a. Find the center of mass of a thin plate of constant density covering the region between the c

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

Key Concepts

Constant Density vs. Variable Density

When a region has constant density, the integrals simplify because the density factor can be factored out of the integral, making it proportional to the geometric area and moments. In contrast, a variable density function requires integrating the density function along with the coordinate factors, leading to different contributions to the mass and moments that reflect the non-uniform distribution of mass within the region.

Moment Integrals

Moment integrals are used to calculate the weighted average of the coordinates in a region. The moments about the y-axis and the x-axis are found by integrating x times the density and y times the density, respectively, over the region. These integrals are then used to determine the center of mass by dividing each moment by the total mass.

Setting Up and Evaluating Integrals

A key part of solving center of mass problems is correctly setting up the integrals with appropriate limits of integration based on the geometry of the region. This involves identifying the boundaries of the region and writing the integral in a form that accurately represents the area (or volume) and its mass distribution, whether the density is constant or variable.

Center of Mass

The center of mass of a continuous object is the point at which the entire mass of the object can be assumed to be concentrated. In the context of planar regions or thin plates, it is computed by taking the first moments of the mass distribution about the coordinate axes and then dividing by the total mass. This point provides valuable insights into the balance and distribution of mass within the region.

Mass of a Planar Region

In mass distribution problems, the total mass of a region is determined by integrating the density function over the region. For a region in the plane, this typically involves setting up a single or double integral, depending on the description of the region, and it accounts for how mass is distributed across the area.

Transcript

00:01 So for this question, we're given that y is equal to 1 over the square root of x, and we're looking from x equals 1 to x equals 16.

00:08 So for part a, we want to find the center of mass within plate of constant density covering this region.

00:15 So some things that we need to remember that we've learned before is the formulas for center of mass.

00:25 So x bar, we know it's just going to be m y over m, y, bar is mx over m and all these ms what they're equal to is mx that's just going to equal the integral from a to b of one half times our function f of x squared you would normally have a minus another function g of x squared but in this problem we are going around the x axis so we really only have one function and then m y is from a to b x times our function these have a d x so our first step oh and then m is just going to be our areas a to b of f x d x so our first step is to find what these three are and then just to plug it into here and the this x bar and y bar are going to be the coordinates of our center of mass so starting with m y it's just going to be we're going to be we're going to from 1 to 16 of x times our function 1 over the square root of x d x.

02:04 Well, if i simplify that, i just have 1 to 16 of x to 1 half.

02:14 And when i integrate that, i get 2 thirds x to 3 halves from 1 to 16.

02:25 And if you calculate that out, you should get 42.

02:33 Now, mx is going to be from 1 to 16 of 1.

02:39 One half times our function squared so this function squared this one half is just a constant so i can pull it out to the front and then this square is just going to be one over x dx and we learn from this chapter that the integral one over x is just going to be the natural log of x and then if you calculate that out you should get the natural log of four and lastly finding my m that's just going to be my area, so 116 of 1 over the square root of x, the x.

03:32 And if i integrate that, i get 2x to 1 1 1 1 .16.

03:40 And when you calculate that out, you should get 6.

03:45 So that means, from my x bar, which is, again, equal to m y over m, that's going to end up being 42 divided by 6, which is 7.

04:02 And then y bar which is equal to mx over m is just going to be the natural log of 4 over 6 so that's my answer for part a now for part b we're going to find this under mass if instead of being constant the density function is actually going to be x is equal to 4 over the square root of x so really the only thing that's going to to be changing here is instead of just having this we're going to be multiplying by our density out here so that means for m y it's going to be the same thing as before so x times my function 1 over the square root of x now i'm also multiplying by this delta x so if i simplify that out and multiply across, i just get 4x over square root of x times square root of x is x, which is just equal to 4 times integral from 1 to 1 .16 of 1 dx.

05:41 And then when i integrate that, i get 4 times x from 1 to 16, which is just equal to 60.

05:53 And then my mx is going to be very similar.

05:55 So i have 1 to 16.

05:58 Of the same thing...

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