hi!
May i know how to solve this question?
and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?
http://i752.photobucket.com/albums/x...0/DSCF4517.jpg
Printable View
hi!
May i know how to solve this question?
and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?
http://i752.photobucket.com/albums/x...0/DSCF4517.jpg
How about splitting $\displaystyle \vec{F}(x,y)$, into a much simpler component $\displaystyle \vec{F}_1(x,y):=(-y,0)$ and a component $\displaystyle \vec{F}_2(x,y):= (y\sin(xy),x\sin(x,y))$ that happens to be a conservative field like thisQuote:
Originally Posted by alexandrabel90
$\displaystyle \vec{F}(x,y)=\vec{F}_1(x,y)+\vec{F}_2(x,y) := (-y,0)+\nabla \Phi(x,y)$, where $\displaystyle \Phi(x,y) := -\cos(xy)$.
Since the curve is closed, the conservative part can be dropped from the line integral: you only need to integrate $\displaystyle F_1(x,y)=(-y,0)$ along that curve.
Are you asking what happens if you reverse the direction of the entire curve? - Well, in that case the line integral changes its sign.Quote:
and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?
no. i meant just reversing the direction of the curve on the left hand side, while the direction on the right hand side remains the same.
In that case, things just might get a little more complicated (to put it politely): because in that case, the line integral for the remaining (non-conservative) component $\displaystyle \vec{F}_1(x,y)=(-y,0)$, may not (due to the symmetry of the curve) evaluate to 0 anymore...Quote:
Originally Posted by alexandrabel90
may i know how you got F1 and F2 from your reply above?
from how i see it, you split the curves up so that it will be a simple( non intersecting) curve?
sorry im still very confused how to solve this current question.
and for the question where if the direction on the left hand side changes, i was thinking that since both the arrows will now oppose each other, they will cancel out and hence the line integral will be 0? but i guess i cant say it that way right?
No my separating $\displaystyle \vec{F}(x,y)$ has got nothing to do with the curve, it is simply a splitting up of the vector $\displaystyle \vec{F}(x,y)$ itself into a non-conservative and a conservative part. Integration along a closed curve allows you to just drop the conservative part and concentrate on the non-conservative part exclusively.Quote:
Originally Posted by alexandrabel90
You can say it that way, alright, but you can't write it down like this: just take the intuition it provides to split the overall line integral into two line integrals (over parts of the curve) that cancel each other. To my eyes at least it seems that the line integral over the part of the curve in quadrants IV and I should cancel against the line integral over the part of the curve in quadrants II and III. If that hypothesis happens to be correct (I haven't checked), you just have to juggle the parametrizations of these two line integrals in such a way that their cancelling each other out is made obvious (without any need for actually calculating the values of the two line integrals themselves).Quote:
Originally Posted by alexandrabel90
thanks for the explanation.
by the way, in this case, we cant apply green's theorem because the curve is not simple right?
