00:01
I want to use an iterated integral in this question to find the area that is enclosed by these curves.
00:08
We've got x times y equals 9.
00:11
We've got y equals x.
00:14
We've got x equals 0.
00:18
And we have y equals 9.
00:20
So let's start by making a sketch of this region.
00:25
So if i start with x times y equals 9, that's going to look like y equals 9.
00:33
9 over x and at least in the first quadrant which i think is where i'm concerned this time y equals 9 over x is going to look something like this i also have y equals x that is this diagonal line slope 1 passing through the origin x equals 0 x equals 0 that is just the y axis.
01:05
So here is x equals zero and y equals nine that's a horizontal line.
01:15
I believe that horizontal line is going to be up here so that roughly speaking i am getting a region that looks something like this okay now in order to figure out the easiest way to do this we're going to need our intersection point.
01:38
Okay? so let's see where y equals x and y equals 9 over x intersect.
01:45
That'll happen when 9 over x is equal to x, when x squared is equal to 9, and since my region is in the first quadrant, i believe i have x equals 3.
01:59
So this is going to be the point 3.
02:05
Now, what i can do is to figure out this area, split this into two regions.
02:14
Okay? i think i want to split this here along the line, y equals three.
02:22
Now, what can i do? i can figure out the area of my bottom region.
02:28
By the way, this is just a triangle.
02:31
So if you wanted to do geometry for that, you could.
02:36
This will be a double integral of the function 1, dx, d, y.
02:43
Yes? so my x is here start at zero while they end when i hit that diagonal line of x equals y.
02:56
And i'm going to do that from y equals zero to y equals three.
03:04
And so what does that give us? we've got the integral from zero to three.
03:10
My antiderivative of one with respect to x, that's just x, evaluated on the interval from 0 to y, d .y.
03:25
That's giving us the integral from 0 to 3 of plugging in my top limit of integration, y minus my bottom limit of integration, 0, my integral from 0 to 3 of y, dy.
03:44
So how do we evaluate that integral? well, i find my anti -derivative.
03:49
My anti -derivative of y with respect to y, be y squared over two...