1 00:00:00,420 --> 00:00:06,870 All right, so now I want to show you this really useful to consider alternative coordinate systems. 2 00:00:07,660 --> 00:00:16,710 So we will do here is we will consider integration using spherical coordinates so you can see a problem 3 00:00:16,710 --> 00:00:19,170 of a Cartesian coordinate system. 4 00:00:19,200 --> 00:00:21,380 So I have discussed this in the previous lecture. 5 00:00:21,810 --> 00:00:28,080 So the problem could, for example, be you want to calculate the volume of a sphere and you realize 6 00:00:28,410 --> 00:00:34,920 that is very difficult to even determine the boundaries of your three dimensional integral, because 7 00:00:34,920 --> 00:00:41,280 what you would have to do is you would have to parametrized this sphere in terms of X, Y and Z, which 8 00:00:41,280 --> 00:00:42,300 is really difficult. 9 00:00:43,250 --> 00:00:50,780 And the whole problem arises because your volume element is a cube, there is your shape that you want 10 00:00:50,780 --> 00:00:52,820 to integrate over is a sphere. 11 00:00:53,390 --> 00:00:56,750 So it's really difficult to fit this cube in a sphere. 12 00:00:57,800 --> 00:01:04,970 And what would be much better is if this volume element would have already the shape of a sphere or 13 00:01:04,970 --> 00:01:07,100 at least some portion of the sphere. 14 00:01:08,450 --> 00:01:14,870 So what we want to do next is we want to derive the volume element in spherical coordinates. 15 00:01:15,500 --> 00:01:18,230 So I can already tell you it's not that simple. 16 00:01:18,230 --> 00:01:20,210 It's not just the extra Dizzie. 17 00:01:21,230 --> 00:01:27,440 It's what we need in this case is we need the partial derivatives of the position vector with respect 18 00:01:27,440 --> 00:01:31,850 to the individual coordinates and you will soon see why this is the case. 19 00:01:33,020 --> 00:01:35,270 So here it's very easy, actually. 20 00:01:35,300 --> 00:01:41,960 So if you if you calculate the derivative with respect to the length, you just get rid of these ares 21 00:01:41,960 --> 00:01:42,300 here. 22 00:01:42,320 --> 00:01:43,850 So this is what you get. 23 00:01:44,570 --> 00:01:51,380 If you calculate the derivative with respect to theta, you have to differentiate sign of theta, which 24 00:01:51,380 --> 00:01:53,810 is cosine of Seeta here the same. 25 00:01:54,320 --> 00:02:00,200 And here you calculate the derivative of cosine of Sleater, which is minus sign of Theta. 26 00:02:01,220 --> 00:02:06,650 And then the last one where you differentiate with respect to the polar angle, you have to differentiate 27 00:02:06,650 --> 00:02:09,890 cosine of PHI, which is minus sign of Phi. 28 00:02:10,430 --> 00:02:15,650 Here it is the derivative of sign-off which is plus cosine of five. 29 00:02:16,400 --> 00:02:19,700 And here you do not have any phi dependent's. 30 00:02:19,700 --> 00:02:22,160 So the partial derivative is zero. 31 00:02:23,660 --> 00:02:27,830 The next thing we need is the length of these partial derivatives. 32 00:02:28,640 --> 00:02:32,080 So what do you see here is of course still a vector. 33 00:02:32,090 --> 00:02:33,890 You still have these three components. 34 00:02:34,640 --> 00:02:40,700 So if we calculate the length of this vector, we have to take the X component squared plus the Y component 35 00:02:40,700 --> 00:02:45,270 squared, plus the Z component square and then calculate the square root. 36 00:02:46,370 --> 00:02:53,690 And while this one here may look difficult, it really turns out to be one that's actually a great exercise 37 00:02:53,690 --> 00:02:55,400 for you if you want to practice this. 38 00:02:56,150 --> 00:03:04,640 And the only thing that you need is that the cosine square plus the square of any angle is always one. 39 00:03:05,120 --> 00:03:06,540 That's the only thing you need for this. 40 00:03:07,730 --> 00:03:08,180 All right. 41 00:03:08,510 --> 00:03:10,910 So we have the length of this derivative to be one. 42 00:03:11,630 --> 00:03:17,260 This one here is R and this one is R times the sign of data here. 43 00:03:17,330 --> 00:03:18,560 You can see it once again. 44 00:03:19,550 --> 00:03:27,650 And now we have to consider that our position vector is not uniform in space, so in our gret it's not 45 00:03:27,650 --> 00:03:28,180 uniform. 46 00:03:28,580 --> 00:03:31,850 And this is expressed by these lengths of the partial derivatives. 47 00:03:32,120 --> 00:03:38,150 We need to account for these when we determine the volume integral, but also the line integral and 48 00:03:38,150 --> 00:03:38,840 other things. 