
\int_C f ds = \int_a^b f \big( \mathbf{r}(t) \big) \vert \mathbf{r}' (t) \vert dt  = \int f\big(\mathbf{r}(s) \big) ds.
\end{equation}
\medskip

\end{minipage}
}

\medskip


You can see several applications of this formula in the Worked Problems below.
A few important remarks regarding scalar line integrals:


\begin{itemize}
\item If $C:[a,b]\to\rr^n$ is a smooth curve, and if $f$ is continuous on some domain containing $C$, then the limit above exists. (We do not prove it here, but this is similar to the existence of Riemann integrals of continuous functions on bounded intervals.) This does not always happen for arbitrary curves $C$ and functions $f : D \rightarrow \mathbb{R}^n$. In cases where the assumptions above do not hold, the limit may still exist, but we need to justify it on a case-by-case basis.

\smallskip
\item When $f(x,y,z)=1$ for all $x,y,z$ on the curve, these are the same Riemann sums that we used to define arc length of the curve. Hence, $\int_C 1\, ds$ (or $\int_C\, ds$ for short) is just the length of the curve.

\smallskip

\item We can extend arc length integrals to piecewise smooth curves by integrating on each smooth piece and then adding up the integrals. Similarly, improper line integrals (e.g. line integrals over curves $C$ of infinite length) can be evaluated by integrating on a finite part of the curve and then taking the limit (if it exists).  
\end{itemize}



\subsection{A Physical Interpretation of Line Integrals}
Line integrals often have natural interpretations in physical space. 
For example,  if $f : D \rightarrow \mathbb{R}^3$ is some continuous function on a smooth arc $C \subset D$,  we can interpret the line integral $\int_C f ds$ as:
\begin{itemize}
\item The mass of a wire with shape given by the curve $C$, if we make the assumption that the density at point $\mathbf{r}$ is given by $f(\mathbf{r})$. So, the line integral becomes an expression for \textit{total mass} over the wire.

\smallskip

\item If $C$ lies in the $xy$-plane and $f(x,y)\geq 0$,  we can view the line integral $\int_C f\, ds$ as the area of a curvilinear fence that has height $f(x,y)$ at every point $(x,y)$ of $C$. One example is shown in the following figure.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .6\linewidth]{figures/23fig2.png}
\end{minipage}
\end{figure}


\smallskip

\item The \textit{average} of a function $f (\mathbf{r})$ over the curve $C$ is given by the expression:
$$
\frac{\int_C f ds}{\int_C ds}.
$$
Since $\int_C ds$ represents the length of $C$,  this is similar to taking the average of a single-variable function $f: I \rightarrow \mathbb{R}$ over an interval $I \subset \mathbb{R}$.
\end{itemize}











\subsection{Vector Line Integrals} We now define line integrals of vector fields. In this type of integrals, we are \underline{not} just integrating each component of $\mathbf{F}$ separately with respect to arc length: the integral also takes into account how well $\mathbf{F}$ is aligned with $C$.

Again, we assume that $C$ is a smooth curve, parametrized by $\mathbf{r}(t): [a,b]\to \RR^n$.  We further assume that $\mathbf{F} : D \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous on some domain $D$ which contains $C$.  We then define the \textbf{line integral} of $\mathbf{F}$ over $C$ as
\begin{equation}\label{23eq-lineintegralvector}
\int_C \mathbf{F} \cdot d \mathbf{r} = \int_C \mathbf{F} \cdot \mathbf{T} ds
\end{equation}

The function $\mathbf{F} \cdot \mathbf{T}$ is equal (up to $\pm$ signs) to the length of the tangential component $\mathbf{F}_T$ of $\mathbf{F}$ along $C$--a scalar quantity.  This is depicted in the following diagram.


