STOKES’ THEOREM BEFORE YOUR FINAL… you NEED this one | Calculus 3

Added: 2026-05-14

[00:00:00]Welcome to Math with Professor V. In this video, we're going to go over Stokes theorem, which is probably one of the hardest topics for students at the end of calculus 3. Now, Stokes theorem tells us that the work done by a force field along a closed curve is equal to the surface integral where S is the surface bounded by C of the curl of F. N DS.

[00:00:26]Okay, I'll break it down for you step by step, but here's the thing. It's not that using the line integral or using the surface integral is always easier.

[00:00:34]Stokes theorem just tells us that they're going to be equal. So, it really depends on the surface, the setup, and honestly, sometimes just what your instructor wants you to do. So, I'm going to first do one example both ways to show that they're equal. And then after that, we're going to focus on surface integrals because usually my students struggle with that one more and it's a little bit more popular to ask examples that deal with surface integrals. Okay, just some conditions to make sure that are satisfied. S has to be peacewise smooth. That's usually not a problem for the, you know, the way that examples that you're going to run into. C has to be a simple closed curve.

[00:01:08]That's the boundary of S. So C bounds S.

[00:01:12]And then here's the most important thing. C has positive orientation. So make sure it follows the right- hand rule and your normal N points outward from the surface. This is key. So, if you're specifically going to evaluate the surface integral, here's the steps that I outlined for you guys to kind of help break it down. So, take a screenshot if you want. Um, first, find the curl of f. If you forgot how to do that, you take d operator, cross it with f. I'll remind you, and I'll do examples right now. Graph s, your surface, and then find n. This is your normal vector.

[00:01:47]Normal to what? The surface. And it has to point outward from the surface. I have a slick little stick figure trick to help you never get mixed up about this. Then figure out what the curl of fn is. Decide your integration region.

[00:02:02]Notice we're doing a surface integral.

[00:02:04]That's a double integral. So we're going to integrate in a 2D region. Yes. So we're either going to be rectangular or we're going to be polar. Don't even think about cylindrical spherical. Those are threedimensional coordinate systems.

[00:02:18]This is a 2D problem. So yes, we should only have two variables in this integral. Decide the integration region.

[00:02:24]Graph if necessary. Determine the limits. And boom, just integrate. Okay, here we go. Verify that Stokes theorem is true for the given vector field and surface S. So let me first do it as a surface integral. Here's my vector field. Doesn't look too intimidating. S is the cone. Z ^2= X^2 + Y^2. They tell me Z is between 0 and four inclusive.

[00:02:50]Okay, so we're just dealing with the little piece of the upper half of the cone and orientation is downward, which is standard. So they didn't even need to tell me that. I would have known. Let's first do the surface integral. So Stokes theorem helps us evaluate work. Okay, work in this chapter. You're either finding work or flux most of the time.

[00:03:10]And that's the other thing that confuses students. If you need a good little recap summary, I have this video where I make a nice little chart going through all the theorems in this unit in four minutes. I'll link it in the description. Four minutes, I tell you.

[00:03:25]Okay, so first step, let's find the curl of f.

[00:03:29]So, how would you do that? Okay, so you put your vector field in the bottom row -y x -2.

[00:03:37]And we're going to cross that vector field with the dell operator. So d x d y d z. Okay. So let's see here. So eliminating column one, row one, I need partial with respect to y of -2. That's going to be zero. And the partial derivative of x with respect to z, that's also zero.

[00:04:00]And then the next component, let's see here. I always do the opposite order to absorb the negative. So this is 0 0.

[00:04:09]I mean, if we get all zeros, then we're done with the problem, right? The work's going to be zero. That would be exciting. Oh, no. This is going to be a one minusgative one. So, two. Okay, it's not zero. We must continue. Step two, let's graph the surface. And then we're going to determine what our normal vector is. Normal to the surface. So, we have the cone z^2 = x^2 + y^2. Only the portion where z is between 0 and 4. So, we're above the xy plane and it's not going to extend forever upwards. Here's x. Here's y. Here's z.

[00:04:47]So, here's my cone coming in hot.

[00:04:53]Beautiful.

[00:04:55]Okay. And then I'll label so you guys know that's specifically right where Z equals 4. Okay.

[00:05:03]Great. Just color it in that way you know. Okay. This is a surface. So basically like the shell the outside.

[00:05:09]It's not solid inside. And this is at Z equals four. Okay, perfect. Now they told me it's the surface, right? Our orientation is downward. So what that means is if you were to walk on the surface, remember downward outward outward is really the more general term. If you're walking on the surface, your head would be pointing down, right? When you get to the very tip here, outward is our standard orientation.

