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# Green's theorem example 2

## Video transcript

let's say have a path in the XY plane that's essentially the unit circle this is essentially the unit circle so it's my y-axis this is my x-axis so my x-axis and our path is going to be the unit circle it's going to be the unit circle and we're going to traverse it just like that I'm going to traverse it clockwise o'clock 1c I think you get the idea and so because it's equation it's the unit circle so the equation of this is x squared plus y squared is equal to 1 has a radius of one unit circle and what we're concerned with is the line integral the line integral over this curve C over this curve C it's a closed curve C it's a closed curve C and it's actually going in that direction of 2y DX minus minus 3 X minus 3 X D Y so we are probably tempted to use greens theorem and why not so let's try so this is our path so Green's theorem tells us that the integral of some curve f dot dr f dot er over some path where F is equal to where F is let me write it a little bit neater where f of X Y is equal to P of X Y I plus Q of X Y J that F dot that this integral is equal to it is equal to the double integral over the region this would be the region under question in this example over the region of the partial of Q with respect to X the partial of Q with respect to X minus minus the partial P with respect to Y all of that da the differential of area and of course the region is that what I just want to show you now you may or may not remember I did I made what well what there's a slight subtle thing in this which would give you the wrong answer in the last video we said that greens theorem applies we're going counterclockwise counterclockwise notice even on this little thing on the integral I made it go counterclockwise in our example the curve goes clockwise the region is to our right Green's theorem this applies when the region is to our left region to left so in this situation when the region is to our right and we're going o'clock so this is counter clockwise counter clockwise so in our example where we're going clockwise the region is to our right Green's theorem is going to be the OP is going to be the negative of this so in our example we're going to have the integral of C and we're going to go in the clockwise direction so maybe I'll draw it like that of F dot d R this is going to be equal to the double integral over the region and we could just swap these two of the partial of P with respect to Y minus the partial of Q with respect to X da so let's do that so this is going to be equal to in this example the integral over the region let's just keep it abstract for now we could start setting the boundaries so let's just keep the region abstract and what is the partial of P with respect to let's just remember this right here is our I think we could recognize right now that this if we take F dot d R we're going to get this the dr contributes those components the the f contributes these two components so this is P of XY that is P of XY and then this is Q of XY and if we've seen I don't want to go into the whole dot d R and take the dot product over and over again I think you can see that this is the dot product of two vectors this R this is the X component of F Y component of F this is the X component of dr y component of the are so let's take the partial of P with respect to Y you take the derivative of this with respect to Y you get two derivative of 2y is just 2 so you get 2 and then minus the derivative of Q with respect to X derivative of this with respect to X s minus 3 so we're going to get minus 3 and then all of that d a and this is equal to the integral over the region over the region what's this is 2 minus minus 3 that's the same thing as 2 plus 3 so it's the integral over the region of 5 da 5 just a constant so we can take it out of the integral so this is going to turn out to be quite a simple problem so this is going to be equal to 5 times the double integral of over the region R da now what is this thing what is this thing right here it looks very abstract but we can solve this this is just the area area of the region that's what that double integral represents you just sum up all the little DA's right that's a da that's the da you sum up the infinite sums of those little DA's over the region well what's the area of this unit circle here we just break out a little bit of ninth-grade would actually even earlier than that pre-algebra or middle school geometry area is equal to PI R squared what's our radius it's a unit circle our radius is 1 our radius is one length this one so the area here is PI so this thing over here that whole thing is just equal to PI so the answer to our line integral is just 5 pi which is pretty straightforward I mean we could have taken the trouble of setting up a double integral where we take the antiderivative with respect to Y first and write you know Y is equal to the negative square root of 1 minus x squared Y is equal to the positive square root X goes from minus 1 to 1 but that would have been super hairy and a bunch of a huge pain we just have to realize no this is just the area and and the other interesting thing is I challenge you to solve the same integral without using greens theorem doing this a way we you know actually take get generating a parameterization for this curve going in that direction taking the derivatives of X of T and y of T multiplying by the appropriate thing and then taking the antiderivative way hairier way hairier than what we just did using greens theorem to get 5 pi and remember the reason why it wasn't minus 5 PI here is because we're going in a in a clockwise direction if we were going in a counterclockwise direction we could have applied the straight-up greens theorem and we would have gotten minus 5 pi anyway hopefully found that useful