Find the mass and center of mass of the lamina that occupies the region and has the given densit

Find the mass and center of mass of the lamina that occupies...

Key Concepts

Mass of a Lamina

The mass of a lamina with a variable density is determined by setting up and evaluating a double integral over the region occupied by the lamina. This integral takes the density function as its integrand, thereby capturing how variations in density contribute to the overall mass.

Center of Mass

The center of mass of a lamina is the point that represents the average position of its mass, considering its density distribution. It is computed using moment integrals (weighted averages) with respect to the coordinate axes, where the density function weights the positions.

Double Integrals

Double integrals are a fundamental tool in calculating quantities over two-dimensional regions. They allow the integration of a function of two variables over an area, which is essential for computing mass, moments, and other properties in a variable density context.

Region of Integration

A region of integration specifies the area over which the integration is performed. Its boundaries are defined by curves or functions, and correctly determining these limits is critical when setting up the corresponding double integrals for computing mass and center of mass.

Transcript

00:01 Okay, so for this problem, we are going to have to find the mass, and then we're going to use that to find the center of mass.

00:08 So what we are given is that this is bounded by y equals x squared and y equals x plus two.

00:21 So if we were just to quickly draw a graph, we're going to know that this is going to be approximately y equals x plus two.

00:28 And then our y equals x squared is going to look something kind of like this.

00:32 So we're looking for this area that is shaded, but we want to find out where they intersect.

00:38 So obviously, yeah, so what we need to do is figure out.

00:41 So we have x squared equals x plus two.

00:45 So i'm going to go ahead and put these both on the same side and equal it to zero.

00:51 And if i were to factor, i would get x minus 2 and x plus 1.

01:04 So it lets me know that x would equal 2 and x could equal negative.

01:07 1.

01:08 Okay.

01:11 So what i want to do is i want to go ahead and set up my double integral to find my mass.

01:22 And so i know that my x plus 2 is up further on the graph than the x squared.

01:29 So if i were to subtract those two, i would get a greater value with the x plus 2.

01:33 And then for my x, i know that it's going to be this.

01:35 So d y will be first and then d x.

01:39 And then what i know is that my my function is kx.

01:46 So first thing i want to do is integrate between, for y.

01:52 So if i do this, it'll be just kxy.

01:56 For x plus 2, x squared, dx.

02:02 So this is going to end up being k, and then x times x plus 2 minus x times x squared.

02:24 So i'm going to have, i'm actually going to put k in the front because it's considered to be a constant.

02:29 So i'm going to have x squared plus 2x minus x to the third.

02:35 And now i'm ready to integrate in terms of x.

02:40 So i'm going to have one third x to the third plus x squared minus one fourth x to the third, i'm sorry, x to the fourth, from negative 1 to 2.

02:56 So this is going to leave me with 8 thirds plus 4 minus 4, and then minus negative 1 3rd plus 1, minus 1 4th.

03:21 Okay? so by the time i add, subtract everything together, i end up getting 9 4ths k...

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