[00:05:38]Because the cones opening upward, then standard orientation, positive orientation would be downward. So, what does that mean for us? That tells me how to decide my normal vector. You have two options.

[00:05:50]This surface here, this cone, since it's just the upper half, we can express it as a function of x and y. I could say Z = square<unk> of X^2 + Y^2. Right? Since we're only dealing with the positive portion. So since Z is a function of X and Y, my normal vector has two options.

[00:06:10]Either these first two partials are positive and the last one's negative. Or depending on the surface, my normal vector would be negative partial with respect to X, negative partial with respect to Y, positive one. So this last component always has the opposite sign of the other two um which involve the partial derivatives. Okay. If your surface Z is a function of X and Y. So when you're choosing between the two, look at the very tip or the very extreme point of your surface. Here you are.

[00:06:40]Yes, it's you. Okay. In order for you to not be in the surface cuz standard orientation is outward, your head has to be going downward. Downward. Which means if you were to express a vector, the normal vector to the surface specifically right here, in order for it to point outward, the Z component would be negative one. So this is the normal vector that we're going to use, normal to the surface. In other problems, you might pick the other one. Okay? Yeah, we'll do some where you pick the other one. That way you can see how it's going to work. All right. Now that that's done, let's go ahead and figure out what the curl of f is dotted with n cuz that's what we're going to actually integrate right here. And look, we kind of lucked out because see how these are zeros? Then I'm not even going to bother taking the partial derivatives because I know when I dot them, they'll just end up giving me zero. So that's relaxing.

[00:07:30]Let's just write this out for the people. The curl of f dotted with n would be 0 * zx + 0 * z y + 2 * -1, which is -2. Okay, perfect. Now go back.

[00:07:47]Look at the formula. Here we have to decide what plane, right? What our integration region is going to be for this surface. Well, since I already said my surface is a function of x and y, I'm going to integrate over the xy plane.

[00:08:00]So, you're going to project the surface onto the xy plane. And that's our integration region. Well, if I project this cone along or onto the xy plane, notice here you're just going to have a circle radius how much?

[00:08:16]radius 4. Right? Look at this equation.

[00:08:20]If Z is four, the circle is going to be radius 4. So, let me write it out for you if you like.

[00:08:26]So, we're going to integrate over the XY plane.

[00:08:31]All of these surface integrals are double integrals. So, you're either integrating over XY or maybe occasionally you'll switch to polar.

[00:08:38]That's it. That's it for Stokes theorem.

[00:08:40]Okay? So, it's not any more complicated than that. This is four. This is four.

[00:08:45]Since the equation is x^2 + y^2 = 4^2 and I'll color this in the same color so you know oh that's the projection of the surface onto the xy plane we may need to go polar right in which case we would figure out okay the limits of integration if it's the entire circle theta would go 0 to 2 pi and r would go 0 to 4 but in this problem we don't even have to do that much so the work I always tell my students write down what it is because you're going to get confused double integral / s curl of f do n d s in this case it's double integral over d that's my region d what was curl of f dotted with n -2 and then now d s becomes da once we've written up the double integral oh but since I'm just integrating a constant I don't even have to actually evaluate the integral this is just -2 * the area of the region d we know the area of this region Yes, it's a circle. So, the area is p<unk> r^ 2. So, that's -2 * 16<unk> which is going to be 32 pi. Bada bing, bada boom. There you go. How was that?

[00:10:01]Not too bad. Once you do enough of them, you'll get the hang of it. Okay, you guys? So, don't panic. Now, let's try finding the work using um a line integral. Okay, so we have line integral along C. C is a closed curve. So that's why I put that little closed circle F dot DR. So in this case, what we need to do is parameterize the curve C. That is the boundary of S. So where's my curve C? It's right here.

[00:10:30]And we need to be careful so that the orientation follows the right hand rule.

[00:10:35]So this little person right here, that's the direction your thumb should be pointing in. So take your right hand and curl your fingers along this pink boundary. I hope you're doing it over there. And make sure your thumb points downward. Which way are we walking along the curve? We have to walk this way, right?

[00:10:56]Okay. Another way to think about it is if you're actually walking on the curve, the surface has to be on your left. So imagine you're going surface is on my left. That's the correct standard orientation. Okay. So let's figure out how to parameterize or represent parametrically this curve here. It's a circle radius 4 and okay so let's figure out how to parametrically represent the equation of this curve. It's a circle radius 4.

