WEBVTT 00:00:07.220 --> 00:00:07.690 JOEL LEWIS: Hi. 00:00:07.690 --> 00:00:09.360 Welcome back to recitation. 00:00:09.360 --> 00:00:10.560 In lecture, you've been learning 00:00:10.560 --> 00:00:11.600 about Stoke's Theorem. 00:00:11.600 --> 00:00:13.980 And I have a nice question here for you that can put 00:00:13.980 --> 00:00:15.580 Stoke's Theorem to the test. 00:00:15.580 --> 00:00:17.880 So what I'd like you to do is I'd like you to consider this 00:00:17.880 --> 00:00:22.910 field F. So its components are 2z, x, and y. 00:00:22.910 --> 00:00:27.450 And the surface S that is the top half of the unit sphere. 00:00:27.450 --> 00:00:31.070 So it's the sphere of radius 1 centered at the origin, but 00:00:31.070 --> 00:00:32.080 only its top half. 00:00:32.080 --> 00:00:34.540 Only the part where z is greater than or equal to 0. 00:00:34.540 --> 00:00:37.600 So what I'd like you to do is to verify Stoke's Theorem for 00:00:37.600 --> 00:00:38.540 this surface. 00:00:38.540 --> 00:00:43.950 So that is, I'd like you to compute the surface integral 00:00:43.950 --> 00:00:46.470 that comes from Stoke's Theorem for this surface, and 00:00:46.470 --> 00:00:48.980 the line integral that comes from Stoke's Theorem for the 00:00:48.980 --> 00:00:50.770 surface, and check that they're really 00:00:50.770 --> 00:00:52.030 equal to each other. 00:00:52.030 --> 00:00:55.320 Now, before we start, we should just say one brief 00:00:55.320 --> 00:00:57.610 thing about compatible orientation. 00:00:57.610 --> 00:01:00.130 So I didn't give you any orientations, but of course, 00:01:00.130 --> 00:01:02.240 it doesn't matter as long as you choose ones that are 00:01:02.240 --> 00:01:03.170 compatible. 00:01:03.170 --> 00:01:06.130 So if you think about your rules that you have for 00:01:06.130 --> 00:01:06.620 finding them. 00:01:06.620 --> 00:01:09.540 So if you imagine yourself walking along this boundary 00:01:09.540 --> 00:01:12.940 circle with your left hand out over that sphere. 00:01:12.940 --> 00:01:18.270 So you'll be walking in this counterclockwise direction 00:01:18.270 --> 00:01:22.040 when your head is sticking out of the sphere. 00:01:22.040 --> 00:01:22.470 All right? 00:01:22.470 --> 00:01:27.040 So in other words, the outward orientation on the sphere is 00:01:27.040 --> 00:01:30.640 compatible with the counterclockwise orientation 00:01:30.640 --> 00:01:33.130 on the circle that is the boundary. 00:01:33.130 --> 00:01:36.190 So let's actually put in a little arrow here to just 00:01:36.190 --> 00:01:38.835 indicate that is our orientation for the circle. 00:01:41.730 --> 00:01:44.010 And our normal is an outward-pointing normal. 00:01:44.010 --> 00:01:47.060 And let's call our circle C, and our S is our 00:01:47.060 --> 00:01:48.620 sphere is our surface. 00:01:48.620 --> 00:01:49.230 OK. 00:01:49.230 --> 00:01:52.810 So just so we have the same notation. 00:01:52.810 --> 00:01:53.200 Good. 00:01:53.200 --> 00:01:55.950 So why don't you work this out, compute the line 00:01:55.950 --> 00:01:58.890 integral, compute the surface integral, come back, and we 00:01:58.890 --> 00:02:00.140 can work them out together. 00:02:08.180 --> 00:02:10.300 Hopefully you had some luck working on this problem. 00:02:10.300 --> 00:02:12.260 We have two things to compute. 00:02:12.260 --> 00:02:16.500 I think I'm going to start with the line integral. 00:02:16.500 --> 00:02:18.960 So let me write that down: line integral. 