00:01
In this problem, our job is to find the centroid of the given region.
00:06
So we have the picture here, and we notice that this is the top quarter, if you will, the circle with equation x squared plus y squared equals a squared.
00:21
So that tells us that y can be written as the square root of a squared minus x squared, since we're in the first quadrant there.
00:29
The bottom boundary is a line that contains the point a comma 0 and the point 0 comma 2.
00:40
So it has slope negative 1 1 1ā2, and we also know its y intercept is a over 2.
00:47
So its equation is y equals negative 1 1ā2 plus a over 2.
00:53
So we have the top boundary of our region and the bottom boundary of our region.
00:59
So we will use those two boundaries defined first of all, the area of the region, then we will find the x coordinate of the centroid, and finally the y coordinate of the centroid.
01:10
So step one is to find the area of that region, and ordinarily we would integrate, and we would do top minus bottom.
01:18
But we notice that we have familiar shapes here.
01:21
The area at this region, we can take the quarter circle under the top boundary, and has radius a, so that area is one -fourth times pi times a squared.
01:34
Then we can subtract the area of the triangle underneath that.
01:40
It's a nice right triangle and its area is one half times its base times its height.
01:45
We'll do the height here times the base.
01:48
And so we see we can factor out a 1 fourth and an a squared and we are left with pi minus 1.
01:57
So we could have integrated to find that, but it's much easier just to use geometry to find that area.
02:03
And we will need that area to find both the x and y quarter.
02:06
Of the centroid.
02:09
So next let's move on and find the x coordinate of the centroid.
02:17
That next coordinate, we'll call that x bar, is 1 over a from part 1, times the integral from 0 to a of we have to take x times the top boundary.
02:32
So we'll use our square root of x squared minus, square of a squared minus x squared, excuse me, minus we also have to multiply x times our bottom boundary.
02:43
So minus x times our negative 1 half x plus a over 2.
02:50
And we need to integrate that with respect to x from 0 to a.
02:56
So the first part of the integral can be done with a substitution, letting u equal a squared minus x squared so that du is negative 2x d x.
03:08
So we see we need to adjust by a two -thirds and a negative one -half.
03:15
So the bottom line is we get negative one -third here times a squared minus x squared to the three -halfs, where we've skipped a few steps there.
03:27
Minus, now we still have to go back and tackle the second part of the integral.
03:33
Let's distribute the x.
03:35
So here we have negative 1 -half -x squared, plus a over 2 times x.
03:44
This is an a back here.
03:46
There we go.
03:49
So with the first part of our calculation, when we substitute a, here we get zero.
03:57
When we substitute zero, we get a squared to the three a halfs, which is a cubed...