[00:11:38]So x^2 + y^2 = 4^2 and it's at the height where z is equal to 4. So basically the easiest thing to do would be let x= 4 cosine t. y = 4 sin t. We're going to have to check this in just a minute. And let z equal 4. Right? It lives where z is 4 and it just goes around and around and around. And then t would be between 0 and 2 pi.

[00:12:03]Now I'm not sure if this is going to traverse the curve in the correct direction. So we're going to check right now. Okay. Now remember your parameter t is always increasing. So say I plugged in t equals 0. What ordered triple would it give me? So 4 * cosine of 0. That would be 4. 4 * s of 0 is 0. And then z would be 4. So where is 4 0 4? It's right here. This is where t equals z.

[00:12:33]Then if t is plug in something nice, pi / 2, 4 * cosine pi / 2, that's 0. This would be four. And then z is four. So where does that take me to? 04.

[00:12:46]That takes me here. That's the wrong way. We went this way. Uh-oh. Uh-oh. So how do I fix it? Well, I just want to reverse the order. Let me add a minus sign in front of y. Now this is probably the most difficult part for students.

[00:13:01]Okay. So what my recommendation is is plug in two values right for your parameter and see which way the curve is getting traced out and if it's going the wrong way then just add a minus sign here on the y component or on whatever is attached to sign cuz that'll reverse the order. Also, another way that you could figure out that you need to reverse the sign on y basically is imagine you're standing here looking down at the xy plane. This circle is being traversed clockwise, not counterclockwise. So maybe that would clue you in that you need to switch the sign on y. Anyways, however you get there, get there. So the work let me write down we're finding work is f dr but remember you can just jump to we'll have limits for t in this case but p ddx + qd y + r dz. So f my vector field in case you forgot it was -yx -2. So based on this parameterization y so that would be pos4 sin t x is 4 cossine t and z is -2. This is p qr. Now I need to find dx dy dz from right here.

[00:14:27]Okay let's erase this. So dx is going to be -4 sin t dt.

[00:14:34]dy would be -4 cossine tdt and dz 0 and then these are my limits of integration. This is just a single integral, right? Single variable. So it's going to be relaxing. So here we go. 0 to 2 pi p dx 4 sin t dx is -4 sin t dt plus q d y 4 cossine t * -4 cossine tdt plus r dz well dz is zero so just plus z okay let's see what we got here now so 0 to 2 pi this is -16 sin^ 2 t - 16 cosine^ 2 t dt.

[00:15:22]Well, if I take out -16, then I just have sin^ square * cosine^ 2, which is going to be 1. Very good. So, this is 0 to 2<unk>i -16 dt, which is -16 * 2<unk>i, which is -32 pi. Yes, Stokes theorem works. That's what we got with the surface integral. And I wouldn't say one method was particularly easier or harder than the other. It just really depends what you're more comfortable setting up, but so that's why I'm just saying ask your instructor and you know, are they going to be picky that you do surface integrals or that you should be ready to do both ways depending on the problem. Just pay attention to the directions. I think that's where things can get a little tricky. Okay, I'm going to do two more examples. We're only going to do the surface integral version. If you want, I would say try doing it as a line integral too on your own and verify Stokes theorem. Okay. The directions say find the flux of the curl of field f through the shell f. I wanted to just show you a different way the directions could be phrased cuz depending on the textbook, depending on your teacher, depending on where you are, they might ask for the same thing different ways. This is still asking for work. Okay, flex of the curl. Flex of the curl is not flex, it's work. Okay, here's my vector field. S is the portion of the paraboid Z = 2 - X^2 - Y^2 that lies above the XY plane. Okay, so let's just follow the steps that I had outlined for you guys. First thing we're going to do is find the curl of f.

[00:16:56]Okay, so curl of f. I'm going to put the vector field in the bottom row first. So this is -5x^2 y 5x y^2 z 5th then we have ihat jhat khat d x d y dz okay so here we go we need that's gone that's gone partial of z to the 5th with respect to y is zero this is also zero next component That partial zero. That partial zero.

[00:17:33]Last component. Okay. Now we get some nonzero stuff. This is 5 y^2 minus 5x^2. So 5 y^2 + 5x^2.

[00:17:45]Okay. Second, let's graph the surface.

[00:17:49]So it's the portion of the paraboid z = 2 - x^2 - y^2 that lies above the xy plane.

[00:17:59]Okay, z x y. So the z intercept is going to be at 0 02.

[00:18:09]And then when the paraboid like hits the xy plane, it's going to make a circle of radius rad 2. Right.

[00:18:18]I need more. Yes, there we go. So where it hits the xy plane, this is at rad two. Rad two.

[00:18:28]Here's the paraboid.

