In this quick video, I just want to do an example where we find the area enclosed by two curves, but we have to do it with respect to y. Let's jump right to it, here's the problem. Find the area between the following, the line y equals x minus 1 and the parabola y squared equals 2x plus 6. Notice in the question they don't tell you that you're going to go dx to dy, that's for you to decide. But you have to since this is obviously the title of the video, we're going to go dy, but let's pretend for a second we didn't know that. What should we do first? Always draw the picture just to help you out, so I'll draw a nice big picture off on the side. Let's draw the line y equals x minus 1, it's at this line of positive slope pulled down one, there's some line over here, y equals x minus 1 is now drawn. Then let's draw the parabola, y squared equals 2x plus 6 and be careful, this is now your up-down parabola because it's y squared. When you have y squared instead of x, y equals x squared. This is the parabola that opens to the right or to the left depending on the thing. Since it's a positive coefficient, we're going to open to the right. If you haven't drawn these one thing you can always do is of course, plug-in a few numbers, for example, when x is one, then you can you get say, eight and you can plug in some points. Where this thing crosses, you don't need to be the world's most accurate, you just get the general shape. But it's some parabola that opens to the right and you can plot some dots to see that and get some more numbers. But the point is I have some region, that is enclosed and that's what the area that I'm after. Take as much time as you need to graph this thing to get a handle on it. The region in question is there. Now not surprising the bounds are not given, expect them not to be given, and then just be pleasantly surprised when they're. The bounds are not given. We know how to deal with that, we know how to find these bounds, these points we're going to have to equate the two curves together. I can get the x values or the y values, it doesn't matter. But before I go forward with that, let's just think about this for a second. I want to find the areas, I want to find the area of this thing. It makes a little more sense in this case, I have a top curve and a bottom curve normally is what I'm looking for. But it gets a little weird towards the left of the y-axis. Where if I really wanted to use dx, if I want to start making horizontal slices, I have to break this region up into two curves, it just gets a little messy. It doesn't matter how you slice something you can cut it across or cut it up down horizontally, vertically, you're going to get the same area. Let's cut this thing using lines slices that are parallel to the x-axis. I'm going up and down the y-axis, I'm going to go dy. In this case to me it's a little more obvious, it's a little easier to do to think of as the area from whatever the two bounds are going to be, we'll find that in a second. But I look at this picture, there's a very clear right curve and there's a very clear left curve. I don't want to break this region up in terms of top and bottom, I'd have to split this and be mess. When you do the right curve minus the left curve when it's that picture, then you go dy, keep that in mind. We're going to go dy on this example because there's clear left curve, the parabola. When we go find our bounds, let's go ahead and find our bounds for a second, I'm really after the upper and lower bound on y. What is our upper and lower bound on y? Put these things together, it doesn't matter how you do it. Honesty, you can set them solve for x, you get x equals y plus 1, then over here you get x equals y squared minus 6, all divided by 2. You can set these equal to each other or you could solve for x and plug in substitute, there's many ways to do this. However, you do this though, pause the video if you want and go soft at these things. But when you put these things together and do some algebra, maybe we'll do that on the side over here, you're going to get two numbers, the x or y, you can find one, you could find the other. Let's go ahead and set these things together. You have y plus 1 is equal to y squared minus 6, all over 2. Multiply the two cross 2y plus 2 is y squared minus 6, move everything to one side, let's keep the y squared positive y squared minus 2y and then let's subtract 2x from both sides you get minus 8 equals 0. That will factor and you get y minus 4, then y plus 2, y plus two is 0, and that gives our values of positive four and negative two. According to our picture, the y value is positive four and negative two. Those are our bounds of integration just be careful, you're used to solving for x, now you need y, if you found x, of course, you can always go plug in your y too, it doesn't matter. Let's work this out, the integral becomes minus 2-4 of the right curve, that's the line x minus 1 subtracted from the left curve, which is 2x plus 6 dy. Now, I hope you're yelling at me on the screen because this is wrong, this is a common mistake people do. Let's stare at this for a second, how do you have x as your variable, but then try to do things dy? It doesn't make any sense, but you're so used to putting xs in there that you got to be careful. Let's just fix this so that everything is right except for the inside. I want to function of y, my area is a function of y dy. Let's go ahead and do that, let's write this again. I have the integral from minus 2-4, and now remembers the right curve but with a y as a function, you solve for x and we did that already. Up here becomes y plus 1 minus y squared minus 6 over 2 dy. Now it makes more sense, I have y's inside, I'm integrating the start to why the bounds are with respect to y, everything is good. You can work this out at this point now becomes just working out the integral, I leave that to you as an exercise but if you do this right, you should get 18. Again, the hard part of this thing is usually just setting up the integral, I leave it as an exercise to work this out. But final answer should be 18. Be ready to switch back and forth between dy and dx as you go through these problems. Good job. See you next time.
