00:01
Okay, so before we go on with the solution of this exercise, let's find the parametrization of alpha.
00:07
Well, this is easy.
00:08
We can just set x of t equal to cosine of t and y of t equal to sine of t.
00:18
This is a parametrization of alpha with t belonging to the interval 0 ,2pi.
00:27
Well, with this parametrization, our ds is just 2pi multiplied by dt.
00:45
Perfect.
00:46
Now, let's compute our integral along alpha of xy minus x plus y in ds.
00:56
Well, this one is what? this one is going to be 2pi multiplied by an integral from 0 to 0.
01:05
If t belongs to this interval, ds is just dt.
01:11
Perfect.
01:11
This one is just 1 multiplied by dt, so that the integral from 0 to 2pi of ds is equal to 2pi.
01:21
Perfect.
01:22
Okay, we just need to be careful with the parametrization.
01:25
Okay, that being said, so here we have just 1...