Find the center of mass of a thin plate of constant density...

Find the center of mass of a thin plate of constant density ? covering the region in the first and fourth quadrants enclosed by the curves y = 3 / (1 + x^2) and y = -3 / (1 + x^2) and by the lines x = 0 and x = 1. x? = , y? = (Type exact answers, using ? as needed.)

Transcript

00:01 Hello, hope you're doing well.

00:02 So we have our two functions here that's graphs form the boundary of our flat or thin plate.

00:10 Then our thin plate is also bounded by x is equal to zero and x equal to one.

00:15 So if we draw that out here, the graph is going to look something like that.

00:25 And then our boundaries of x equal to zero and x is equal to one.

00:31 So this kind of right here is our area of our thin plate.

00:36 So that means that our f of x function, if we define our f of x function is the function on top and g of x function is the function on the bottom.

00:46 This right here, this function is equal to f of x.

00:51 And this second function here is equal to g of x.

00:54 All right.

00:56 So first things first, we want to find the mass of this flat plate or this thin plate.

01:03 So to do that, we are going to calculate our mass that's going to be equal to the integral of our density times f of x minus g of x, dx.

01:18 It's going to go from a to b.

01:20 Just bring our density, which is just a constant out front, it's the integral from, you know that it's going to be between x is equal to zero and x is equal to one times f of x is 1 is 1 plus x squared.

01:33 It's going to be minus negative 1 over 1 plus x squared.

01:43 This just becomes a plus here, dx.

01:46 So combining these two, you get delta times integral from 0 to 1 of 2 over 1 plus x squared dx.

01:56 We can bring the 2 out front because it's a constant.

01:58 So that's 2 times delta times the integral from 0 to 1 for 1 over 1 plus x squared dx.

02:05 So taking this integral, this is going to delta, the integral of 1 over 1 plus x squared is arc tan of x, the inverse tan.

02:16 We evaluate this from 0 to 1.

02:20 It's going to be equal to 2 delta times arc tangent of 1.

02:27 Sorry, let me make that a little bit neater.

02:29 Arc tangent of 1 minus arc tangent of 0, which goes to 0.

02:37 And then arc tangent of 1, that is equal to pi over 4.

02:41 So that means our mass is equal to 2 delta pi over 4.

02:46 It means our mass is equal to delta pi over 2.

02:52 So that is our mass right there.

02:54 So now we want to find our x and y values of our center of mass.

02:59 So our x value for our center of mass is going to be equal to 1 over our mass, times the integral from a to b are delta x times f of x minus g of x, dx.

03:15 So since our mass is delta pi over 2, 1 over mass is going to be 2 over delta pi.

03:22 You can bring out this delta right here back out front.

03:27 So these just cancel it go to 0.

03:30 It's going to be integral from 0, i'm sorry, yes, from 0 ,000.

03:35 To 1, and we've got x times, we already know from before that f of x minus g of x gives us 2 over 1 plus x squared squared d x.

03:49 We can bring the constant 2 out front.

03:52 So it gives us 4 over pi times the integral from 0 to 1 of x over 1 plus x squared d x.

04:01 So now we can use, we need to use substitution to solve this integral.

04:06 So you can use u is equal to this 1 plus x squared down here.

04:12 I'm sorry, something messed up with my connection.

04:18 There we go.

04:19 Should be back to normal...

Question

Find the center of mass of a thin plate of constant density ? covering the region in the first and fourth quadrants enclosed by the curves y = 3 / (1 + x^2) and y = -3 / (1 + x^2) and by the lines x = 0 and x = 1. x? = , y? = (Type exact answers, using ? as needed.)

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