49 00:03:39,920 --> 00:03:46,310 This is because, for example, if you express a line integral for a line integration, you write down 50 00:03:46,310 --> 00:03:47,860 that this is just D.R. 51 00:03:48,440 --> 00:03:53,210 But now of course, the question is what does the R look like in our coordinate system? 52 00:03:54,470 --> 00:04:00,650 Well, in any cordin system, what you can do is you can use such a chain rule, so to say. 53 00:04:01,490 --> 00:04:04,410 So in very rough terms, this is not correct. 54 00:04:04,430 --> 00:04:09,770 So please be careful about in very rough terms, you see that you could cancel this one in this one 55 00:04:09,770 --> 00:04:11,480 out, so you would still have D-R. 56 00:04:11,810 --> 00:04:16,100 But this is just like to memorize and to verify that this is true here. 57 00:04:16,730 --> 00:04:18,230 So what do you have here? 58 00:04:18,230 --> 00:04:22,220 Is the the chain rule basically of differentiation? 59 00:04:22,610 --> 00:04:27,530 What you have to do is here you have to differentiate your position vector with respect to every single 60 00:04:27,530 --> 00:04:33,170 coordinate and then have this differential here and use some of all the three components. 61 00:04:34,040 --> 00:04:39,020 So in our case, this would be here, the R d to define. 62 00:04:40,230 --> 00:04:45,420 And now, since this is a vector, you can also express it in terms of the length of times the unit 63 00:04:45,420 --> 00:04:45,900 vector. 64 00:04:47,430 --> 00:04:52,980 So we now need to use this one does and designer of sinus Tito. 65 00:04:54,270 --> 00:04:58,980 And these unit vectors are E.R. Zitter and IFI. 66 00:05:00,520 --> 00:05:08,230 So what we get in our case is that the ah is the scale of the ah times factor here plus ah there's one 67 00:05:08,230 --> 00:05:16,580 here times Dr. Elita plus our science theater times DFI is fine. 68 00:05:17,380 --> 00:05:18,700 This is our line element. 69 00:05:20,050 --> 00:05:22,750 And similarly you can consider the surface elements. 70 00:05:23,290 --> 00:05:28,900 So here, if you want to consider a surface element along the radial direction, then you have to of 71 00:05:28,900 --> 00:05:30,930 course multiply the. 72 00:05:31,720 --> 00:05:31,890 Yeah. 73 00:05:32,170 --> 00:05:34,450 The other two perpendicular components. 74 00:05:34,460 --> 00:05:36,430 So this would be the DFI. 75 00:05:36,820 --> 00:05:41,140 But you also have to take into account that this base is not uniform anymore. 76 00:05:41,440 --> 00:05:46,180 So you have to take into account these two absolute values of these derivatives. 77 00:05:46,720 --> 00:05:50,390 So this gives us, in this case, our times, our times sign of detail. 78 00:05:52,470 --> 00:05:55,090 And for the volume element, it's also very similar. 79 00:05:55,620 --> 00:06:03,630 So here we have to multiply the three individual yeah, these are the three to divide elements, defi 80 00:06:03,720 --> 00:06:04,120 elements. 81 00:06:04,170 --> 00:06:07,290 Sorry, but we have to account for these lengths. 82 00:06:07,530 --> 00:06:13,500 So we have to multiply all of these three terms here, which gives us an additional factor of our square 83 00:06:13,500 --> 00:06:14,490 sign of theta. 84 00:06:16,520 --> 00:06:23,570 So most important for our purposes is probably this the development here, then sometimes also this 85 00:06:23,570 --> 00:06:31,370 D-R element and sometimes just DSR element, which is the surface for surface integration. 86 00:06:33,180 --> 00:06:40,470 And so what this means is we now have found a way to express our volume elements. 87 00:06:41,010 --> 00:06:46,620 We do not have a cube anymore, but we have something that looks like a part of our sphere and we know 88 00:06:46,920 --> 00:06:48,010 how to express this. 89 00:06:49,140 --> 00:06:52,080 So here is what our volume element looks like. 90 00:06:52,990 --> 00:07:00,820 So you see here, it's really not surrounded by straight lines, it's really taking a part of the surface 91 00:07:00,820 --> 00:07:08,710 of the sphere and then extruding it towards the radio direction so you can really build your whole sphere 92 00:07:08,860 --> 00:07:12,210 by these volume elements that are no cubes anymore. 93 00:07:13,450 --> 00:07:16,280 And the same applies to this, to the surface. 94 00:07:16,660 --> 00:07:20,280 You really have a surface element that is not a square anymore. 95 00:07:20,290 --> 00:07:25,440 And you can take multiple of these elements and fill up the whole sphere. 96 00:07:26,410 --> 00:07:33,310 But please also consider that your element and also this element will look different for different positions 97 00:07:33,310 --> 00:07:36,820 in space, which is due to the non uniformity of our coordinate system. 