00:01
Okay, we are going to find the perimeter of the region bounded by the graphs, y equals 3 and x equals 2, and y equals x.
00:11
So to find out where these are equal, we can either replace our y or replace our x, but they can be replaced with each other because of the y equals x.
00:22
So we can say y to the third equals y squared.
00:25
If we subtract y squared from both sides, we can factor out of y squared, and we see that we get y squared times y minus one.
00:36
So that means the two places the intersect are at y equals zero and y equals one.
00:43
So we'll be integrating from zero to one, and we are going to be in terms of y for this.
00:50
Now, the reason is if i wanted to solve and have this in terms of x, it would be y equals x to the two -thirds, but when i take my derivative, i'm going to not have continuity at zero.
01:05
So if i do it in terms of y, i'm going to be okay.
01:09
Okay, so we're going to have the square root of, so our derivative is the three halves come down, and then we're to the one half power.
01:17
Now we're going to square it as we're putting it in here.
01:20
So we have four nines y plus one and we're in terms of y.
01:25
Okay, so i want you to kind of think about this.
01:32
One of them is the y equals x.
01:35
So when you have y equals x, its length is just a hypotenuse, right? it's a straight line, and it's got a side one and a height one.
01:43
So one squared plus one squared, and then you take the square root, you get a square root of two.
01:47
So at some point we will add a square root of two.
01:50
But first we're going to find the piece that we have that is on the, the curve y to the third equals x to the half...