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .4\linewidth]{figures/23fig3.png}
\caption{The vector field $\mathbf{F}$ and its tangential component $\mathbf{F}_T$ along some smooth arc. The dashed line reminds you that  $\mathbf{F}_T$ is the orthogonal projection of $\mathbf{F}$ onto the unit tangent vector $\mathbf{T}$ of $\mathbf{r}$.}
\end{minipage}
\end{figure}

\noindent
\fbox{
\begin{minipage}{\textwidth}
\begin{center}

\medskip

 \textbf{Evaluating Vector Line Integrals}
\end{center}

\vspace{.2cm}

Recall that $\mathbf{T} = \frac{\mathbf{r}'}{\vert \mathbf{r}' \vert}$ and that the arc-length differential satisfies $ds = \vert \mathbf{r}' \vert dt$. Using this,  the line integral of a vector field $\mathbf{F}$ is then:
\begin{equation}\label{23eq-lineintegralvectorformula}
\int_C \mathbf{F} \cdot d \mathbf{r} = \int_a^b \mathbf{F} \cdot \frac{\mathbf{r}'}{\vert \mathbf{r}' \vert} \vert \mathbf{r}' \vert dt = \int_a^b \mathbf{F} \cdot \mathbf{r}' dt.
\end{equation}
In terms of components, this gives the equation
\begin{eqnarray*}
\int_C \mathbf{F} \cdot d \mathbf{r} & = & \int_a^b F_1 \, x' (t) dt + \int_a^b F_2 \, y' (t) dt + \int_a^b F_3 \, z' (t) dt \\[1ex]
\textrm{(Alternative Notation)} & = & \int_C F_1 dx + F_2 dy + F_3 dz.
\end{eqnarray*}
So, to evaluate the line integral of a vector field $\mathbf{F} : D \rightarrow \mathbb{R}^3$, it suffices to evaluate the component definite integrals.
\end{minipage}
}

\medskip
Line integrals $\int_C \mathbf{F} \cdot d \mathbf{r}$ of vector field $\mathbf{F}$ also have natural interpretations:
\begin{itemize}
\item The work done by a force $\mathbf{F}$ in moving a particle along a path $\mathbf{r}(t)$. (Note: if $\mathbf{F}\cdot \mathbf{T}$ is negative, so that the motion is against the direction of the force, this means that the force has opposed the motion and therefore made a negative contribution to it.)

\smallskip

\item The \textit{circulation} of $\mathbf{F}$ along $C$--this is especially true in the case where $C$ is a closed curve.  This is shown in the picture below, with the circulation measured in the positive (counterclockwise) direction.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .25\linewidth]{figures/23fig4.png}
\end{minipage}
\end{figure}
\end{itemize}

\section*{Worked Problems}

\begin{enumerate}[1.]
\item
{\it Find $\int_C z \, ds$ if $C$ is the arc of the helix $\mathbf{r}(t) = \cos (t)\mathbf{i} + \sin (t) \mathbf{j} + t \mathbf{k}$ over the interval $t \in I = [0, 2 \pi]$.}

\medskip

One readily calculates that $ds = \sqrt{2} dt$, whereas $z (t) = t$ along the helix.  Hence, the line integral formula \eqref{23-lineintegralcalc} gives
$$
\int_C z ds = \int_0^{2\pi} t \sqrt{2} dt = \frac{t^2 \sqrt{2}}{2} \big\vert_0^{2 \pi} = 2 \sqrt{2} \pi^2.
$$

\medskip

\item{\it Find $\int_C x^2 y\, ds$ if $C$ is the line segment from $(0,1)$ to $(1,3)$ in $\mathbb{R}^2$.}

\medskip

We parametrize $C$ as $\mathbf{r} (t) = (t, 2t + 1)$ for $t \in I = [0,1]$.  Then,  $\mathbf{r}' (t) = (1,2) \Rightarrow ds = \sqrt{5} dt$.   Hence, applying the line integral formula \eqref{23-lineintegralcalc} gives
\begin{eqnarray*}
\int_C x^2 y ds &=& \int_0^1 t^2 (2t + 1) \sqrt{5} dt = \int_0^1 \sqrt{5} (2 t^3 + t^2) dt \\
\quad & = & \sqrt{5} \bigg( \frac{t^4}{2} + \frac{t^3}{3} \bigg)\big\vert_0^1 \\
\quad & = & \frac{5 \sqrt{5}}{6}.
\end{eqnarray*}

\medskip


\item{\it Evaluate the line integral ${\int_C (x-y) ds}$, where $C$ is the curve parametrized by $x=t^2+t$, $y=t^2-t$, $z=\frac{t^2}{2}+1$, from $t=0$ to $t=1$.
}