00:02:23.240 --> 00:02:26.370 So what I need to do to compute the line integral is I 00:02:26.370 --> 00:02:37.170 need to compute the integral over the curve C of F dot dr. 00:02:37.170 --> 00:02:41.290 And so I know what F is on that circle. 00:02:41.290 --> 00:02:43.680 So I need to know what dr is. 00:02:43.680 --> 00:02:44.610 So I need to know what r is. 00:02:44.610 --> 00:02:46.400 I need a parametrization of circle. 00:02:46.400 --> 00:02:48.870 Well, you know, that is a pretty easy circle to 00:02:48.870 --> 00:02:49.440 parametrize. 00:02:49.440 --> 00:02:51.800 It's the unit circle in the xy plane. 00:02:51.800 --> 00:02:58.690 So we have C. And we're wandering around it 00:02:58.690 --> 00:02:59.760 counterclockwise. 00:02:59.760 --> 00:03:01.670 So it's our usual parametrization. 00:03:01.670 --> 00:03:02.360 It's the one we like. 00:03:02.360 --> 00:03:10.030 So we have x equals cosine t, y equals sine t-- 00:03:10.030 --> 00:03:15.660 where t goes from 0 to 2 pi-- 00:03:15.660 --> 00:03:18.200 and this is in three dimensions, so the other part 00:03:18.200 --> 00:03:20.730 of the parametrization is z equals 0. 00:03:20.730 --> 00:03:26.310 So this is my parametrization of this circle. 00:03:26.310 --> 00:03:29.370 OK, so let's go ahead and put that in. 00:03:29.370 --> 00:03:41.460 So the integral over C of F dot dr is the 00:03:41.460 --> 00:03:45.370 integral from 0 to 2 pi. 00:03:45.370 --> 00:03:48.390 So we've got three parts. 00:03:48.390 --> 00:03:53.030 So F is 2z, x, y. 00:03:53.030 --> 00:03:57.690 So it's 2z dx plus x dy plus y dz. 00:03:57.690 --> 00:04:00.900 But z is 0 on this whole circle. 00:04:00.900 --> 00:04:03.190 So that piece just dies. 00:04:03.190 --> 00:04:05.470 And dz is also 0, so that piece just dies. 00:04:05.470 --> 00:04:07.490 So we're just left with x dy. 00:04:07.490 --> 00:04:11.850 So this is equal to the integral x dy. 00:04:11.850 --> 00:04:12.020 Oh. 00:04:12.020 --> 00:04:14.620 So I guess this is not from 0 to 2 pi. 00:04:14.620 --> 00:04:18.570 This is still over C. Sorry about that. 00:04:18.570 --> 00:04:18.910 OK. 00:04:18.910 --> 00:04:25.805 And now I change to my parametrization. 00:04:25.805 --> 00:04:26.170 OK. 00:04:26.170 --> 00:04:26.730 Yes. 00:04:26.730 --> 00:04:26.940 Right. 00:04:26.940 --> 00:04:32.430 So this is still in dx dy dz form, so it's still over C. 00:04:32.430 --> 00:04:34.700 Now we switch to the dt form, so now t is 00:04:34.700 --> 00:04:36.230 going from 0 to 2 pi. 00:04:36.230 --> 00:04:37.550 OK, so now we have x dy. 00:04:37.550 --> 00:04:40.750 So x is cosine t, and dy-- 00:04:40.750 --> 00:04:42.430 so y is sine t-- 00:04:42.430 --> 00:04:44.680 so dy is cosine t dt. 00:04:44.680 --> 00:04:51.090 So this is cosine t times cosine t is cosine squared t. 00:04:51.090 --> 00:04:51.820 dt, gosh. 00:04:51.820 --> 00:04:55.040 So now you have to remember way back in 18.01 when you 00:04:55.040 --> 00:04:56.920 learned how to compute trig integrals like this. 00:04:56.920 --> 00:04:59.660 So I think the thing that we do when we have a cosine 00:04:59.660 --> 00:05:01.410 squared t is we use a half-angle formula. 00:05:01.410 --> 00:05:03.900 So let me come back down here just to finish 00:05:03.900 --> 00:05:06.190 this off in one board. 