integrals

**alexandrabel90** · April 23rd 2010, 06:59 AM

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?

**Failure** · April 24th 2010, 11:24 AM

Originally Posted by alexandrabel90

hi!

May i know how to solve this question?

How about splitting $\vec{F}(x,y)$ , into a much simpler component $\vec{F}_1(x,y):=(-y,0)$ and a component $\vec{F}_2(x,y):= (y\sin(xy),x\sin(x,y))$ that happens to be a conservative field like this

$\vec{F}(x,y)=\vec{F}_1(x,y)+\vec{F}_2(x,y) := (-y,0)+\nabla \Phi(x,y)$ , where $\Phi(x,y) := -\cos(xy)$ .

Since the curve is closed, the conservative part can be dropped from the line integral: you only need to integrate $F_1(x,y)=(-y,0)$ along that curve.

and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?

Are you asking what happens if you reverse the direction of the entire curve? - Well, in that case the line integral changes its sign.

**alexandrabel90** · April 24th 2010, 02:02 PM

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.

**Failure** · April 24th 2010, 10:53 PM

Originally Posted by alexandrabel90

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 $\vec{F}_1(x,y)=(-y,0)$ , may not (due to the symmetry of the curve) evaluate to 0 anymore...

**alexandrabel90** · April 24th 2010, 11:16 PM

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?

**alexandrabel90** · April 24th 2010, 11:18 PM

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?

**Failure** · April 24th 2010, 11:28 PM

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?

No my separating $\vec{F}(x,y)$ has got nothing to do with the curve, it is simply a splitting up of the vector $\vec{F}(x,y)$ 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.

**Failure** · April 24th 2010, 11:38 PM

Originally Posted by alexandrabel90

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?

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).

**alexandrabel90** · April 25th 2010, 12:36 AM

thanks for the explanation.

by the way, in this case, we cant apply green's theorem because the curve is not simple right?

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