\medskip


We have $\bfr'(t)=(2t+1)\ii+(2t-1)\jj+t\kk$, so that 
$$
|\bfr'(t)|^2=(4t^2+4t+1)+(4t^4-4t+1)+t^2=9t^2+2.
$$
Therefore
$${\int_C (x-y) ds }  = \int_0^{1} 2 t\sqrt{9t^2 +2 } dt .
$$
Substituting $u=9t^2+2$, $du=18t$, $2\leq u\leq 11$, we get
\begin{align*}
{\int_C (x-y) ds } & = \int_2^{11} \sqrt{u } \ \frac{du}{9}
\\
&=  \frac{2}{3}\left[ \frac{u^{3/2}}{9}\right]_2^{11} = \frac{2}{27}(11^{3/2}-2^{3/2})
\end{align*}





\medskip

\item {\it Evaluate the integral
$\int_C xy\,ds$, where $C$ is the part of the curve of
intersection of the surfaces $x^2+y^2=1$ and $z=2y$ where $x\geq 0$ and $y\geq 0$.}

\medskip

We use the parametrization $x=\cos t$, $y=\sin t$, $z=2\sin t$, $0\leq t\leq \pi/2$.
Then ${\bf r}'(t)=-\sin t\, \ii +\cos t\, \jj +2\cos t\,\kk$, so that
$$
|\bfr'|=\sqrt{\sin^2 t + \cos^2 t + 4\cos^2 t} = \sqrt{1+4\cos^2 t},
$$
$$
\int_C xy\,ds = \int_0^{\pi/2} \cos t \sin t \sqrt{1+4\cos^2 t}\, dt.
$$
Letting $u=1+4\cos^2 t$, $du=-8\cos t \sin t \, dt$, we get
$$
\int_C xy\,ds = -\frac{1}{8} \int_5^1   \sqrt{u}\,du = \frac{1}{8}\cdot\frac{2}{3} u^{3/2}\Big|_1^5
= \frac{1}{12}\left( 5^{3/2}-1\right).
$$




\medskip



\item
{\it 
(\textbf{Warning: Long Calculation!}) The mass of a wire in the shape of a parametrized curve $\bfr(t)$ with density $f(x,y,z)$ is given by
$$
M=\int_C f ds.
$$
The center of mass of the same wire has coordinates $(\frac{M_x}{M}, \frac{M_y}{M}, \frac{M_z}{M})$, where
$$
M_x=\int_C x f(x,y,z)ds,\ \ M_y=\int_C y f(x,y,z)ds,\ \ M_z=\int_C z f(x,y,z)ds.
$$
Find the mass and the center of mass of a wire in the shape of the helix
$x=\cos t$, $y=\sin t$, $z=at$, $0\leq t\leq \pi$, if the density of the
wire at a point $(x,y,z)$ is $f(x,y,z)=x^2+y^2+z^2$.}


\medskip

We have $f(\bfr(t))=\cos^2t+\sin^2t+a^2t^2=1+a^2t^2$, $\bfr'(t)
=(-\sin t,\cos t,a)$, $|\bfr'(t)|=\sqrt{1+a^2}$.  The mass of the 
wire is
$$M=\int_0^\pi f(\bfr(t))|\bfr'(t)|dt
=\int_0^\pi (1+a^2t^2)\sqrt{1+a^2}dt$$
$$=\sqrt{1+a^2}\left(t+\frac{a^2t^3}{3}\right)\Big|_0^\pi 
=\sqrt{1+a^2}\left(\pi+\frac{a^2\pi^3}{3}\right).$$
To find the center of mass, we evaluate $M_x,M_y,M_z$ as follows:
$$M_x=\int_0^\pi x(t)f(\bfr(t))|\bfr'(t)|dt
=\int_0^\pi (\cos t) (1+a^2t^2)\sqrt{1+a^2}dt$$
$$=\sqrt{1+a^2}\Big(\sin t|_0^\pi + \int_0^\pi a^2t^2 \cos tdt\Big)$$
$$=\sqrt{1+a^2}\Big(a^2t^2\sin t|_0^\pi - \int_0^\pi 2a^2t\sin tdt\Big)$$
$$=-2a^2 \sqrt{1+a^2}\int_0^\pi t\sin tdt$$
$$=2a^2\sqrt{1+a^2}\Big(t\cos t|_0^\pi - \int_0^\pi \cos tdt\Big)$$
$$=2a^2\sqrt{1+a^2}\Big(-\pi- \sin t|_0^\pi\Big)
=-2a^2\pi\sqrt{1+a^2},$$