00:05:06.190 --> 00:05:11.570 OK, so cosine squared t is the integral from 0 to 2 pi. 00:05:11.570 --> 00:05:20.410 So cosine squared t is 1 plus cosine 2t over 2, dt. 00:05:20.410 --> 00:05:22.110 And now cosine 2t-- 00:05:22.110 --> 00:05:24.720 as t goes between 0 and 2 pi-- 00:05:24.720 --> 00:05:27.600 well, that's two whole loops of it. 00:05:27.600 --> 00:05:27.880 Right? 00:05:27.880 --> 00:05:30.650 Two whole periods of cosine 2t. 00:05:30.650 --> 00:05:32.280 And it's a trig function. 00:05:32.280 --> 00:05:33.680 It's a nice cosine function. 00:05:33.680 --> 00:05:35.670 So the positive parts and the negative parts cancel. 00:05:35.670 --> 00:05:39.200 The cosine 2t part, when we integrate it from 0 to 2 pi, 00:05:39.200 --> 00:05:40.830 that gives us 0. 00:05:40.830 --> 00:05:45.550 So we're left with 1/2 integrated from 0 to 2 pi, and 00:05:45.550 --> 00:05:49.202 that's just going to give us 1/2 of 2 pi, so that's pi. 00:05:49.202 --> 00:05:50.980 All right. 00:05:50.980 --> 00:05:51.480 So good. 00:05:51.480 --> 00:05:52.700 So that was the line integral. 00:05:52.700 --> 00:05:54.540 A very straightforward thing. 00:05:54.540 --> 00:05:57.910 We had our circle back here. 00:05:57.910 --> 00:05:59.150 We had our field. 00:05:59.150 --> 00:06:02.530 So we parametrized the curve that is the 00:06:02.530 --> 00:06:04.500 circle that is the boundary. 00:06:04.500 --> 00:06:07.110 And then we just computed the line integral, and it was a 00:06:07.110 --> 00:06:08.750 nice, easy one to do. 00:06:08.750 --> 00:06:10.740 You had to remember one little trig identity 00:06:10.740 --> 00:06:12.170 in order to do it. 00:06:12.170 --> 00:06:13.230 All right. 00:06:13.230 --> 00:06:14.360 That's the first one. 00:06:14.360 --> 00:06:18.290 So let's go on to the surface integral. 00:06:25.760 --> 00:06:28.870 So the surface integral that you have to compute in Stoke's 00:06:28.870 --> 00:06:34.120 Theorem is you have to compute the double integral over your 00:06:34.120 --> 00:06:43.090 surface of the curl of F dot n with respect to surface area. 00:06:43.090 --> 00:06:47.610 So this is the integral we want to compute here. 00:06:47.610 --> 00:06:48.210 So OK. 00:06:48.210 --> 00:06:50.580 So the first thing we're going to need is we're going to need 00:06:50.580 --> 00:06:55.270 to find the curl of F. Let me just write it here so we don't 00:06:55.270 --> 00:06:56.920 have to walk all the way back over there. 00:06:56.920 --> 00:07:04.030 So F is 2z, x, y. 00:07:04.030 --> 00:07:05.270 So curl of F-- 00:07:05.270 --> 00:07:07.950 OK, you should have lots of experience 00:07:07.950 --> 00:07:10.500 computing curls by now-- 00:07:10.500 --> 00:07:12.330 is going to be-- 00:07:12.330 --> 00:07:16.400 I always think of it as these little 2 by 2 determinants 00:07:16.400 --> 00:07:19.230 with the partial derivatives in them, but most of those are 00:07:19.230 --> 00:07:20.360 going to be 0. 00:07:20.360 --> 00:07:25.480 We've got a dx x term that's coming up in k, and a dy y 00:07:25.480 --> 00:07:30.250 term that's coming up in i, and a dz 2z term that's 00:07:30.250 --> 00:07:32.520 coming up in j. 00:07:32.520 --> 00:07:32.970 So OK. 00:07:32.970 --> 00:07:35.370 So almost half the terms are 0. 00:07:35.370 --> 00:07:37.160 The others are really easy to compute. 00:07:37.160 --> 00:07:42.320 I trust that you can also compute and get that the curl 00:07:42.320 --> 00:07:44.300 is 1, 2, 1 here. 00:07:44.300 --> 00:07:47.120 OK, so this is F. This is curl of F. Great. 00:07:47.120 --> 00:07:47.390 So OK. 00:07:47.390 --> 00:07:48.260 So that's curl of F. 00:07:48.260 --> 00:07:51.390 So now we need n. 00:07:51.390 --> 00:07:52.000 Well, let's think. 