anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Consider a region R covered from x=a to x=c and y=y1(x) to y=y2(x) let \[\frac{\partial \phi(x,y)}{\partial y}=\frac{\partial \phi}{\partial y}\] be continuous on it Then, \[\int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi }{\partial y}dy=\phi(x,y_{2})-\phi(x,y_{1})\] But how?? Here is what I think is how it's supposed to be.. \[d\phi=\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial x}dx\] Integrating while keeping x constant we get \[\int\limits_{\phi(x,y_{1})}^{\phi(x,y_{2})}d\phi=\int\limits_{y_{1}}^{y_{2}}\frac{\partial \phi}{\partial y}.dy+\int\limits_{x}^{x} \frac{\partial \phi}{\partial x}dx\] Now \[\int\limits_{a}^{a}f(x)dx=0\] so we get \[[\phi]_{\phi(x,y_{1})}^{\phi(x,y_{2})}=\int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi}{\partial y}dy+0\]\[\implies \int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi}{\partial y}dy=\phi(x,y_{2})-\phi(x,y_{1})\]
zzr0ck3r
  • zzr0ck3r
FTC?
Jhannybean
  • Jhannybean
Seems like it.

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Jhannybean
  • Jhannybean
I would like a solid example of this so i can picture how phi is moving though. :\
anonymous
  • anonymous
Similarly we can get \[\int\limits_{a}^{c} \frac{\partial \phi}{\partial x}dx=\phi(c,y)-\phi(a,y)\]
IrishBoy123
  • IrishBoy123
this is a really powerful statement, you made it right at the start \[\int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi }{\partial y}dy=\phi(x,y_{2})-\phi(x,y_{1})\] here, you are taking a partial in one term and generating a potential function for a conservative field, all in one go.
IrishBoy123
  • IrishBoy123
and one you have a conservative potential, you can pretty much switch your brain off
anonymous
  • anonymous
What if both x and y simultaneously varied? \[\int\limits_{\phi(a,y_{1})}^{\phi(c,y_{2})}d\phi= \int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi}{\partial y}dy+\int\limits_{a}^{c}\frac{\partial \phi}{\partial x}dx\]\[\phi(c,y_{2})-\phi(a,y_{1})=\phi(x,y_{2})-\phi(x,y_{1})+\phi(c,y)-\phi(a,y)\] \[\phi(c,y_{2})-\phi(a,y_{1})=(\phi(x,y_{2})+\phi(c,y))-(\phi(x,y_{1}+\phi(a,y))=\phi_{\max}-\phi_{\min}\]
anonymous
  • anonymous
Interesting stuff, the 1st post is actually part of Green's Theorem
Jhannybean
  • Jhannybean
Now I see a relation t that. Im understanding your forms, but not so much of whats going on in the problem.
anonymous
  • anonymous
I wanted to know how I arrived at \[\int\limits_{y_{1}}^{y_{2}}\frac{\partial \phi}{\partial y}dy=\phi(x,y_{2})-\phi(x,y_{1})\] was correct
IrishBoy123
  • IrishBoy123
here's some background
1 Attachment
IrishBoy123
  • IrishBoy123
i agree nish, this is really cool stuff!
anonymous
  • anonymous
Cheers!
zzr0ck3r
  • zzr0ck3r
I hate this stuff :) I like my math to be completely useless.
IrishBoy123
  • IrishBoy123
@zzr0ck3r lol!!! that is hilarious!!
IrishBoy123
  • IrishBoy123
i'm wondering, what does this actually mean?! \[\int\limits_{\phi(a,y_{1})}^{\phi(c,y_{2})}d\phi= \int\limits_{y_{1}}^{y_{2}} \frac{\partial \phi}{\partial y}dy+\int\limits_{a}^{c}\frac{\partial \phi}{\partial x}dx\] looks like a line integral, but the limits remind me of a area integral, where you actually do an iterated integral or an area, as opposed to a line integral between distinct points or along a distinct path and line integrals are integrals in 1 variable only, always. so you paramaterise or you find the relationship between say, x and y, and you build that into the 1 integration in 1 variable.
IrishBoy123
  • IrishBoy123
so "What if both x and y simultaneously varied?" you can't!

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