$$M_y=\int_0^\pi y(t)f(\bfr(t))|\bfr'(t)|dt
=\int_0^\pi (\sin t) (1+a^2t^2)\sqrt{1+a^2}dt$$
$$=\sqrt{1+a^2}\Big(-\cos t|_0^\pi + \int_0^\pi a^2t^2 \sin tdt\Big)$$
$$=\sqrt{1+a^2}\Big(2-a^2t^2\cos t|_0^\pi + \int_0^\pi 2a^2t\cos tdt\Big)$$
$$=\sqrt{1+a^2}\Big(2+a^2\pi^2+2a^2t\sin t|_0^\pi-\int_0^\pi 2a^2\sin tdt\Big)$$
$$=\sqrt{1+a^2}\Big(2+a^2\pi^2+2a^2\cos t|_0^\pi\Big)$$
$$=\sqrt{1+a^2}(2+a^2\pi^2-4a^2),$$

$$M_z=\int_0^\pi z(t)f(\bfr(t))|\bfr'(t)|dt
=\int_0^\pi at(1+a^2t^2)\sqrt{1+a^2}dt$$
$$=\sqrt{1+a^2} \int_0^\pi (at+a^3t^3)dt
=\sqrt{1+a^2} \Big(\frac{at^2}{2}+\frac{a^3t^4}{4}\Big)\Big|_0^\pi$$
$$=\sqrt{1+a^2} \Big(\frac{a\pi^2}{2}+\frac{a^3\pi^4}{4}\Big).$$

Hence the center of mass is $(\bar x,\bar y,\bar z)$, where
$$\bar x=\frac{M_x}{M}=\frac{-2a^2}{1+\frac{a^2\pi^2}{3}},\ 
\bar y=\frac{M_y}{M}=\frac{2+a^2\pi^2-4a^2}{\pi+\frac{a^2\pi^3}{3}},$$
$$\bar z=\frac{M_z}{M}=\frac{
\frac{a\pi^2}{2}+\frac{a^3\pi^4}{4}
}{1+\frac{a^2\pi^2}{3}}
=\frac{6a\pi+3a^3\pi^3}{12+4a^2\pi^2}.$$

\medskip

\item{\it Find $\int_C \mathbf{F} \cdot d \mathbf{r}$ if $\mathbf{F} = z \mathbf{i} + 2 y \mathbf{j}$ and $\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}$, for $t \in [0,2]$.}

\medskip
Along the curve parametrized by $\mathbf{r}(t)$, we have
$$
\mathbf{F}(\mathbf{r}(t)\big) = z(t) \mathbf{i} + 2 y(t) \mathbf{j} = t^3 \mathbf{i} + 2 t^2 \mathbf{j}.
$$
Moreover, 
$$
\mathbf{r}' (t) = \mathbf{i} + 2 t \mathbf{j} + 3 t^2 \mathbf{k}.
$$
Hence, the line integral is given by
\begin{eqnarray*}
\int_C \mathbf{F} \cdot d \mathbf{r} & = & \int_0^2 \big(t^3 \mathbf{i} + 2 t^2 \mathbf{j} \big) \cdot \big(\mathbf{i} + 2 t \mathbf{j} + 3 t^2 \mathbf{k} \big) dt = \int_0^2 \big(t^3 + 4 t^3 \big) dt \\
\quad & = & \int_0^2 5 t^3 dt = \frac{5 t^4}{4} \big\vert_0^2 \\[1ex]
\quad & = & 20.
\end{eqnarray*}

\medskip

\item{\it Find $\int_C \mathbf{F} \cdot d \mathbf{r}$ if $\mathbf{F} = xy \mathbf{i} + 2 \mathbf{k}$ and $\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k}$ for $t \in I = [0, 2\pi]$.}