00:07:52.000 --> 00:07:55.880 So we need the unit normal to our surface. 00:07:55.880 --> 00:07:58.580 So back at the beginning before we started, we said it 00:07:58.580 --> 00:08:00.780 was the outward-pointing normal. 00:08:00.780 --> 00:08:02.460 So we need the outward-pointing normal. 00:08:02.460 --> 00:08:04.360 Well, this is a sphere, right? 00:08:04.360 --> 00:08:07.790 So the normal is parallel to the position vector. 00:08:07.790 --> 00:08:13.520 So that means n should be parallel to the 00:08:13.520 --> 00:08:17.850 vector x, y, z. 00:08:17.850 --> 00:08:20.620 So n should be parallel to this vector x, y, z, but in 00:08:20.620 --> 00:08:22.590 fact, we're even better than that. 00:08:22.590 --> 00:08:24.380 We're on a unit sphere. 00:08:24.380 --> 00:08:27.230 So the position vector has length of 1. 00:08:27.230 --> 00:08:30.490 So n should be pointing in the same direction as this vector, 00:08:30.490 --> 00:08:32.610 and they both have length 1, so they had better be equal to 00:08:32.610 --> 00:08:34.460 each other. 00:08:34.460 --> 00:08:36.520 Great. 00:08:36.520 --> 00:08:40.130 So this unit normal n is just this very simple 00:08:40.130 --> 00:08:41.570 vector x, y, z. 00:08:41.570 --> 00:08:44.330 If it had been a bigger sphere, then you would have to 00:08:44.330 --> 00:08:47.910 divide this by the radius to scale it appropriately. 00:08:50.770 --> 00:08:51.280 All right. 00:08:51.280 --> 00:08:54.770 So we've got curl F. We've got n. 00:08:54.770 --> 00:09:02.420 So the integral that we want is this double integral over 00:09:02.420 --> 00:09:05.400 the surface of curl F dot n. 00:09:05.400 --> 00:09:14.290 So that's x plus 2y plus z, with respect to surface area. 00:09:14.290 --> 00:09:14.840 OK. 00:09:14.840 --> 00:09:16.790 Well, now we've just got a surface integral. 00:09:16.790 --> 00:09:19.800 It's over a hemisphere. 00:09:19.800 --> 00:09:21.990 Not a terrible thing to parametrize. 00:09:21.990 --> 00:09:22.740 So that's what we should do. 00:09:22.740 --> 00:09:25.250 We should go in, we should parametrize it, and then we 00:09:25.250 --> 00:09:28.280 should just compute it like a surface integral, like 00:09:28.280 --> 00:09:29.140 we know how to do. 00:09:29.140 --> 00:09:31.830 So before we start though, I want to make one little 00:09:31.830 --> 00:09:32.640 observation. 00:09:32.640 --> 00:09:34.430 Well, maybe two little observations. 00:09:34.430 --> 00:09:36.370 We can simplify this. 00:09:36.370 --> 00:09:37.040 All right? 00:09:37.040 --> 00:09:39.180 x. 00:09:39.180 --> 00:09:42.880 We're integrating x over the surface of a hemisphere 00:09:42.880 --> 00:09:44.820 centered at the origin. 00:09:44.820 --> 00:09:47.150 This hemisphere is really symmetric. 00:09:47.150 --> 00:09:52.090 And on the back side-- the part where x is negative-- 00:09:52.090 --> 00:09:54.690 we're getting negative contributions from x. 00:09:54.690 --> 00:09:57.170 And on the front side-- where x is positive-- 00:09:57.170 --> 00:09:59.310 we're getting positive contributions from x. 00:09:59.310 --> 00:10:02.250 And because this sphere is totally symmetric, those just 00:10:02.250 --> 00:10:04.440 cancel out completely. 00:10:04.440 --> 00:10:13.380 So when we integrate x over the whole hemisphere, it just 00:10:13.380 --> 00:10:14.060 kills itself. 00:10:14.060 --> 00:10:15.710 I mean, the negative parts kill the positive parts. 