Thread: Line integral

Evaluate the line integral, where C is the given curve.

$\displaystyle \int_C xy~dx + (x-y)~dy$ C consists of line segments from (0,0) to (2,0) and from (2,0) to (3,2)

can someone help me set this up properly? I thought I was doing it right, but seem to be getting incorrect answers.

Also, latex doesn't seem to like this format or something, but the actual integral is: the integral over the curve of xy dx + (x-y) dy

Originally Posted by angel.white

Evaluate the line integral, where C is the given curve.

$\displaystyle \int_C xy~dx + (x-y)~dy$ C consists of line segments from (0,0) to (2,0) and from (2,0) to (3,2)

can someone help me set this up properly? I thought I was doing it right, but seem to be getting incorrect answers.

Also, latex doesn't seem to like this format or something, but the actual integral is: the integral over the curve of xy dx + (x-y) dy

Since latex isn't working, I wrote it out...



Hopefully this is still legible... :-|

Can you take it from here?

--Chris

Originally Posted by Chris L T521

Since latex isn't working, I wrote it out...



Hopefully this is still legible... :-|

Can you take it from here?

--Chris

Thank you very much, Chris. However, as I understand it, this problem only has lines from (0,0) to (2,0) and from (2,0) to (3,2) but not from (0,0) to (3,2)

Originally Posted by angel.white

Thank you very much, Chris. However, as I understand it, this problem only has lines from (0,0) to (2,0) and from (2,0) to (3,2) but not from (0,0) to (3,2)

Alright, I think we can fix that...



I think I got this...

I hope you can follow this!

--Chris

Originally Posted by Chris L T521

Alright, I think we can fix that...



I think I got this...

I hope you can follow this!

--Chris

Thank you ^_^

I'm not sure where my discrepancy has been, perhaps in converting to vectors, I had just kept everything in terms of x, that could be it. Or maybe my mind is disorganized and I errored somewhere along the line.

Also, for C2, I got \int_0^1 xy dx = 8/3
then 0+0+8/3 + 3 = 17/3

Which my book likes.

I appreciate it, there is always 1 or 2 problems which just drive me nuts. I think if I solve it wrong the first time, I'm more and more likely to solve it wrong the subsequent times.

Originally Posted by angel.white

Thank you ^_^

I'm not sure where my discrepancy has been, perhaps in converting to vectors, I had just kept everything in terms of x, that could be it. Or maybe my mind is disorganized and I errored somewhere along the line.

Also, for C2, I got \int_0^1 xy dx = 8/3
then 0+0+8/3 + 3 = 17/3

Which my book likes.

I appreciate it, there is always 1 or 2 problems which just drive me nuts. I think if I solve it wrong the first time, I'm more and more likely to solve it wrong the subsequent times.

Woops! I forgot that 8/3 part [you tend to overlook things when you're dead tired. >_> ] ...at least you got the answer now!

--Chris