Stokes' Theorem Example // Verifying both Sides // Vector Calculus

in this video i'm going to do an example that uses stoke's theorem we were first introduced to stoke's theorem in actually the previous video in my vector calculus playlist the link to that is down in the description but in this video i actually want to use stoke's theorem to do some computations so let's first try to sketch the region that we're talking about and see how and why stokes's theorem might apply so what we have is some particular half sphere x squared plus y squared plus z squared equal to one but only the portion which is greater than zero so i'm going to go and sketch a half of a sphere and then through that surface we have this vector field y and the i hat minus x and the j hat and zed in the k hat just some vector field doesn't matter too much for the purposes of our story here now i actually have to tell you one more thing in the example and i'm going to tell it to you pictorially i have to tell you an orientation for this surface and for the boundary curve that should be compatible with each other so what i'm going to imagine is the boundary curve is rotated around this way in the counterclockwise direction when viewed from above this is compatible with having our normal vectors to the surface be pointing outwards at any particular point and that satisfies the right hand rule that when you take your fingers and go along the boundary the normal would be sticking upwards so that's my choice of orientation and i'm going to impose that as part of the problem now stoke's theorem is an equality between two relatively different things on the left hand side it's the circulation around the curve so maybe i'll just take the base curve and i'll draw it in on blue here to indicate that's one question i could ask what is the circulation around that curve the other question i could ask would be to integrate well the curl in the normal direction the curl of f dotted with n and integrate that over the entire region so a surface integral that would be something i'd be doing over the entire region now often one of those two things is easier and the other is going to be harder so the types of problems you might use doses theorem for are often when one of these is doable and the other is not in this example actually both of them are going to be doable both sides of this equation so i'm really going to use this example to make sure i can do both sides of the computation and note that they are indeed the same as stoke's theorem is going to get us and then we're gonna have a really cool point at the end where we say well hold on i could even replace this surface with a completely different surface that still has the same boundary and we'd get the same result there as well but it will be just vastly easier okay so which side should we do first how about the left side the circulation the question of what is the circulation along this boundary curve of this field is a problem we could have done back earlier in the course and indeed the most important thing is to come up with a parameterization for the curve so i'm going to do that i'm going to have a parameterization for that boundary curve i'm going to parameterize it into theta the boundary curve is just this unit circle and so i'm going to write this as cosine of theta with radius one in the i hat sine of theta in the j hat and then zero in the k hat because that boundary circle is in the x y plane it has no k hat component the next thing i need to do to compute a flow interval or a circulation interval like this is to compute the derivative so i'm going to compute r prime of theta and i'm going to get minus sine of theta in the i hat cosine of theta in the j hat and still zero in the k hat so now i'm actually going to compute that line integral along the curve of f dot dr having established my r and my r prime and what this is is the integral from zero up to two pi i by the way added in that the theta values were between zero and two pi as you always tell the endpoints of your parametrization as well okay so then what we're going to be integrating is the field f dotted with r prime dt and so what i have is my field components well the first field component is y which in this parameterization is sine of theta so we sometimes call that the m if the field is written in m n and p components anyway sine of theta times the minus sine of theta the i hat component of my r prime then to that i'm going to add the minus x which written in terms of our parameterization is going to be minus cosine so minus cosine of theta that is our n multiplied by the j hat component of r prime which is cosine of theta and then plus zero because the r prime has zero and the k hat so i'll just come there and put in my dt minus sign squared plus minus cos squared is just minus one and so this is going to be minus two pi all right so there's our result for the circulation minus 2 pi not so bad so if i scroll back up to stokes's theorem that was the left-hand side but i also want to do the right-hand side and i'm hoping that i can do this and also get the value of minus 2 pi and i have to do the curl of f and then i'm going to have to dot it with n and then take a d sigma so working from the inside out the first thing for me to do is figure out my curl of f and then i'm going to go and dot that with the normal vector and i'm going to figure out what both of those things are going to be all right so let's do that computation the curl of f is equal to the determinant of i hat j hat k hat then the partial with respect to x the partial with respect to y and the partial with respect to z all goes along the second row and then finally the components of the field f itself y minus x and z go in the third row you'll notice here that i'm not memorizing the larger formula for the curl that we've introduced i'm preferring to think of it in the determinate way this is perfectly fine for all the computations that i'm going to be doing okay so i hat component first is going to be the partial respect to y of z zero the partial with respect to z of x which is zero so just zero in the i hat the minus j hat is going to be the partial of x with respect to z again zero the partial of z respect to y again zero nothing there as well and then finally i'm going to have a k hat component i think that's non zero thankfully so the k hat is the partial with respect to x of minus x which is minus one minus the partial respect to y of y so in other words another minus 1 this is equal to minus 2 k hat the next thing i have to do is compute the normal vector and as we've seen when we're doing surface intervals before there's actually three different ways that we might want to try and approach this issue of the normal vector and the d sigma we could