[00:18:31]Beautiful. Now we need to figure out our orientation for N. Okay. Now look, my surface Z is already a function of X and Y so beautifully. So we have to decide is N going to be ZX Z Y -1 like last time or is Z going to be or is N going to be negative ZX negative Z Y positive one? Well, remember standard orientation is outward, outward, outward, outward.

[00:18:58]Look at the absolute max or min on the surface. If you're standing there, your head has to point in the direction away out from the surface. You can't be going in the surface. And your head will give you the z component in this case, which is up. It's positive. So, we're going to be going with this version for our normal vector. Okay. Normal to the surface. So always use the surfaces equation to find n. Okay. So n is going to be negative partial with respect to x. That would be -2x. So positive 2x.

[00:19:32]Similarly, this would be pos2 y and positive 1. Okay, we're so positive today. I love it. Step three, we're going to figure out curl of f dotted with n. Well, since the first two components are zero, that'll give me zero. those products and then I just have 1 * 5 y^ 2 + 5x^2.

[00:19:57]Okay, now we're going to determine what our integration region is. Well, since our surface is a function of x and y, we're going to integrate over the xy plane.

[00:20:07]What does it look like if I project the surface over the xy plane? Well, we already talked about it a little bit.

[00:20:13]It's a circle of radius rad 2. Yes.

[00:20:19]rad 2 and it's the whole circle. So the sounds like we should go polar and the limits of integration shouldn't be too complicated just so long as you don't forget that there's a rad 2. There's a rad 2. So we're going to go polar.

[00:20:35]Theta should go from 0 to 2 pi and then r should go from 0 to rad 2. Okay, fabulous.

[00:20:44]Now let's set up this integral and bring it on home. So the work which is going to be double integral over s of the curl of f n ds is going to be integral 0 to 2 pi 0 to rad 2. What was the curl of f n?

[00:21:06]Didn't I do it? Oh yeah, here 5 y^2 + 5 x^2. But we're going to go polar. So that's 5 r 2. Very good. d s becomes da which is impolar r d r d theta. Okay, beautiful, beautiful. Now from here, what do we do? Well, notice this is kind of nice. Um there's no thetas in the problem. So I can just slap a 2 pi on the outside.

[00:21:33]In fact, take the five out as well. Then we have 0 to rad 2 r cub dr. So then now I have this is 10 pi. We can integrate this is going to be 1/4 r 4th evaluated from 0 to rad 2 rad 2 to 4th.

[00:21:50]Just think rad 2^ squared twice in a row. I know. So it's going to become four which will cancel with the 1/4. So then this is just 10 pi. That was fabulous. Good job you guys. Okay, let's do one more.

[00:22:06]Find the flux of the curl. So again find work because I see the word curl, right?

[00:22:11]I know it's not just flux, different definition, different thing of field through the shell S. So here's my vector field. S is the portion of the cone Z = 6 rad X^2 + Y^2 below the plane Z = 2.

[00:22:30]H, how peculiar. Okay, so really we're below Z equals 2. The six in the front, you know, I just started drawing it cuz I do that first. We should find the curl, but whatever. I'm excited to draw this cone. The six in the front. Let me get back to my thoughts. Um, it just makes it more steep, right? Normally, we don't have a constant out there, I'd say, for the majority of the cones that we graph. So, just adding a six is a vertical stretch. It's really not a big deal. Pay attention to the fact that we have to be below the plane Z equals 2.

[00:23:01]Okay, no problem, Professor V. So, here's our cone. Here's our cone. Come on.

[00:23:11]And the equation is z = 6 rad x^2 + y^2.

[00:23:18]Okay, forgive me for drawing it first and not going in order.

[00:23:23]I just got excited.

[00:23:25]Okay.

[00:23:27]Okay. First step, let's find the curl of f. Yes. Yes.

[00:23:31]So curl of f. So put your vector field in the bottom row. You know want to know why I do that first? So I have enough space for everything. Cuz when you just write I hat, jhat, k hat, sometimes you write them too close and then you run out of room. I know. Look at these tips and tricks I'm giving you guys. Okay, so this is going to come out to let's see oo four minus 0. So four next component seven. Yeah. And then last component six. Okay. I graphed the surface already. Now we need to determine n. Is n going to be zx z y -1 or is n going to bez y pos1? How to tell? Remember standard orientation outward outward outward.

[00:24:33]Here's the min on the surface. Your head has to be out of the surface outward.

[00:24:38]Well, your head tells you the direction of the Z component of the normal. Notice it's pointing downward. So, this is the normal that we're going to use.

