# 6.5 Vector fields, differential forms, and line integrals  (Page 2/3)

REMARK There is no doubt that the integral in this definition exists, because $P$ and $Q$ are continuous functions on the compact set $C,$ hence bounded, and ${\gamma }^{\text{'}}$ is integrable, implying that both ${x}^{\text{'}}$ and ${y}^{\text{'}}$ are integrable. Therefore $P\left(\gamma \left(t\right)\right){x}^{\text{'}}\left(t\right)+Q\left(\gamma \left(t\right)\right){y}^{\text{'}}\left(t\right)$ is integrable on $\left(0,L\right).$

These differential forms $\omega$ really should be called “differential 1-forms.”For instance, an example of a differential 2-form would look like $R\phantom{\rule{0.166667em}{0ex}}dxdy,$ and in higher dimensions, we could introduce notions of differential forms of higher and higher orders, e.g., in 3 dimension things like $P\phantom{\rule{0.166667em}{0ex}}dxdy+Q\phantom{\rule{0.166667em}{0ex}}dzdy+R\phantom{\rule{0.166667em}{0ex}}dxdz.$ Because we will always be dealing with ${R}^{2},$ we will have no need for higher order differential forms, but the study of such things is wonderful.Take a course in Differential Geometry!

Again, we must see how this quantity ${\int }_{C}\omega$ depends, if it does,on different parameterizations. As usual, it does not.

Suppose $\omega =Pdx+Qdy$ is a differential form on a subset $U$ of ${R}^{2}.$

1. Let $C$ be a piecewise smooth curve of finite length contained in $U$ that joins ${z}_{1}$ to ${z}_{2}.$ Prove that
${\int }_{C}\omega ={\int }_{C}P\phantom{\rule{0.166667em}{0ex}}dx+Q\phantom{\rule{0.166667em}{0ex}}dy={\int }_{a}^{b}P\left(\phi \left(t\right)\right){x}^{\text{'}}\left(t\right)+Q\left(\phi \left(t\right)\right){y}^{\text{'}}\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt$
for any parameterization $\phi :\left[a,b\right]\to C$ having components $x\left(t\right)$ and $y\left(t\right).$
2. Let $C$ be as in part (a), and let $\stackrel{^}{C}$ denote the reverse of $C,$ i.e., the same set $C$ but thought of as a curve joining ${z}_{2}$ to ${z}_{1}.$ Show that ${\int }_{\stackrel{^}{c}}\omega =-{\int }_{C}\omega .$
3. Let $C$ be as in part (a). Prove that
$|{\int }_{C}P\phantom{\rule{0.166667em}{0ex}}dx+Q\phantom{\rule{0.166667em}{0ex}}dy|\le \left({M}_{P}+{M}_{Q}\right)L,$
where ${M}_{P}$ and ${M}_{Q}$ are bounds for the continuous functions $|P|$ and $|Q|$ on the compact set $C,$ and where $L$ is the length of $C.$

The simplest interesting example of a differential form is constructed as follows. Suppose $U$ is an open subset of ${R}^{2},$ and let $f:U\to R$ be a differentiable real-valued function of two real variables; i.e., both of its partial derivatives exist at every point $\left(x,y\right)\in U.$ (See the last section of Chapter IV.) Define a differential form $\omega =df,$ called the differential of $f,$ by

$df=\frac{tialf}{tialx}\phantom{\rule{0.166667em}{0ex}}dx+\frac{tialf}{tialy}\phantom{\rule{0.166667em}{0ex}}dy,$

i.e., $P=tialf/tialx$ and $Q=tialf/tialy.$ These differential forms $df$ are called exact differential forms.

REMARK Not every differential form $\omega$ is exact, i.e., of the form $df.$ Indeed, determining which $\omega$ 's are $df$ 's boils down to what may be the simplest possible partial differential equation problem.If $\omega$ is given by two functions $P$ and $Q,$ then saying that $\omega =df$ amounts to saying that $f$ is a solution of the pair of simultaneous partial differential equations

$\frac{tialf}{tialx}=P\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\frac{tialf}{tialy}=Q.$

See part (b) of the exercise below for an example of a nonexact differential form.

Of course if a real-valued function $f$ has continuous partial derivatives of the second order, then [link] tells us that the mixed partials ${f}_{xy}$ and ${f}_{yx}$ must be equal. So, if $\omega =Pdx+Qdy=df$ for some such $f,$ Then $P$ and $Q$ would have to satisfy $tialP/tialy=tialQ/tialx.$ Certainly not every $P$ and $Q$ would satisfy this equation, so it is in fact trivial to find examples of differential forms that are not differentials of functions.A good bit more subtle is the question of whether every differential form $Pdx+Qdy,$ for which $tialP/tialy=tialQ/tialx,$ is equal to some $df.$ Even this is not true in general, as part (c) of the exercise below shows. The open subset $U$ on which the differential form is defined plays a significant role, and, in fact, differential forms provide a way of studyingtopologically different kinds of open sets.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
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what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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