00:10:15.710 --> 00:10:16.880 We just get 0. 00:10:16.880 --> 00:10:20.430 Similarly, this hemisphere is symmetric between its left 00:10:20.430 --> 00:10:24.510 side and its right side, and so the parts where y are 00:10:24.510 --> 00:10:26.530 negative cancel out exactly the 00:10:26.530 --> 00:10:28.770 parts where y are positive. 00:10:28.770 --> 00:10:31.430 So as a simplifying step, we can realize right at the 00:10:31.430 --> 00:10:34.190 beginning, that this is actually just the integral 00:10:34.190 --> 00:10:39.040 over S of z with respect to surface area. 00:10:39.040 --> 00:10:42.350 Now, if you didn't realize that, that's OK. 00:10:42.350 --> 00:10:44.180 What you would have done is you would have done the 00:10:44.180 --> 00:10:46.280 parametrization that we're about to do. 00:10:46.280 --> 00:10:49.220 And in doing that parametrization, you would 00:10:49.220 --> 00:10:51.320 have found that you were integrating something like 00:10:51.320 --> 00:10:55.600 cosine theta between 0 and 2 pi, or something like this. 00:10:55.600 --> 00:10:57.110 And that would have given you 0. 00:10:57.110 --> 00:11:02.090 So you would have found this symmetry even if you don't 00:11:02.090 --> 00:11:03.290 realize it right now. 00:11:03.290 --> 00:11:06.030 You would have found it in the process of computing this 00:11:06.030 --> 00:11:09.410 integral, but it's a little bit easier on us if we can 00:11:09.410 --> 00:11:10.910 recognize that symmetry first. 00:11:10.910 --> 00:11:13.960 Now, notice that z doesn't cancel, because this is just 00:11:13.960 --> 00:11:16.930 the top hemisphere, so it doesn't have a bottom half to 00:11:16.930 --> 00:11:18.030 cancel out with. 00:11:18.030 --> 00:11:18.250 Right? 00:11:18.250 --> 00:11:21.930 So the z part we can't use this easy analysis on. 00:11:21.930 --> 00:11:24.400 If we integrated this z over the whole sphere-- 00:11:24.400 --> 00:11:26.450 if we had the other half of the sphere-- well, then that 00:11:26.450 --> 00:11:28.550 would also give us 0. 00:11:28.550 --> 00:11:31.980 But we only have the top half of the sphere. 00:11:31.980 --> 00:11:34.315 So it's going to give us something positive, because z 00:11:34.315 --> 00:11:35.940 is always positive up there. 00:11:35.940 --> 00:11:39.470 OK, so let's actually set about parametrizing it. 00:11:39.470 --> 00:11:41.800 We want to parametrize the unit sphere. 00:11:41.800 --> 00:11:42.440 Well, OK. 00:11:42.440 --> 00:11:44.830 So we have our standard parametrization that comes 00:11:44.830 --> 00:11:46.220 from spherical coordinates. 00:11:46.220 --> 00:11:48.120 So rho is just 1. 00:11:48.120 --> 00:11:49.370 Right? 00:11:54.990 --> 00:11:55.500 You know what? 00:11:55.500 --> 00:11:57.610 I always get a little confused, so I'm just going to 00:11:57.610 --> 00:12:01.320 check carefully that I'm doing this perfectly right. 00:12:01.320 --> 00:12:06.660 x is going to be cosine theta sine phi. 00:12:06.660 --> 00:12:07.520 Good. 00:12:07.520 --> 00:12:14.370 y is going to be sine theta sine phi. 00:12:14.370 --> 00:12:20.250 And z is going to be cosine phi. 00:12:20.250 --> 00:12:22.140 So that's our parametrization. 00:12:22.140 --> 00:12:26.240 But we need bounds, of course, on theta and phi in order to 00:12:26.240 --> 00:12:28.470 properly describe just this hemisphere. 00:12:28.470 --> 00:12:29.000 So let's think. 