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6.2 H - Scalar line integrals
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[6.2] - Scalar Line Integrals
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<ol>
<li>Let $f(x,y,z)=x-3y^2+z$.<br>
Integrate $f$ over the line segment $C$ joining the origin $(0,0,0)$ and the 
point $(1,1,1)$.<br>
<img src="0g/13.2.scalar1.jpg"><br>
<br>

<li>Let $f(x,y,z)=x-3y^2+z$.<br>
Integrate $f$ over the 2-segment path shown joining the origin $(0,0,0)$ and the 
point $(1,1,1)$.<br>
<img src="0g/13.2.scalar2.jpg"><br>

</ol>

<h3 style="page-break-before:always" id="A">Answers</h3>
<p>1.) 0;  2.) $\frac{-3-\sqrt 2}2$
<p><a href="#A" onclick="toggle_visibility('aA');">show / hide</a> the details...
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<ol>
<li>The line segment $C$ can be specified by $\myv r(t)=t\uv i+t\uv j+t\uv k$.
<p>$\myv r'(t)=\langle 1,1,1\rangle \Rightarrow |\myv r'(t)|=\sqrt 3$. And so,
$$\begineq \int_C f(x,y,z)\,dx&=&
\int_0^1 (t-3t^2+t)\,\sqrt 3\,dt\\
&=&\sqrt 3\int_0^1(2t-3t^2)=\sqrt 3(1^2-1^3)=0\endeq$$
The answer is <span class="answer">0</span>.

<li>The path consists of 2 segment, I'll call these $C_1: \ (0,0,0)\to(1,1,0)$ 
and $C_2:\ (1,1,0)\to(1,1,1)$.
Then $$\int_C f\,ds=\int_{C_1}f\,ds+\int_{C_2}f\,ds.$$

<h4>$C_1$:</h4>
$\myv r(t)=\langle t,t,0 \rangle$ with $0\leq t\leq 1$<br>
$\myv r'(t)=\langle 1,1,0\rangle$ and $|\myv r'(t)|=\sqrt 2$.<br>
$f(x(t),y(t),z(t))= t-3t^2$<br>
$$\int_{C_1}f\,ds=\int_0^1 (t-3t^2)\,\sqrt 2\,dt=\left(\frac{1^2}2-(1)^3\right)\sqrt 2=-\frac{\sqrt2}2.$$
<h4>$C_2$:</h4>
$\myv r(t)=\langle 1,1,z \rangle$ with $0\leq z\leq 1$<br>
$\myv r'(t)=\langle 0,0,1\rangle$ and $|\myv r'(t)|=1$.<br>
$f(x(t),y(t),z(t))= 1-3(1)^2+t=t-2$<br>
$$\int_{C_2}f\,ds=\int_0^1 (-2+z)\,dz=\left(-2*(1)+\frac12(1)\right)=-\frac32.$$
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<h4>$C=C_1+C_2$:</h4>
$$\int_C f\,dS=\frac{-\sqrt 2}{2}-\frac32=\frac{-3-\sqrt 2}2.$$
The answer is <span class="answer">$\frac{-3-\sqrt 2}2$</span>.
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