00:01
Locate the centroid of the following shape.
00:04
The centroid of a two -dimensional figure is described by two coordinates.
00:11
In this case, one along x and one along y.
00:17
Let's recall the formula for the centroid.
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Along x, x bar, the x -centroid is given by one over the area of our shape, times the integral, the double integral of x, d a, over our shaded region.
00:36
Where this here is r.
00:42
Similarly, y bar is given by 1 over the total area, time the double integral of ydar.
00:57
So we have three integrals to evaluate.
01:00
The double integral of xda, the double integral of y da, and the integral that will give us a, the integral over da.
01:10
Let's start by finding a.
01:16
This is given by the integral of if we integrate with respect to what, we can take this curve and subtract this curve.
01:25
So we're integrating 2 minus y minus y square the y for y range between 0 and 1.
01:35
Integrating term by term we obtain 2y minus y square over 2 minus y cube over 3 for y range even 0 and 1.
01:46
And evaluating this we obtain 7 over 6.
01:58
Now let's evaluate.
02:03
I'll call this i 1 or i well actually i can mean something else.
02:12
I'll call this a x.
02:15
The double integral of x d a let's write da as d x times the y or x will range between y square to 2 minus y or rather the opposite 2 minus y to y square and y range between 0 and 1.
02:44
Integrating with respect to x first we obtain x squared over 2 and then evaluating the integration limits.
03:00
And let's go step by step.
03:01
So we obtain the integral of x squared over 2 for x ranging between these upper or lower bound functions...