[00:24:46]And look, since the curl has no zero components, I do actually have to figure out these derivatives. Okay, so here's Z partial with respect to X.

[00:24:58]So I always tell my students you take when you have a radical and you're doing chain rule you take 2 * the original six is just coming outside 2 * the original radical in the denominator and then you do chain rule. So derivative of the inside goes in the numerator this time derivative with respect to x would be 2x. So the twos cancel. So partial with respect to x is 6x over rad x^2 + y^2.

[00:25:29]And then similarly partial with respect to y would be the same thing. You would just have a 6 y up top, right?

[00:25:36]Okay, perfect. So then my normal vector is going to be 6 x over red x^2 + y^2 6 y over red x2 + y^2 and 1. Perfect. So that's our normal.

[00:25:57]Okay. What do we do next? Step three.

[00:25:59]Take the curl of f and dot it with n. So curl of f dotted with n. What was the curl of f? It's right here. 4 -76. So if I dot it with n, it's just going to change what the constants are in the numerator. It'll be 24x over rad x^2 + y^2 - 42, right? Yes. - 42 y over this one's a little nasty. Brad x2 + y^2 and then - 6, right? Right. Was the last Uh-huh.

[00:26:36]Perfect. Okay. Save this. Save this.

[00:26:38]Save this. Now, we're going to decide.

[00:26:41]Are we integrating over the xy plane?

[00:26:43]Are we going polar? Let's see. If I project this surface over the xy plane, yes, indeed. I get a circle. I get a circle. H. What's the radius of the circle? Well, I have to see where this cone intersects z= 2 to figure out what the radius is going to be. So let's see. I'm going to set 6 rad x^2 + y^2 = 2, which means rad x^2 + y^2 = 2 over 6, which is a 3. So x^2 + y^2 is 1/3 squared. So we have a circle radius 1/3 centered at the origin. That was kind of a tricky little moment. I have to say this though, we did this problem in class. I had the students do it on their own after I went over some examples.

[00:27:35]They did it just fine. They got that intersection. I was so proud. So here's 1/3.

[00:27:43]Yeah, that's what we're going to be integrating over. This is that base region. Okay, radius 1/3. Let me go like this. So sounds like we should go polar, right? Go polar.

[00:27:56]Theta is going to go from 0 to 2 pi. And then r would go from 0 to 1/3. Then let me make sure this is all in polar as well. Right? So this would be 24 x is r cosine theta. Right? x^2 + y^2 is just r - 42 r sin theta / r. So that will cancel - 6. So this cancels. This cancels. Okay. You guys all right? Good. Here we go. ready to bring it on home. So step five, the work is going to be surface integral curl of f n ds. So it's going to be integral 0 to 2<unk>i 0 to 1/3.

[00:28:45]Now I have all of this here time rd r d theta. So 24 cossine thet - 42 sin thet - 6 * r d r d theta. Okay, this one's a little nasty. It sure is. Notice here though, you could think of it like this is all a function of theta and this is a function of r. And since my limits are constant, I can split this. I can do integral 0 to 2 pi. That would be my theta integral. 24 cossine thet - 42 sin thet - 6 d thet right that's one integral and then I can multiply that by the integral from 0 to 1/3 of r d oh that's that makes it better okay then let's just go ahead integrate term by term so this would be 24 sin theta + 42 cossine thet - 6 thet evaluated from 0 to 2 pi and then this would be 12 r^ 2 evaluated from 0 to 1/3 and then here what do we get? So 24 * 0 + 42 - 12<unk>i that's the upper limit - 0 + 42 - 0.

[00:30:08]So what happened? I just have a -2 pi, right?

[00:30:14]-12<unk>i.

[00:30:16]Okay. And then what's this guy? 12 * 1 9th.

[00:30:22]So 118th.

[00:30:26]So I can cancel out a 6 -2<unk> over 3. Wow. Wow. Wow. Oh no.

[00:30:37]Really irritates me when it doesn't finish nicely like that. Okay, perfect.

[00:30:41]How do you guys feel? I hope this made Stokes theorem a little like less intimidating before your final, but know that it's not easy. This is not easy stuff. No one came out of the womb finding work using surface integrals, okay? This is not something you do day one. And quick reminder, I'm going to be working on building out my calculus 3 course on my website this summer along with differential equations, pre-calc, linear algebra is coming soon. So, if you want more structured notes, quizzes, practice problems like I'm doing with you guys right now, definitely keep an eye out for that. Thank you guys so much for watching. Don't forget to give the video a thumbs up, subscribe if you haven't already. You can follow me on Instagram and Tik Tok, Math with Professor V. And I will be back sooner than later. Hope your finals go