98 00:07:38,170 --> 00:07:40,210 OK, but that was enough theory. 99 00:07:40,210 --> 00:07:41,380 Let's use this. 100 00:07:41,410 --> 00:07:43,370 Let's see what we can do with it finally. 101 00:07:44,020 --> 00:07:44,990 So, yes, yes. 102 00:07:45,040 --> 00:07:48,640 It's also, again, the equation for the surface element from the previous slide. 103 00:07:49,660 --> 00:07:52,710 So now we take these two elements and we integrate over them. 104 00:07:53,410 --> 00:07:58,760 So we want to calculate the volume of this of the sphere, which was our problem from the beginning. 105 00:07:59,410 --> 00:08:05,860 So what we do is we integrate over the sphere with a radius capital R and we integrate over our volume 106 00:08:05,860 --> 00:08:06,360 elements. 107 00:08:07,150 --> 00:08:08,420 So we take this one here. 108 00:08:09,220 --> 00:08:10,960 So, yeah, that's very simple. 109 00:08:11,890 --> 00:08:17,380 Now, what we have to consider is that this is actually a three dimensional integral so we can split 110 00:08:17,380 --> 00:08:23,920 it up into three individual integrals and we have to make sure that we get the correct boundaries. 111 00:08:24,520 --> 00:08:31,180 So for the angle of five, we have to boundaries zero to two PI because this is the full circle and 112 00:08:31,180 --> 00:08:31,630 the pull up. 113 00:08:31,630 --> 00:08:38,770 Right then we have for the star angle from zero to PI, which is from the positive direction to the 114 00:08:38,770 --> 00:08:39,940 negative C direction. 115 00:08:40,720 --> 00:08:47,830 And for the radius we go from zero to the Capitol to our radius, which is the radius of us of our sphere. 116 00:08:49,060 --> 00:08:55,060 And here you can see already what is so nice about spherical coordinates and what is so useful. 117 00:08:55,990 --> 00:09:01,600 You can see that our boundaries do not depend on any of these quantities here. 118 00:09:02,290 --> 00:09:07,540 For example, if we would calculate the same thing in Cartesian coordinates and our boundaries would 119 00:09:07,540 --> 00:09:13,810 be X, Y and Z dependent so we could never write it down in terms of a product as I did here. 120 00:09:15,040 --> 00:09:20,400 So what this means is we have now three independent, independent, one dimensional problems. 121 00:09:20,950 --> 00:09:24,280 We can first calculate this integral than this one than this one. 122 00:09:24,760 --> 00:09:28,090 And that is something that we can do that's very easy to do, actually. 123 00:09:29,500 --> 00:09:36,230 So let's go ahead and calculate the in the find an integral integral of these things here. 124 00:09:36,730 --> 00:09:39,550 So here we would write one time as defined. 125 00:09:40,000 --> 00:09:42,610 And the integral of one is. 126 00:09:43,800 --> 00:09:47,230 Fight, if you will, right here, the X then would be X. 127 00:09:47,670 --> 00:09:48,740 So here it's five. 128 00:09:48,990 --> 00:09:55,770 And now you have to take into account is the integration boundaries, which goes from zero to two PI. 129 00:09:56,700 --> 00:10:03,480 Here we have to integrate a sign of theta, which gives us a minus cosine of Seeta, because the opposite 130 00:10:03,720 --> 00:10:10,020 of the opposite way would mean which function do you have to take and differentiate to get the sign 131 00:10:10,020 --> 00:10:10,650 of theta? 132 00:10:10,920 --> 00:10:15,660 Well, of course you have to take minus cosine of data and then here the boundaries. 133 00:10:16,680 --> 00:10:21,220 And for the last integral we have to think about which function do we have to differentiate to get on 134 00:10:21,260 --> 00:10:21,750 square. 135 00:10:21,780 --> 00:10:27,810 And this is of course, one of the three are to the power of three because then we would get three times 136 00:10:27,810 --> 00:10:28,620 this whole thing. 137 00:10:28,620 --> 00:10:31,470 But here with the power of two, which is exactly how square. 138 00:10:33,120 --> 00:10:39,980 So now let's take into account our duration boundaries, so here we get to PI, minus zero is two PI. 139 00:10:40,860 --> 00:10:48,750 Here we get the cosine of Pi is minus one, but we have another minus sign here, which is plus one. 140 00:10:49,530 --> 00:10:55,590 And then we have to calculate minus minus the cost of zero, which is one. 141 00:10:55,590 --> 00:10:58,710 So we have one minus minus one, which is two. 142 00:10:59,670 --> 00:11:00,780 And he is very simple. 