A heat integral (sometimes dubbed a path integral) of ascalar-valued role can be assumed of as a generalizationof the one-variable integral of a function over an interval,where the interval can be shaped right into a curve.A simple analogy that catches the essence of a scalar heat integralis that of calculating the fixed of a wire from that density.

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If the thickness of a cable were constant, climate one could calculateits fixed by multiply the thickness by the arc lengthof the curve.If, top top the other hand, the thickness varies along the wire, one can use a procedure similar to thatof calculating the arc lengthto derive a formula because that the density. This formula defines of the line integral end the wireof the duty giving the density.

Extending the slinky example used to present arc length,we allow one coil the the slinky it is in parametrized by $dllp(t) = (cos t, sin t, t)$, because that $0 le t le 2pi$,and imagine the the thickness of the slinky in ~ the suggest $(x,y,z)$is given by some role $dlsi(x,y,z)$.This function describes how the slinky could be more thick in someparts and also thinner in others. Because that a given value the $t$, apoint ~ above the slinky is $dllp(t)$, and also the thickness of the slinkyat that point is$dlsi(dllp(t))$, as shown below.


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Helix v variable density. The duty $dllp(t) = (cos t, sin t, t)$, because that $0 le t le 2pi$ parametrizes a helix. The blue curve representing the helix is drawn with a variable width, representing the varying thickness $dlsi(x,y,z)$ that the helix at each suggest $(x,y,z)=dllp(t)$ follow me its length. Through dragging the cyan point representing $t$ ~ above the slider, one can move the red suggest $dllp(t)$ follow me the helix.

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To calculation the mass of the slinky, us repeat the procedure us usedto calculation its length. Us pretend the slinky is created by a bunch of directly line segments. Moreover, we assume each line segment has actually constantdensity, therefore the massive of the line segment is just its size times itsdensity. Us let the thickness of each line segment it is in the density ofthe slinky at the allude where the upper finish of present segmenttouches the slinky. (In the listed below applet, we usage width the the curves to stand for density.) because that a provided discretization $Delta t$, we estimate the mass of the slinky together the totalmass of the heat segments (labeled as mass($Delta t$)).


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Line integral the helix mass from density. The role $dllp(t) = (cos t, sin t, t)$, because that $0 le t le 2pi$ parametrizes a helix. The blue curve representing the helix is attracted with a change width, representing the varying density $dlsi(x,y,z)$ the the helix in ~ each suggest $(x,y,z)=dllp(t)$ along its length. The green line segment of consistent width represent segments of constant density that approximate the helix. The discretization size of the heat segments $Delta t$ deserve to be readjusted by relocating the cyan allude on the slider. Together $Delta t o 0$, the massive of the heat segment approximation, labeling mass($Delta t$), viewpoints the actual fixed of the helix.

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We can increase the variety of line segments (decreasing the size ofeach line segment) so that the full mass the the heat segments becomesa much better estimate of the slinky mass. As the discretization dimension $Delta t$approaches zero, the size of every line segment shrinks towardzero, the number of line segments increases, and also the density of eachline segment much more closely approximates the density of the slinky.Since the total length of the heat segments ideologies the slinkylength, the total mass the the heat segments philosophies the mass of theslinky.

The source of the heat integral formula is similar tothe procedure we used to estimate theslinky length.The size of the $i$th heat segment can be created as$|dllp(t_i) - dllp(t_i-1)|$. The density of the heat segment issimply $dlsi(dllp(t_i))$ since $dllp(t_i)$ is the point where theupper finish of the line segment touches the slinky. Themass the the heat segment is its thickness times that is length:$dlsi(dllp(t_i))|dllp(t_i) - dllp(t_i-1)|$. We attain the complete mass of the heat segments bysumming over every $n$ line segments:eginalign* sum_i^n dlsi(dllp(t_i)) | dllp(t_i) - dllp(t_i-1)|.endalign*

The just difference in between the above expression and the onewe acquired forarc size calculationis the $dlsi(dllp(t_i))$ factor. To rotate this into an integral, us repeat the same measures we did on the page.We again allow $Delta t_i = t_i - t_i-1$, multiply and divide each term by $Delta t_i$, and also obtain the more complicated looking expression.eginalign* sum_i^n dlsi(dllp(t_i))| dllp(t_i) - dllp(t_i-1)| &= sum_i^ndlsi(dllp(t_i)) left| fracdllp(t_i-1 + Delta t_i) - dllp(t_i-1)Delta t_i ight| Delta t_i.endalign*Since the ugly expression in the $| cdot |$ is the amount inlimit meaning of the derivative of $dllp(t)$, once we let $Delta t_i o 0$ and $n o infty$, the above Riemann amount converges to the integraleginalign* int_a^b dlsi(dllp(t))| dllp,"(t) | dt,endalign*where $a=0$ and also $b=2pi$. This is dubbed the heat integral of $dlsi$over the curve parametrized by $dllp$.

We often use $als(t)$ to be the length of the slinky from thebeginning allude $dllp(a)$ approximately the allude $dllp(t)$. If friend thinkof $dals$ together being the length of a tiny heat segment approximatingthe slinky, climate the fixed of the tiny line segment is $dlsi$ times$dals$. For this reason, us often represent the integral representingthe fixed of the slinky aseginalign* slintdllpdlsi =int_a^b dlsi(dllp(t))| dllp,"(t) | dt.endalign*

The notation $slintdllpdlsi$ indicates that the integralof $dlsi$ is over the parametrized curve $dllp$.But, if the integral can really represent the fixed of a wirewith thickness $dlsi$, climate the mass have to not count on just how wechose to parametrize the wire. In fact, the does not, and also you can read around how the heat integralis independentof the parametrization supplied to stand for the curve.

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You can additionally read part examplesof calculating line integrals that scalar functions.


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Nykamp DQ, “Introduction come a line integral the a scalar-valued function.” From math Insight. Http://barisalcity.org/line_integral_scalar_function_introduction

Keywords:integral, heat integral, parametrized curve

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