Find the center of mass of a thin plate of constant density δcovering the given region. The regi

step 1 Hello, hope you're doing well. So we've got our two equations here. So we're going to figure out

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

Key Concepts

Center of Mass (Centroid)

The centroid or center of mass of a region is the point where the mass of an object or area can be considered to be concentrated. When the density is constant, the coordinates of the center of mass are found by dividing the first moments (integrals of x and y over the area) by the total area. This concept is widely used in both physics and engineering to predict the behavior of objects under various forces.

Double Integration

Double integration is a technique used to integrate functions over a two-dimensional region. It allows the calculation of total mass, area, and moments about the coordinate axes by setting up integrals with proper limits that reflect the boundaries of the region. This concept is fundamental in calculus and is essential for finding centers of mass for irregular regions.

Moments of a Region

The moments of a region about the x-axis and y-axis are integrals that capture the distribution of the area relative to these axes. They are calculated by integrating y and x multiplied by the differential area element respectively. These moments are used along with the total area to determine the coordinates of the center of mass.

Constant Density Assumption

Assuming a constant density simplifies the calculation of the center of mass because the density factor can be factored out and typically cancels when computing the ratio of moments to total mass. This allows the problem to reduce to purely geometric calculations, relying solely on the area's shape and distribution.

Transcript

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

00:02 So we've got our two equations here.

00:04 So we're going to figure out where these equations intersect to try and figure out our boundary, like of our region of our thin plate.

00:13 So since both of these functions are equal to x, we're just going to set both of these functions equal to each other.

00:19 We're going to solve for y by subtracting y from both sides.

00:22 We get y squared minus 2y is equal to zero.

00:26 Factoring this, we get y times y minus 2 is equal to zero.

00:30 So this means that y is equal to 0 and y is equal to 2.

00:35 So these are kind of our boundaries as far as the y values for our thin plate.

00:40 So we want to try and visualize this a little bit better by graphing it.

00:44 So i'm just going to make a quick sketch of this.

00:52 So our function x is equal to y squared minus y.

00:55 It's going to have intersections at these two axes.

01:00 And then our, i'm sorry, it's going to intersect the, y -axis set, y is equal to 1 and y is equal to 0.

01:07 So the graph's going to go look something like this.

01:11 Then x is equal to y.

01:13 The graph is going to look about like that.

01:18 So these are our two intersection points that we got.

01:21 This kind of is our region right here, where our thin plate is.

01:26 So i'm going to kind of zoom in to that region where our thin plate is.

01:33 And we're going to solve this problem using horizontal strips.

01:37 So that's a horizontal strip right there.

01:39 So the height of this horizontal strip is just going to be d .y.

01:43 Now the length of the horizontal strip, we're going to figure out what x distance that is.

01:47 So it's essentially going to be the exposition of this right -hand function minus the exposition of this left -hand function.

01:54 So our right -hand function is x is equal to y, and our left -hand function is x is equal to y -squared minus y.

01:59 So it's going to be equal to y minus this side there.

02:08 It's y minus y squared minus y.

02:12 It's equal to y minus y squared plus y.

02:16 It's equal to 2y minus y square.

02:20 Meaning our area of this strip da is going to be this times our height, d .y is equal to 2y minus y squared dy.

02:29 All right, so we can use this value to find the mass of our thin plate.

02:34 So our mass is equal to the integral of dm, which is the integral of delta, which is our density function, da.

02:43 So that's, in this case, it's going to be, we can take out our delta, our density function, put it out in front.

02:50 This range is going to go from, as we can see here, zero to two.

02:55 We go from zero to two.

02:56 Our da function is 2y minus y squared dy.

03:01 We can solve this.

03:03 So we've got our delta times taking this integral, we get 2 y squared divided by 2 is just y squared, minus 1 3rd y cubed.

03:12 This is evaluated from 0 to 2.

03:15 That's equal to delta times 2 squared minus 1 3rd, 2 cube.

03:24 This is equal to delta times 2 squared is 4, minus two cubed is eights, it's eight thirds, getting a common denominator of three, that's delta times 12 thirds minus eight thirds.

03:37 This just goes to four thirds, meaning our mass is equal to four delta over three.

03:45 So this is our value for our mass of our thin plate right here.

03:49 So now i can use this value to figure out our center of mass.

03:53 So we know that our x value for our center of mass is going to be our mi or our moment over n, and y value is going to be an x over m.

04:03 So we got our x values, i'm sorry, our m values.

04:07 Now we're going to figure out nx and ny.

04:10 So we're first going to start with my.

04:12 So my is going to be equal to the integral of our x bar, our central, or i'm sorry, center of mass of our horizontal strip, and to dm, which is equal to the integral of x bar, delta, so we want to figure out what our x bar is...

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