143 00:11:00,780 --> 00:11:05,430 We just have one third times capital R to the power of three minus zero. 144 00:11:06,260 --> 00:11:08,790 So these are our three individual integrals. 145 00:11:09,420 --> 00:11:16,830 And so our result is that the volume of a sphere is four over three times pi times are to the power 146 00:11:16,830 --> 00:11:17,310 of three. 147 00:11:19,020 --> 00:11:26,610 So maybe you see now why this is really, really great, but maybe you may also say now that is trivial. 148 00:11:26,760 --> 00:11:29,310 Everybody knows what the volume of a sphere is. 149 00:11:30,120 --> 00:11:38,100 But the truth is, without these three dimensional integration in spherical coordinates, no one would 150 00:11:38,100 --> 00:11:40,230 be able to tell about the volume of a sphere. 151 00:11:40,890 --> 00:11:46,860 I mean, you get this taught in school that you have to know what the volume of the sphere is, but 152 00:11:46,860 --> 00:11:49,380 you never learn where this comes from. 153 00:11:49,800 --> 00:11:51,300 And now you know where this comes from. 154 00:11:51,450 --> 00:11:57,820 It comes from the spherical coordinates and from the integration of for the surface elements. 155 00:11:57,820 --> 00:11:59,680 The whole procedure is very similar. 156 00:12:00,510 --> 00:12:05,430 So this time we calculate the surface of the boundary of atmosphere. 157 00:12:05,760 --> 00:12:12,420 So we have a closed integral over this boundary, which I just indicate indicated here by this partial 158 00:12:12,420 --> 00:12:13,050 symbol. 159 00:12:13,050 --> 00:12:19,320 And then here we have again the radius of our sphere, which is the capital R, and we integrate over 160 00:12:19,680 --> 00:12:26,750 the, um, the the surface area element along the radio direction. 161 00:12:28,350 --> 00:12:36,210 So we have to take this one here and use Capital R instead of this normal R here and write this down 162 00:12:36,210 --> 00:12:37,500 here under the integral. 163 00:12:39,120 --> 00:12:44,540 And now what you can see is we have an integral over to two angles, Theta and Phi. 164 00:12:45,010 --> 00:12:52,080 We have no integration over the radios because we know already that we only want to consider a certain 165 00:12:52,080 --> 00:12:53,500 radius of our sphere. 166 00:12:54,660 --> 00:12:58,610 So this means there's our square can be pulled out of the integration. 167 00:12:58,620 --> 00:13:05,640 So we have our square times and then we have these two integrations here that we can factories because 168 00:13:05,640 --> 00:13:09,270 the boundaries do not depend on any of these variables here. 169 00:13:10,320 --> 00:13:15,420 So this means we have a product of our squared times, this integral times, this integral here. 170 00:13:17,220 --> 00:13:22,740 And this is, again, very simple because we already know what this one is here is the same as that 171 00:13:22,740 --> 00:13:23,020 one. 172 00:13:23,190 --> 00:13:25,050 So we get to pie in this case. 173 00:13:25,650 --> 00:13:30,120 And here we have to do it in a very similar way as here. 174 00:13:30,150 --> 00:13:32,040 In fact, it's, again, the same thing. 175 00:13:32,370 --> 00:13:37,440 So we get minus cosine theta as the undefined integral. 176 00:13:37,800 --> 00:13:40,890 And this gives us a two if we consider the boundaries. 177 00:13:41,730 --> 00:13:45,210 So what we get is our squared times two pi times two. 178 00:13:45,720 --> 00:13:53,340 So the area of a sphere is four PI Times Square, which is also something that you probably have learned 179 00:13:53,340 --> 00:13:56,660 already in school in the very early year. 180 00:13:57,060 --> 00:14:00,540 But you were never really told where this comes from. 181 00:14:00,690 --> 00:14:06,990 And in fact, it's not possible for a student in school to know where this comes from because you need 182 00:14:06,990 --> 00:14:14,070 to consider integration and it's very much easier to consider spherical coordinates. 183 00:14:14,440 --> 00:14:19,350 So I hope that now you see that sometimes it can really be useful to consider alternative coordinate 184 00:14:19,350 --> 00:14:26,660 systems and you will find the use of these coordinate systems, especially in physics, in many problems. 185 00:14:27,270 --> 00:14:29,480 I've already introduced two of them to you. 186 00:14:29,520 --> 00:14:35,850 We have a wire where we calculate the magnetic field that we typically use cylindrical coordinates, 187 00:14:36,660 --> 00:14:42,330 or we could also calculate the electric field or four point charge where we would use the spherical 188 00:14:42,330 --> 00:14:46,470 coordinate system because our whole problem is rotational symmetric.