Stokes Theorem

all right let's get stoked with stoke's theorem so today i would like to present you the dark side of stoke's theorem where you use it to calculate a line integral instead of a surface integral and in fact today we'll calculate the line integral of fdr where f is this adorable vector field x plus y squared y plus z squared z plus x squared and where c is the curve of intersection of two surfaces on the one hand the hemisphere z equals square root of five minus x squared minus y squared and the plane z equals one so what this surface is you simply take a hemisphere and you cut it off by a plane and we'll show later that in fact this is a circle and this circle will call it c again claro que si and and what we would like to do we would like to calculate the line integral of f over this circle and orientation matters here so we orient it counterclockwise if you look at it from a bird's eye view so if we look at it from above it is counterclockwise all right and so what does stoke's theorem say it simply says the following it says that the line integral of f so the stuff we want to calculate is just the same thing as the surface integral of the curl of f in other words the integral of a function is the same as the double integral of a derivative so it is like the fundamental theorem of calculus but here what it says is simply the following the line integral so an easy integral of a complicated function becomes a surface integral so a complicated integral of an easy function so in some sense this is the yin yang of math easy of heart becomes hard of easy so it can balances out and for this now the first step is of course to calculate the curl of f so the curl of f i like to remind you is just the gradient crossed with your vector field and what it becomes is just a big determinant i j k of again partial over partial x partial over partial y partial over partial z and you put your vector field so which i like to remind you is just x plus y squared y plus z squared and then z plus x squared so all we need to do is to calculate this determinant so let me remind you how to do this so you start with the first component and bomberman this and what you get is simply partial over partial y of that minus partial over partial z of this component let me write that down so partial over partial y of z plus x squared minus partial over partial z y plus z squared and then the second one some in red remember because you're doing determinants it becomes plus minus so that is minus thing here and what this becomes is not this times this but minus this times this so minus partial over partial x z plus x squared and then minus minus becomes plus so plus partial over partial z x plus y squared and last but not least we have the very last component this one here which becomes partial over partial x of this so partial over partial x of sorry of this sorry of y plus z squared minus partial over partial y of x plus y squared now i know this looks complicated but look it simplifies so beautifully this thing disappears because there is no y this thing just becomes minus 2 z now this thing becomes minus 2x and this disappears okay you're the weakest link goodbye and then um this disappears because there's no x and lastly you get minus 2y so let me ask you this which is easier to integrate x plus y squared y plus z squared z plus x squared or hm minus 2 z minus 2x minus 2y it's like this meme of this person that's like uh you know sounds disgusting but now this is more and more delicious yeah all right so all we need to do again instead of doing a line integral of this hard thing you just need to do the surface integral of this easier thing so win-lose situation but now the next question is what is the surface what is this and for this let's figure out first of all what the curve is all right so again let me remind you the curl of f was this easy vector field and c from the very beginning was just a curve of intersection of those two surfaces so as i said our goal is ultimately to find s but before we do that let's first of all find the curve find c but that's not very hard because all you need to do is intersect those two things so that set this equal to this so square root of 5 minus x squared minus y squared equals 1 and then we get 5 minus x squared minus y squared equals 1. and if you put this on the right hand side and this on the left you ultimately get x squared plus y squared equals 4. so what this is is that c is a circle okay it's a circle of radius 2 but just be a little bit careful it's not the one in the x y plane it's the circle in the plane z equals 1. so again just like our picture showed c is a circle but in the plane as z equals 1. so c is again just a circle of radius two in the plane z equals one now here's the thing the question is what is the surface you might be tempted to just say s is just the upper hemisphere and then parameterize the upper hemisphere and then cry because you would have to do you know cross products etc etc but actually none of that thing because here's a beautiful thing about stoke's theorem stokes theorem does not depend on what the surface is so if this is the curve c no matter which surface you pick if you are oriented upward or something you still get the same result so you could pick s to be this weird surface you could pick s to be the hemisphere or you could even pick s to just be the inside of that circle it would all give you the same result so why not just be lazy and pick the easiest version which here is just this flat surface so again just to reiterate what is s well because the result doesn't depend on which surface we have let's just pick the easiest one which is just the inside of this circle all right so now that we figured out that it's the inside of the circle the question is how do you parameterize it well here you have many different choices you could use polar coordinates or even lazier just pick x y so if this is x y this surface is just parameterized by x y and one because remember this is in everything is in the plane z equals one so kind of the shadow is x y but if you go up you can just parameterize this with x y comma one so parameterize s and so simply we just choose r of x y to be x y comma 1. and by the way very soon i will write d well d what it is is just a space where x and y lies in and in this case what it is as we found is just a disk of radius 2. in other words the shadow under s and so the question is then well now that we found a parametrization how can we calculate the surface integral remember it's just finding the outward pointing normal vector which you can just get by calculating the derivatives of r and crossing them so let's calculate rx so you just take the x derivative of each component so 1 0 0 you calculate ry which is 0 1 0. and then you cross them i know physicists are like obviously it's k but just in case you have a harder problem at some point you calculate the normal vector so let's i called it n n-hat which is rx cross ry and that becomes again i j k 1 0 0 0 1 0 and we get in the end 0 0 1. but remember you always need to check if the orientation makes sense so here if the curve is counterclockwise you need to make sure that the normal vector faces up and indeed it faces up because the last component is non-negative i guess it's positive in this case so this is the correct normal vector and then the question is how do you evaluate the surface integral well you just take you curl and you dot it with this normal vector and you integrate the resulting thing so last but not least we arrived at our final step all right and now we've just arrived at our final step so let's actually calculate the line integral so remember by stoke's theorem the line integral of f is the same thing as the surface integral of the curl of f dotted with the s now i would like to remind you the curl was this easy vector phil that was just i keep forgetting minus 2 z minus 2x minus 2y and we had this parametrization so r x y equals xy1 and the normal vector was just zero zero one so i would like to remind you how do you calculate the surface integral of a vector field all you do you take your vector field but replace x y z with the parametrization so here all you do you replace c by 1 and you dot this with the normal vector 0 0 1 and you integrate the resulting thing so what this becomes is then it's the double integral over this shadow region where x and y lie in of again curl of f but here it becomes minus 2 times 1 minus 2x minus 2y so again that's just curl of f dotted with that normal vector 0 0 1 and you integrate this with respect to x and y so the nice thing is in the end this simplifies tremendously just to become the double integral over d of minus 2 y d x d y and lastly i would like to remind you what was d it was just the disc centered at zero in radius two so again it was just a shadow under your curve and for d you can just parametrize this with polar coordinates so this becomes the integral from 0 to 2 pi integral from 0 to 2 of minus 2 r sine of theta and again r d r d theta and i know some of you already know the answer but in case you don't let's just split it up so we get then the following thing so this becomes on the one hand the integral from 0 to 2 of minus 2r squared dr and the integral from 0 to 2 pi of sine of theta d theta but the point is without even calculating the first thing an anti-derivative here is minus cosine of theta from 0 to 2 pi and then if you evaluate this in the end you get 0. so at the very end you get the answer is zero just like my previous stokes video so maybe not a satisfying answer but i think the method is quite satisfying and never in my life would i have thought that i would make a video on stoke's theorem because when i took multi-variable calculus i didn't know how to do any problems about stoke's theorem so this is very very cool all right i hope you like this if you want to see more math 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Channel: Dr Peyam

Views: 21,320

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Id: L3fWasM-F5E

Length: 16min 14sec (974 seconds)

Published: Mon Jan 18 2021