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12.1 - Double integrals from data
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<h2>Lab 09 - Vector Line Integrals</h2>
<h3>[Names]</h3>

<p>Writing a force field as \(\overset{\rightharpoonup }{F}=P \hat{i}+Q\hat{j}\), the vector line integral over some path \(C\) can be written in at
least two ways:
<ul>
<li>\(\begin{array}{c}
 \int _C\overset{\rightharpoonup }{F}\cdot d\overset{\rightharpoonup }{r}=\int _CPdx+\int _CQdy \\
\end{array}\)

<li>\(\begin{array}{c}
 
\begin{array}{c}
 
\begin{array}{c}
 \int _C\overset{\rightharpoonup }{F}\cdot d\overset{\rightharpoonup }{r}=\int _C\overset{\rightharpoonup }{F}(x(t),y(t))\cdot \overset{\rightharpoonup
}{r}'dt \\
\end{array}
 \\
\end{array}
 \\
\end{array}\)

<p>where \(\overset{\rightharpoonup }{r}'=\left\langle \frac{\partial x}{\partial t},\frac{\partial y}{\partial t}\right\rangle\). 
</ul>

<p>Use whichever
of these expressions is most useful in evaluating the following line integrals. Write down the answers you get here in your 
notebook, but attach paper showing your work (and labelled according to question number) to a printout of the notebook. 
  
<ol>
<li> Consider the force field \(\overset{\rightharpoonup }{F}_1=\overset{\rightharpoonup }{r}-2\hat{i}=x\hat{i}+y\hat{j}-2\hat{i}=(x-2)\hat{i}+y\hat{j}\).
Evaluate the vector line integral of this force field along the path \(C_1\): the straight line from (-1,0) to (0,1). 
<div class="justme"
$\(\pmb{\text{C1}=\text{ParametricPlot}[\{x,x+1\},\{x,-1,0\}]}\)$
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\includegraphics{Lab09-VectorLineIntegrals_gr1.eps}


<li> Evaluate the vector line integral of the same force field, \(\overset{\rightharpoonup }{F}_1,\) but this time along the path \(C_2\): the portion
of a radius one circle, traversed in counterclockwise direction from (-1,0) to (0,1). \\

\includegraphics{Lab09-VectorLineIntegrals_gr2.eps}

<li> Plot the vector field \(\overset{\rightharpoonup }{F}_1\) (use red arrows, see Lab 08) together with the integration paths \(C_1\)and \(C_2\).
<p>Do the *signs* of your vector line integrals match your visual estimation of their signs?

<li>Could \(\overset{\rightharpoonup }{F}_1\)be the gradient of some potential function \(f_1(x,y)\)? 

<p>Carry out the curl test to decide. (Show
the results of your test). 
Would you have guessed this result
(well, whether it's positive, zero or negative) by looking at the graph of the vector field arrows? 
<p>If the field is "irrotational", find
a potential function \(f_1\)such that \(\overset{\rightharpoonup }{F}_1=\overset{\rightharpoonup }{\nabla }f_1\) and make a plot which includes the
integration paths, the vector field of arrows *and* a ContourPlot[...] of \(f_1\)-- all together in the same figure.

<li> If the potential function exists, calculate the surface height difference \(f_1(0,1)-\)\(f_1(-1,0)\).

\includegraphics{Lab09-VectorLineIntegrals_gr3.eps}

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6-10.) Repeat the sections 1-5.) with the same curves \(C_1\)and \(C_2\), but this time with the vector field \(\overset{\rightharpoonup }{F}_2=\left(-y-\frac{1}{2}\right)\hat{i}+x^3\hat{j}\).
{ }(And try to find a potential called \(f_2\) if possible.)

11.) Back to the first vector field, \(\overset{\rightharpoonup }{F}_1\): { }Carry out a vector line integration of this vector field from the origin
to a point with coordinates \(\left(x_0,y_0\right)\) along a path with two segments: { }\\
\\
First go along the \(x\) axis from (0,0) to \(\left(x_0,0\right)\). { }\\
\\
Next go from (\(x_0\),0) to \(\left(x_0,y_0\right)\). { }This second path is parallel to the \(y\) axis. \\
\\
Now, replace \(x_0\rightarrow x\) and \(y_0\rightarrow y\). { }Comment on the result.

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