Line Integrals (I'm missing something here)

AI Thread Summary
The discussion focuses on evaluating the line integral of the vector field F(x,y) = <3xy, 8y^2> along the curve C defined by y = 8x^2 from (0,0) to (1,8). The initial approach considered checking if the vector field is conservative, but it was determined that it is not due to the inequality of mixed partial derivatives. The participants discussed the need to parameterize the curve and suggested using x as the parameter to express dy in terms of dx. The integral can then be evaluated by taking the dot product of F with the differential vector dr along the parameterized curve. Overall, the conversation emphasizes understanding parameterization and the evaluation process for line integrals in vector calculus.
FrogPad
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Ok the question is:

\vec F(x,y)=&lt;3xy,8y^2&gt;\,=3xy\,\hat i + 8y^2\,\hat j
C: y=8x^2
Where C joins (0,0)\,,\,(1,8)

Evaluate \int_{C} \vec F \cdot d\vec r

I'm unsure how to evaluate this. I really do not want just an answer, I want to know how to solve it.
My first thought was to check if it is conservative then I can simply let \vec r(t) be a line between the point. Or, (if conservative) I could use the fundamental theorem for line integrals. But:

\frac{dP}{dy} \neq \frac{dQ}{dx}

...so it is not conservative.

My last idea is that I need to paramterize y=8x^2 and this is where I get kind of shaky as far as my abilities with vector calculus. (We just started this chapter). So any insight on this problem, or parameterizing functions would be awesome. Thanks in advance.
 
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Frogpad,

I'm not familiar with the notation in your first equation. What does it mean?
 
its just some vector notation, it is the same thing as:

F(x,y) = 3*x*y (i) + 8*t^2 (j)

where i and j are the unit vectors in the x and y directions
 
He's just using pointed brackets instead of rounded ones to signify vectors. Okay, my question is, if the vector field F is only a function of x and y, why does its definition have t in one of the components?
 
Well that would be because I made an error!
hehe, sorry about that.
It is fixed now.
 
As far as I remember,

\int{F(x)\bullet d\vec{r} = \int{F(y(x))y&#039;(x)dx}

In case I translated it wrong, for us it was
\int{F\bullet dr} = \int F(r(t))r&#039;(t) dt
 
Ok. You want to take the dot product of F along this curve with the vector dr. So you can just evaluate the integral of

\int_0^1 \vec{F}(x, y = 8x^2)\cdot (dx\vec{i} + dy\vec{j})

taking care to express dy as a function of x and dx.

What I have done here is I have set x to be the parameter.
 
Don't use that notation,please...It usually stands for inner product...


Daniel.
 
Who are you talking about?
 
  • #10
Wasn't it obvious i was referring to the vicious

\vec{F}\left(x,y\right)=\left\langle 3xy,8y^{2}\right\rangle ?

Daniel.
 
  • #11
the < > notation
 
  • #12
it's easier to write, and I am lazy --- so naturally I am ok with your notation :D
 
  • #13
Ok wow, that was easy. Thanks guys.

Yeah my book is all about using <,> notation to refer to vectors. I usually just use a matrix represntation but I didn't want to spend the time looking up the latex formatting. Anyways, thanks for the help.
 
  • #14
Hmm,then your book doesn't treat vector and preHilbert spaces and uses the \cdot notation for the (euclidean) scalar product.

Daniel.
 
  • #15
how big is your brain? seriously... does your neck hurt. :)
 
  • #16
BTW,the latex formatting is

\vec{F}=\left(F_{x},F_{y}\right)

Not too much to look for.

Daniel.
 
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