1 00:00:00,179 --> 00:00:05,000 [MUSIC]. 2 00:00:05,000 --> 00:00:09,410 We've seen that we can use either horizontal strips or vertical strips when 3 00:00:09,410 --> 00:00:14,971 using integrals to calculate area. We end up computing the same number, the 4 00:00:14,971 --> 00:00:18,924 area of this region, whether we use vertical rectangles or horizontal 5 00:00:18,924 --> 00:00:25,270 rectangles, we're getting the same answer, the area of this region. 6 00:00:25,270 --> 00:00:28,490 But if both work, which one should we choose? 7 00:00:28,490 --> 00:00:30,790 The best choice depends on the shape of the region. 8 00:00:30,790 --> 00:00:35,715 I like to think about how many different kinds of edges my rectangles will touch. 9 00:00:35,715 --> 00:00:37,770 Here is an example to demonstrate what I mean. 10 00:00:37,770 --> 00:00:41,800 I've drawn some region in here with three edges, this curved edge, this curved 11 00:00:41,800 --> 00:00:45,605 edge, and this straight edge along the x axis. 12 00:00:45,605 --> 00:00:49,120 Now, I could calculate the area of the enclosed region. 13 00:00:49,120 --> 00:00:55,640 Either by decomposing this into vertical strips or horizontal strips. 14 00:00:55,640 --> 00:00:59,849 The bad news about vertical strips is that there's two different kinds of 15 00:00:59,849 --> 00:01:03,857 vertical strips. There's vertical strips over here, and 16 00:01:03,857 --> 00:01:07,662 there's vertical strips over here. The vertical strips over here touch the 17 00:01:07,662 --> 00:01:10,478 orange and the purple edge, and the vertical strips over here would touch the 18 00:01:10,478 --> 00:01:15,122 orange and the blue edge. Contrast that with the much better 19 00:01:15,122 --> 00:01:20,640 situation that happens if I cut this up into horizontal strips. 20 00:01:20,640 --> 00:01:26,800 All of my horizontal strips have an orange side and a blue side. 21 00:01:26,800 --> 00:01:32,316 I only have one type of horizontal strip. So I should calculate the area of a 22 00:01:32,316 --> 00:01:37,367 region like this using horizontal strips. Let's take this advice and apply it to 23 00:01:37,367 --> 00:01:41,892 solve a specific problem. So here's a bigger, specific version of 24 00:01:41,892 --> 00:01:46,090 this problem. Let's figure out the area of this region. 25 00:01:46,090 --> 00:01:50,034 this orange curve is the curve y equals the square root of x, and this blue curve 26 00:01:50,034 --> 00:01:54,310 is y equals the square root of 2x minus one. 27 00:01:54,310 --> 00:01:57,790 And then here I've just got the purple line along the x-axis. 28 00:01:57,790 --> 00:02:01,020 I've really got a choice as to how to approach this. 29 00:02:01,020 --> 00:02:06,260 I could do this with vertical strips. But then I'd have to do two different 30 00:02:06,260 --> 00:02:09,000 integrals. I'd have to handle the case over here 31 00:02:09,000 --> 00:02:13,250 where rectangles are touching the orange line and the purple line. 32 00:02:13,250 --> 00:02:16,337 And I'd have to do an integral over here where my rectangles are touching the 33 00:02:16,337 --> 00:02:21,301 orange line and the blue line. It's going to be better to use horizontal 34 00:02:21,301 --> 00:02:24,088 strips. In that case, all of the rectangles are 35 00:02:24,088 --> 00:02:27,175 the same kind of rectangle, they're a rectangle with one edge on this orange 36 00:02:27,175 --> 00:02:33,125 curve, and one edge on this blue curve. Let me just draw one of those thin 37 00:02:33,125 --> 00:02:37,541 horizontal rectangles. I have to figure out the size of that 38 00:02:37,541 --> 00:02:40,520 rectangle. Let's suppose this thin strip is at 39 00:02:40,520 --> 00:02:44,060 height y. I know how tall this strip is. 40 00:02:44,060 --> 00:02:49,940 I'll call that dy. But how wide is this green strip? 41 00:02:49,940 --> 00:02:52,860 I should figure out what this point and where this point is. 42 00:02:52,860 --> 00:02:58,074 And if I think about it a little bit, if this is y, I can solve this equation and 43 00:02:58,074 --> 00:03:05,544 find out that in that case x. Is y squared, and I can similarly solve 44 00:03:05,544 --> 00:03:10,880 this equation and find out that if this point has a y coordinate y, the x 45 00:03:10,880 --> 00:03:17,522 coordinate will be y squared plus 1 over 2. 46 00:03:17,522 --> 00:03:26,080 And that tells me the width of this green rectangle, it's this minus this. 47 00:03:26,080 --> 00:03:29,470 I'm going to write down the area of that green rectangle. 48 00:03:29,470 --> 00:03:35,102 So the width is y squared plus 1 over 2 minus y squared, that's how wide this 49 00:03:35,102 --> 00:03:40,708 rectangle is. And the height of that rectangle is dy, 50 00:03:40,708 --> 00:03:44,875 so this product is the area of that green rectangle. 51 00:03:44,875 --> 00:03:48,347 But of course I don't care just about that one rectangle, I want to integrate 52 00:03:48,347 --> 00:03:53,120 so that I'm adding up the areas of all kinds of thin rectangles. 53 00:03:53,120 --> 00:03:56,300 And then in the limit I'm getting the area of this region exactly. 54 00:03:56,300 --> 00:04:00,722 So instead of just this single rectangle, I'm going to integrate y goes from where 55 00:04:00,722 --> 00:04:04,424 to where. Well, I look at my picture here, y can be 56 00:04:04,424 --> 00:04:10,430 as small as zero and as tall as this point up here which is one. 57 00:04:10,430 --> 00:04:14,722 So y goes from zero to one and this interval calculates the area of that 58 00:04:14,722 --> 00:04:18,399 region. Let's do the calculation. 59 00:04:18,399 --> 00:04:22,140 This integral is the integral from 0 to 1. 60 00:04:22,140 --> 00:04:29,516 for I can simplify the integrand a bit. It's 1 half minus y squared over 2 dy . 61 00:04:29,516 --> 00:04:35,010 And now we can write an antiderivative for that. 62 00:04:35,010 --> 00:04:40,232 So I add a derivative of 1 half. One half y and a derivative of y squared 63 00:04:40,232 --> 00:04:46,220 over 2 is y cubed over 6. I'm going to evaluate that at 0 and 1, 64 00:04:46,220 --> 00:04:51,430 take a difference. But when I plug in 0 I just get 0. 65 00:04:51,430 --> 00:04:54,430 So the answer is whatever I get when I plug in 1. 66 00:04:54,430 --> 00:04:58,822 And that's 1 half times 1, minus 1 cubed over 6. 67 00:04:58,822 --> 00:05:05,158 Could simplify that a bit, that's 1 half minus a 6th, and 1 half, well, that's 3 68 00:05:05,158 --> 00:05:14,830 6ths, so 3 6ths minus 1 6th, is 2 6ths, which you might write as, 1 3rd. 69 00:05:14,830 --> 00:05:21,770 So the area of that, shark fin shaped region is a third of a square unit. 70 00:05:21,770 --> 00:05:25,735 Often, it's not there are right ways or wrong ways but there are better ways and 71 00:05:25,735 --> 00:05:29,380 worse ways, right. Ways that are better in that they are 72 00:05:29,380 --> 00:05:34,100 quicker, safer, more elegant. So don't just make good choices. 73 00:05:34,100 --> 00:05:45,688 Make the best choices.