00:01
So in this question we're given a shape in the xy plane, and it looks like this.
00:09
So we've got a length here of b.
00:12
This line is y equals h over a minus b, x minus b.
00:20
This line is y equals h over a x.
00:27
This height is h, and this distance here is a.
00:33
And we want the y -coronitor to the center.
00:37
So the y coordinate of the centroid is going to be the double integral over the area of y, which is the value y, divided by or normalized by the double integral over the area.
00:54
So what we can do is because we're integrating over y, we can think about the width in terms of y.
01:01
So this is going to be the integral d y of y times the width in terms of y, divided by the integral d y, of just the width in terms of y.
01:13
So what's the width in terms of y? well, the width is going to be x1 of y minus x0 of y.
01:21
So we need to invert these two things.
01:24
So let's call this x1 and x0.
01:30
The width in terms of y is going to be x1 in terms of y minus x zero in terms of y.
01:37
So we need to invert these...