 Hi everyone. I have one final example for you of finding the area between two curves. And this one will be different from the others that we've worked through. We are asked to find the area of the region bounded by the graphs of x equals 3 minus y squared and x equals y plus 1. So we do still want to graph these in our calculator to get a good look at them. But hopefully one thing that strikes you as being different about this problem is that our equations are given to us x in terms of y as opposed to y in terms of x that you're used to. So of course some rearranging is involved to get these into our calculator. So this one we could rearrange some things. Obviously we have to take the square root of both sides. But in doing so we do need to remember it's both the positive and negative square root. Now one reason as to why? Take a very close look at the wording of this problem. Notice it says the region bounded by the graphs. It does not say bounded by the functions. If it said functions instead of graph you would only want the positive square root because of course by definition of a function each x can only have one y. But this problem is not implying that they are talking of these curves as functions. So that's something very subtle you need to pick up on. The other one of course is just a simple linear equation so that's nice and easy. So when we go to graph these we're going to have to do it as y1, y2, and the two of those I'm going to have as the square root equation. And then y3 I'll have as my linear equation. So let's take a look at those. So you'll notice y1 as the square root of 3 minus x, y2 is the negative square root of 3 minus x, and finally y3 is x minus 1. So you can zoom 6 to graph it. So notice the sideways parabola opening to the left, the blue and the red ones. Those are the two that comprise that curve x equals 3 minus y squared. Of course your black one there in the center, your linear equation. The region you're trying to find is this one over here kind of in the fourth quadrant and peeking a little bit into the first quadrant. Sort of this like triangular looking part here. That's what you're trying to find the area. So as we've talked about before, one thing you need to decide is the orientation of your representative rectangle. Remember the idea is that you want that rectangle to hit both curves at the same time. So let's go back because I actually have a picture of this on which I can write for you. So you can see the sideways parabola opening to the left. So think about what this might look like. If we did vertical rectangles like we're used to. Alright we're right there you're touching both curves so that's good. You're touching both curves that's good. But what if you drew it all the way over here? Technically you're touching the same curve because both the top and the bottom are touching that sideways parabola which are essentially the same curve. So this is one that maybe it's not the best way to do it as vertical rectangles. Maybe we want to do this as horizontal rectangles instead. Because look what happens when you do horizontal rectangles. No matter where I draw my horizontal rectangle or if I put it up here. Notice on the right side I'm touching the sideways parabola that x equals 3 minus y squared. And on the left I'm hitting the linear equation the x equals y plus 1. This turns it into a dy problem because the width of your rectangles now going vertically along the y axis. So this becomes a dy problem and our integrand as opposed to doing top minus bottom like we're used to doing we're going to do right minus left. So the way we'll set this up and now also our limits of integration need to be y values. So you might have to go ahead and find these. And you'll want once you find those points of intersection you'll want to use the y values. So in this case it's going to be the integral from negative 2 to 1. They're the two y values that you find. What you're hitting on the right is that x equals 3 minus y squared. Remember that it's a dy problem therefore our integrand must be in terms of y. So that's your right minus and then on the left it's hitting y plus 1. So you can do that antiderivative by hand if you wish or you can just go ahead and evaluate it in your graphing calculator. I'm going to simplify it a little bit. So I have to distribute that negative so I'm going to go ahead and put these in order. Negative y squared minus y that would be 3 minus 1 so that becomes plus 2. So you can do the integral of that. Again you can either do it by hand or you can go ahead and evaluate in your calculator which is what I think I shall do. So let's go back to our quit screen. Now here too you also have a few options. Let's set up our limits of integration. For consistency sake some people like to keep this in terms of y because that's what you have the way you wrote it out. You can definitely do that. So you just have to hunt and peck for your y so we need a negative y so it's going to be alpha and then the number 1 squared minus y plus 2. Now just remember it has to be a dy so it looks like that. And you get 4.5. Now your other alternative you easily could have done this problem using your x's instead. The calculator basically does not care what letter you use. It's just going to evaluate the integral. So of course the x's faster to hit because you don't have to hunt and peck for it but you'll see I'm going to prove it to you that you do actually get the same answer regardless of whether you use x's or y's. So it's really your choice as to how you want to do it. The important thing though is how you set this up by hand to have put your integrand in terms of x's would have been incorrect. So the moral of the story with this particular problem, if it's a dy problem in which your rectangles you need them horizontal so that you can be touching both curves, one curve on the right end, one curve on the left end, you're going to have to remember it's a dy problem, your limits of integration are y values and your integrand must be in terms of y.