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

step 1 We have this lamina that is between the regions. The xy plane is bounded by y equals to 0, that

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

Key Concepts

Double Integration

Double integration is the process of integrating a function over a two-dimensional region. It serves as a powerful tool for calculating quantities such as area, mass, and other aggregate measures where the function may vary with position over a plane.

Mass of a Lamina

The mass of a lamina with a variable density is calculated by integrating the density function over the region occupied by the lamina. This ensures that every x-y coordinate’s contribution to the total mass is accounted for, making it a fundamental application of double integration.

Center of Mass

The center of mass is the point at which an object’s mass is considered to be concentrated. For a lamina with variable density, it is determined by computing the weighted average of the coordinates using moment integrals, and then dividing these moments by the total mass.

Moments of a Region

Moments refer to the integrals of the coordinates multiplied by the density over a region, reflecting the distribution of mass relative to an axis. The moments about the x-axis and y-axis are critical in determining the center of mass of a variable density lamina.

Transcript

00:01 We have this lamina that is between the regions.

00:09 The xy plane is bounded by y equals to 0, that is this line.

00:20 And the line y equals cosine, cosine of x.

00:36 That is this line between minus pi halves.

00:46 By halves this is y equals the cross of x so you have this region and for that region you would like to find the center of mass so this center of mass has coordinates c x of y and the all c sub x is equal to 1 over the mass times the intro side of d of x times the density of the d.

01:35 The density here is for this region as density y and c of y is 1 over the mass times y at row so we integrate well we need to find what is the mass so that is going to be the integral density so well y is going to vary between 0 and cosine of x and x goes between minus pi halves and pi halves which is at the points at which is is 0 so now being together of y would be y squared halves and that evaluate it became cost x and 0 so that will be just a cosine square of x halves and then that begin minus by halfs and by halves b halfs vx d x so, cosine square is equal to one half of one plus cosine of 2x.

03:43 And this, there's these two -hand identity so we can put that in there.

03:53 So that is going to be equal to one half of interforminous by halves, minus play halves over one half one plus cosine of 2x x x also for this integral let's see what happens with this integral so we have this interval from minus pi halves not 2 pi halves cosine of 2x so cosine of 2 x was going to be like that region that we double the period so that it does something like that so between minus pi halves and pi halves so that if we integrate these with respect to x his negative side this negative region will cancel with that positive so that that integral is zero so that this part will contribute nothing to the integral and then the integral this integral will be equal to one one half comes one a half pro so one fourth at 12 of d x is just x so x evaluated between pi halves and minus pi halves which is equal to one fourth of pi halves minus pi halves which involved this is equal to one fourth of pi halves minus pi halves which involved this still added our pi so the mass is going to be equal to pi over 4 and is pi over 4 m is equal to pi over 4 and now um all now to find the sis sub x we would need to do the integral of x x comes y y x and so if you can see on cost of x and we can pi halves minus by halfs and pi halves um so this integral we integrate y we get y square halves so it's going to cosine of x squared halves and that times x between minus pi halves and pi halves x also so c sub y would be equal to this quantity divided by one over the mass that is this is equal to the x coordinate for the center of mass c sub x not c sub y but c x so if we consider this function this domain is symmetric this function is anti -symmetric because cosine of something is squared that is symmetric but x is antithmic is odd and this domain is symmetric so that this integral has to be equal to 0 as well we can consider there the same mass that is over this side over the left -hand side is going to be the same of the mass that is on the right -hand side so that this integral mass has to lie at x equals to 0.

08:22 Now for c sub y, we can compute this integral involved there.

08:32 That is going to be minus 5 halves, into 5 halves, the integral from 0 to cosine.

08:40 Then we need to multiply y by the density that is y, so that we have this integral...

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