do it parametrically implicitly or explicitly i have an implicit formula here x squared plus y squared plus n squared equal to 1. and so i'm just going to use the implicit formula that the normal is the gradient of g divided by the length of the gradient of g i should be precise by what do i mean by g let's go all the way back to the beginning i'm calling the g all of that the x squared plus y squared plus z squared so the surface is g is equal to one now i actually sometimes find this easier to go and deal with the normal d sigma all in one go it's has a little bit of cancellation to do the normal and the sigma together that's true for each of parametric implicit and explicit so if i do this then what i would be getting would be the gradient of g divided by the length of the gradient of g dotted with k hat and then a little element of area the shift from a little element of surface area d sigma to d a has this stretching factor the magnitude of the gradient of g divided by the magnitude of the gradient of g dotted with k hat because the surface is in the x y plane and the k x just pointing straight up we've seen that the length of the gradient g just cancels in both cases it's just easier to go with this formula directly okay so let's actually do that computation then so the gradient of g is the vector 2x 2y and 2z the gradient of g dotted with k hat is just the third component of this in other words 2z so i'll divide out by the length of 2z and then i'll still multiply this by the d a all right so now i think i'm finally in a position to come forward and write out my surface integral of the curl of f dotted with n d sigma and multiple things are going to be happening in this formula at once so the first thing is it's going to be a double integral over the region now so this is the region in the x y plane i'm thinking of my surface being above it that is i'm shifting from a d sigma to a d a here then i need to execute the dot product between the curl and this expression that i just had so i need to remind myself what my curl was so if i go up a little bit i'll remind myself that my curl was -2 and the k-hat all right so then here the only k-hat component is the 2-z so what do i have i have a minus 4 z that's the dot product between the minus 2 and the k hat and this 2x 2i 2z all divided by the length of 2z multiplied by da and i still haven't figured out what the limits of my integration are i'll do that in a moment and then now it's just a computational question this integral here is well let's do it again the double integral over the region uh minus 4z divided by 2z z was always assumed to be positive so i can drop those absolute value signs it's just going to be a minus 2 times d a and well what is the region so if i scroll all the way back to the beginning the region was the portion above the circle of unit 1. so in other words it's a circle of radius one it's got an area of pi r squared so if i come all the way back down well i don't even need to bother writing limits of integration here it's just the circle of radius one so the final answer is going to be minus 2 times the area of that circle which is pi r squared r is 1 so just pi minus 2 pi n thank goodness we've gotten the exact same answer that we had before so indeed we've gotten the same answer using both sides of stoke's theorem one computing the circulation directly and secondly doing a surface integral of the curl of f dotted with n now i want to do one final thing here and it's kind of a cool little trick in the exact same example so so my setup was i had this circle down in the xy plane and then my choice of surface was the hemisphere above it however if you have two surfaces they both have the same boundary but the surfaces are completely different then the surface integral of curl of f dotted n over the first surface and the second surface have to be the same because they're both equal to the circulation on the boundaries and so if they have the same boundary they also must have the same surface integral so what would be a simpler surface to deal with what if instead i'm going to get rid of the hemisphere i'm going to keep the boundary being the exact same thing here let's close it up for my erasing what if i made my surface be just a flat disc that's a pretty easy surface in fact it's a very easy surface because we can do almost everything about it for example if i want to figure out that normal vector now the normal vector is just k hat if i use a flat disc and so well let me try and do my surface integral now of the curl of f dotted with n well we've already shown that the curl of f was just minus 2 in the k hat direction so that's what the curl of f is dotted with k hat so that's going to be my normal oh i forgot my d sigma here and then integrated with respect to area and i'm going to put down here the region again which is just the circle in the plane circle of radius 1. the dot product is easy enough so is the double integral over the region of minus 2 d a and i know what that area is it's just pi r squared r is one so which is pi minus 2 pi and so we get the exact same answer a third time so previously when we thought of stokes's theorem we had these two different portions of it one for the boundary one for the surface and basically depending on what you're interested in which of those computations was easier you might do one side or the other to get the opposite side but now we're really opening it up because you don't have to use the surface that you're given you could compute it via a completely different surface as long as it has the same boundary indeed you can just make the easiest surface that you can imagine so for example any time when you've got a planar curve so a curve that's just in the xy plane it often makes sense to have as we just did here the surface be just the flat region inside of that curve in the xy plane not always of course it would depend on the computation but the point is if you can replace the surface that you have that's messy and maybe you don't know how to do the surface integral with the surface you have replace it with an easier one you can sometimes gain a really big computational advantage all right so that was a long example here i hope you enjoyed it if you did please give it a like if you have any questions leave them down in the comments below and we'll do some more math in the next video

Channel: Dr. Trefor Bazett

Views: 122,572

Keywords: Solution, Example, math, Stokes' Theorem, Stokes Theorem, verify, both sides, vector calculus, multivariable calculus, vector field, hemisphere, circulation, line integral, surface integral, curl of a vector field, curl, replace surface, boundary, differential operator

Id: ms4JjH0BANU

Length: 13min 42sec (822 seconds)

Published: Sun Dec 13 2020