00:12:29.000 --> 00:12:33.430 So for phi, we want the hemisphere that goes from the 00:12:33.430 --> 00:12:36.200 z-axis down to the xy plane. 00:12:36.200 --> 00:12:42.130 So that means we want 0 to be less than or equal to phi to 00:12:42.130 --> 00:12:46.150 be less than or equal to pi over 2. 00:12:46.150 --> 00:12:46.430 Right? 00:12:46.430 --> 00:12:48.650 That will give us just that top half. 00:12:48.650 --> 00:12:49.760 And we want the whole thing. 00:12:49.760 --> 00:12:51.060 We want to go all the way around. 00:12:51.060 --> 00:12:55.510 So we want 0 less than or equal to theta less than or 00:12:55.510 --> 00:12:58.970 equal to 2 pi. 00:12:58.970 --> 00:13:01.760 OK, so this is what x, y, and z are. 00:13:01.760 --> 00:13:06.320 These are the bounds for our parameters phi and theta. 00:13:06.320 --> 00:13:08.290 Now, the only other thing we need is we need to 00:13:08.290 --> 00:13:10.150 know what dS is. 00:13:10.150 --> 00:13:15.340 So in spherical coordinates, we know that dS-- 00:13:15.340 --> 00:13:17.420 I'll put it right above here-- 00:13:17.420 --> 00:13:26.530 so dS is equal to sine phi d phi d theta. 00:13:26.530 --> 00:13:29.080 Let me again just double-check that I'm not 00:13:29.080 --> 00:13:30.330 doing anything silly. 00:13:32.580 --> 00:13:39.120 So dS is equal to sine phi d phi d theta. 00:13:39.120 --> 00:13:41.750 So we've got our parametrization. 00:13:41.750 --> 00:13:43.450 We've got our bounds on our parameters. 00:13:43.450 --> 00:13:44.980 We know what dS is. 00:13:44.980 --> 00:13:46.750 And we have the integral that we want to compute. 00:13:46.750 --> 00:13:48.450 So now we just have to substitute everything in and 00:13:48.450 --> 00:13:50.760 actually compute it as an iterated integral. 00:13:50.760 --> 00:13:51.380 Great. 00:13:51.380 --> 00:13:52.420 So let's do that. 00:13:52.420 --> 00:13:55.870 So, this integral that we want, I'm going to write a big 00:13:55.870 --> 00:14:00.870 equal sign that's going to carry me all the way up here. 00:14:00.870 --> 00:14:02.380 That's an equal sign. 00:14:02.380 --> 00:14:02.580 All right. 00:14:02.580 --> 00:14:03.565 So our integral. 00:14:03.565 --> 00:14:08.010 The integral over S of z with respect to surface area. 00:14:08.010 --> 00:14:12.700 So z becomes cosine phi. 00:14:12.700 --> 00:14:14.100 So we've got our double integral 00:14:14.100 --> 00:14:16.070 becomes an iterated integral. 00:14:16.070 --> 00:14:20.670 z becomes cosine phi. 00:14:20.670 --> 00:14:23.940 dS becomes sine phi d phi d theta. 00:14:31.380 --> 00:14:32.430 And our bounds. 00:14:32.430 --> 00:14:36.720 So let's see: phi we said is going from 0 to pi over 2. 00:14:41.450 --> 00:14:46.480 And theta is going from 0 to 2 pi. 00:14:46.480 --> 00:14:47.200 OK. 00:14:47.200 --> 00:14:49.560 So now we just have a nice, straightforward iterated 00:14:49.560 --> 00:14:50.750 integral here to compute. 00:14:50.750 --> 00:14:56.282 So let's do the inner one first. So we're computing. 00:14:56.282 --> 00:15:02.340 The inner integral is the integral from 0 to pi over 2, 00:15:02.340 --> 00:15:08.050 of cosine phi sine phi d phi. 00:15:08.050 --> 00:15:08.500 And OK. 00:15:08.500 --> 00:15:10.770 So there are a bunch of different ways 00:15:10.770 --> 00:15:12.060 you could do this. 00:15:12.060 --> 00:15:14.500 If you wanted to get fancy, you could do a double-angle 00:15:14.500 --> 00:15:16.040 formula here, but that's really more 00:15:16.040 --> 00:15:16.880 fancy than you need. 00:15:16.880 --> 00:15:22.510 Because this is like sine phi times d sine phi, right? 