\medskip

We have
$$
\mathbf{F}(\mathbf{r}(t)) = \cos t \sin t \mathbf{i} + 2 \mathbf{k}
$$
$$
\mathbf{r}' (t) = - \sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k},
$$
so that
\begin{eqnarray*}
\int_C \mathbf{F} \cdot d \mathbf{r} & = & \int_0^{2 \pi} \big[(\cos t \sin t) \cdot (-\sin t) + 2 \cdot 1 \big] dt = \int_0^{2\pi} \big(- \sin^2 t \cos t + 2 \big) dt \\
\quad & = & - \frac{\sin^3 t}{3} + 2 t \big\vert_0^{2\pi} \\
\quad & = & 4 \pi.
\end{eqnarray*}

\medskip


\item {\it Evaluate the line integral 
${\int_C - ydx + x dy + x^2 dz}$, if $C$ is the part of the intersection curve of the 
elliptic cylinder $x^2+\frac{y^2}{4}=1$ and the surface $z=x^3 $ that lies in the region $x\geq 0,y\geq 0$, between the points $(1,0,1)$ and $(0,2,0)$,
oriented in the direction of increasing $y$.}

\medskip

Parametrize the curve by $x=\cos t$, $y=2\sin t$, $z=\cos^3 t$, $0\leq t \leq \pi/2$. (Note that $x=\cos t$, $y=2\sin t$ parametrizes the ellipse $x^2+\frac{y^2}{4}=1$ in the $xy$-plane, since $(\cos t)^2 + \frac{1}{4} (2\sin t)^2 = \cos^2t + \sin^2t =1$.)
Then
\begin{align*}
\int_C - ydx &+ x dy + x^2 dz
 \\
& = \int_0^{\pi/2} \Big[ (-2\sin t)(-\sin t)+ (\cos t)(2\cos t) + (\cos^2 t)(-3\cos^2 t \sin t) \Big] dt
\\
&= \int_0^{\pi/2} [2+3\cos^4 t(-\sin t))]dt
\\
&= 2t+\frac{3}{5}\cos^5 t \Big|_0^{\pi/2}= \pi -\frac{3}{5}.
\end{align*}



\medskip


\item
{\it 
Let $f:\rr^3\to\rr$ be a smooth function with $\nabla f(\bfr) \neq {\bf 0}$ for all $\bfr$.  Let $S=\{\bfr: \ 
f(\bfr)=a\}$ be a level surface of $f$.  Prove that
$$\int_C \nabla f\cdot d{\bf r}=0$$
for any oriented smooth curve $C$ which is contained in $S$.}


\medskip

We have $\int_C\mathbf{F}\cdot d{\bf r} =\int_C\mathbf{F}\cdot{\bf T}\, ds.$ If $C$ lies
on a level surface of $f$, then the vector 
${\bf T}$ tangent to the curve $C$ is also tangent to the surface $S$.
Therefore ${\bf T}$ is perpendicular to 
the vector $\nabla f$ which is normal to $S$.
But this means that $\nabla f\cdot{\bf T}=0$ on $C$, so that
$\int_C \nabla f\cdot d{\bf r}
=\int_C \nabla f\cdot {\bf T}\,ds=0$.

\medskip




\item {\it 
Let $\mathbf{F}:\rr^3\to\rr^3$ be a smooth vector field.  Let $C$
be a smooth curve in $\rr^3$.  Prove that }
$$\Big|\int_C\mathbf{F}\cdot d{\bf r}\Big|
\leq\int_C|\mathbf{F}|ds.$$


\medskip
Since $|{\bf F}\cdot{\bf T}|= |\mathbf{F}||{\bf T}| |\cos\theta| \leq
|{\bf F}||{\bf T}|=|\mathbf{F}|$, where $\theta$ is the angle between $\mathbf{F}$ and ${\bf T}$, we get
$$\Big|\int_C\mathbf{F}\cdot d{\bf r}\Big|
=\Big|\int_C\mathbf{F}\cdot{\bf T} ds\Big|
\leq \int_C|\mathbf{F}\cdot{\bf T}| ds
\leq \int_C|\mathbf{F}|ds.$$
(The absolute value of an integral is always $\leq$ than the integral of the absolute value of the integrand!)
\end{enumerate}










\section*{Practice Problems}
\begin{enumerate}[1.]


\item{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


\item 
{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

\item {
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

\item 
{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

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

\medskip


\item
{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



\item {
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 $(2,0,0)$.
}


\item {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.} 

\item {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.}








\end{enumerate}