00:15:25.130 --> 00:15:27.550 Another way of saying that is you can make the substitution 00:15:27.550 --> 00:15:28.990 u equals sine phi. 00:15:28.990 --> 00:15:33.470 Anyhow, this is all CALC I stuff that hopefully you're 00:15:33.470 --> 00:15:34.650 pretty familiar with. 00:15:34.650 --> 00:15:35.050 So OK. 00:15:35.050 --> 00:15:36.660 So this is equal to-- 00:15:36.660 --> 00:15:43.670 in the end-- we get sine squared phi over 2, between 0 00:15:43.670 --> 00:15:44.310 and pi over 2. 00:15:44.310 --> 00:15:44.520 OK. 00:15:44.520 --> 00:15:45.280 So we plug this in. 00:15:45.280 --> 00:15:49.660 So sine squared pi over 2, that's 1/2, minus-- 00:15:49.660 --> 00:15:52.280 sine squared 0 over 2 is 0 over 2. 00:15:52.280 --> 00:15:54.200 So it's just 1/2. 00:15:54.200 --> 00:15:56.180 So the inner integral is 1/2. 00:15:56.180 --> 00:15:58.906 So let's see about the outer one. 00:15:58.906 --> 00:16:05.410 The outer integral is just the integral from 0 to 2 pi d 00:16:05.410 --> 00:16:08.100 theta of whatever the inner integral was. 00:16:08.100 --> 00:16:10.370 Well, the inner integral was 1/2. 00:16:10.370 --> 00:16:14.660 So the integral from 0 to 2 pi of 1/2 is pi. 00:16:14.660 --> 00:16:15.650 Straightforward. 00:16:15.650 --> 00:16:16.000 Good. 00:16:16.000 --> 00:16:16.490 So OK. 00:16:16.490 --> 00:16:19.490 So that's what the surface integral gives us. 00:16:19.490 --> 00:16:22.090 So let's go back here and compare. 00:16:22.090 --> 00:16:28.530 So way back at the beginning of this recitation, we did the 00:16:28.530 --> 00:16:33.170 line integral for this circle that's the boundary of this 00:16:33.170 --> 00:16:35.360 hemisphere, and we got pi. 00:16:35.360 --> 00:16:38.850 And just now what we did is we had the surface integral-- 00:16:38.850 --> 00:16:40.650 the associated surface integral that we get from 00:16:40.650 --> 00:16:43.570 Stoke's Theorem-- this curl F dot n dS. 00:16:43.570 --> 00:16:47.320 So we computed F and curl F and n. 00:16:47.320 --> 00:16:50.580 And then we'd noticed a little nice symmetry here. 00:16:50.580 --> 00:16:53.070 Although if you didn't notice it, you should have had no 00:16:53.070 --> 00:16:56.180 trouble computing the extra terms in the integral that you 00:16:56.180 --> 00:16:57.240 actually ended up with it. 00:16:57.240 --> 00:17:00.680 It would've been another couple of trig terms there 00:17:00.680 --> 00:17:02.230 after you made the substitution. 00:17:02.230 --> 00:17:04.320 So we parametrized our surface nicely. 00:17:04.320 --> 00:17:07.650 Because it's a sphere, it's easy to do. 00:17:07.650 --> 00:17:10.510 And then we computed the double integral and we also 00:17:10.510 --> 00:17:11.780 came out with pi. 00:17:11.780 --> 00:17:13.970 And we had better of also come out with pi, because Stoke's 00:17:13.970 --> 00:17:15.990 Theorem tells us that the line integral and the surface 00:17:15.990 --> 00:17:18.670 integral have to give us the same value. 00:17:18.670 --> 00:17:19.290 So that's great. 00:17:19.290 --> 00:17:21.790 So that's exactly what we were hoping would happen. 00:17:21.790 --> 00:17:24.060 And now we've sort of convinced ourselves, 00:17:24.060 --> 00:17:28.010 hopefully, that through an example now, we have a feel 00:17:28.010 --> 00:17:30.766 for what sorts of things Stoke's Theorem can do for us. 00:17:30.766 --> 00:17:32.016 I'll end there.