Speaker A
We're asked to find the mass, the moment about the x-axis, the moment about the y-axis, as well as x-bar and y-bar, which should be the center of mass of the lamina bounded by y = -x + 3, y = x divided by 2, and x = 0.
Learn how to find mass, moments, and center of mass of a triangular lamina using double integrals with a given density function.
Key Takeaways
- Double integrals can be used to find mass and moments of lamina with variable density.
- Choosing the correct order of integration simplifies the calculation process.
- Moments about axes help determine the center of mass coordinates.
- The center of mass lies within the bounded region and depends on the density distribution.
- Step-by-step algebraic simplification is crucial for evaluating integrals accurately.
What the video covers
- The video covers finding mass, moments about x and y axes, and center of mass of a lamina bounded by three lines.
- The region R is bounded by y = -x + 3, y = x/2, and x = 0, with density function ρ(x,y) = 3(x + y).
- The order of integration chosen is dy dx due to the shape of the bounded region.
- Limits of integration for y are from y = x/2 to y = -x + 3, and for x from 0 to 2.
- Mass is calculated as a double integral of the density function over region R, resulting in mass = 18.
- Moments about the y-axis and x-axis are computed using double integrals with integrands xρ(x,y) and yρ(x,y) respectively.
- Moment about y-axis is found to be 13.5, and moment about x-axis is 27.
- Center of mass coordinates (x̄, ȳ) are calculated as moments divided by mass, resulting in (0.75, 1.5).
- The center of mass is located approximately within the bounded triangular region R.
- Detailed algebraic steps and integral evaluations are shown to reinforce understanding.
Chapters
- 00:00Problem Introduction and Region Description
- 00:33Setting up the Mass Integral
- 00:48Analyzing the Bounded Region and Integration Order
- 01:06Defining Limits of Integration
- 01:50Integrating with Respect to y
- 02:18Evaluating the Integral and Simplifying
- 05:42Calculating the Mass
- 06:03Setting up Moments about Axes
- 06:25Calculating Moments and Center of Mass
- 07:28Conclusion and Summary
Full Transcript — Download SRT & Markdown
Speaker A
I've already graphed the bounded region below. Let's call this region R. The density function is rho of x,y equals 3 times the quantity x + y, which we will use in the form of 3x + 3y.
Speaker A
Let's begin by determining the mass. The mass is equal to the double integral over the region R, the density function, differential A.
Speaker A
Let's first decide on the order of integration or whether we should let differential A be equal to dy/dx or dx/dy.
Speaker A
So looking at the bounded region R, notice it's bounded below by one line and above by one line.
Speaker A
But if we look at the region from left to right, it's bounded on the left by one line, but on the right it's bounded by two different lines, and therefore it's going to be easier to integrate with respect to y first and then x.
Speaker A
So we'll let differential A be equal to dy/dx. So the mass M is equal to the double integral of the density function in the form of 3x + 3y, dy/dx.
Speaker A
And now, determine the limits of integration. Starting with y, again, the region is bounded below by y = x divided by 2 and above by -x + 3.
Speaker A
The lower limit of integration for y is x divided by 2. The upper limit is -x + 3.
Speaker A
Now to find the limits of integration for x, we do have to find the point of intersection here on the right, which I already found is the point 2,1.
Speaker A
The limits integration for x are from x = 0 to x = 2. And now, we integrate with respect to y treating x as a constant.
Speaker A
The antiderivative is 3xy plus 3 times y squared divided by 2, or 3/2 y squared.
Speaker A
Now we need to find F of -x + 3 minus big F of x divided by 2 by performing substitution for y.
Speaker A
So here's where the algebra will get a little messy. Big F of -x + 3 is equal to 3x times the quantity -x + 3 plus 3/2 times the square of -x + 3.
Speaker A
Then, we have minus big F of x divided by 2, which is 3x times x divided by 2 plus 3/2 times the square of x divided by 2.
Speaker A
And now, we need to do a bunch of simplifying. Let's do some work here on the side.
Speaker A
Distributing 3x, we have -3x squared plus 9x, then we have plus 3/2 times the square of -x + 3.
Speaker A
Well, the square of -x + 3 is x squared minus 6x plus 9. We'll come back to this.
Speaker A
And then we have minus quantity 3x times x divided by 2, which is 3/2 x squared.
Speaker A
And then we have plus 3/2 times the square of x divided by 2, which gives us 3/8 x squared.
Speaker A
So let's go back to our integral. We have the integral from 0 to 2 of we have -3x squared plus 9x.
Speaker A
Let's distribute 3/2 now. We have plus 3/2 x squared, 3/2 times -6 is -9, giving us minus 9x, and then we have plus 3/2 times 9, which is plus 27/2.
Speaker A
And then we have minus, in the parentheses here, we have 3/2 x squared plus 3/8 x squared.
Speaker A
3/2 x squared is equivalent to 12/8 x squared. 12/8 x squared plus 3/8 x squared is 15/8 x squared.
Speaker A
And again, we are subtracting. And now, let's simplify again by combining like terms. Notice we have 9x minus 9x.
Speaker A
Looking at the x squared terms, we have minus 3 x squared plus 3/2 x squared minus 15/8 x squared, which simplifies to minus 27/8 x squared.
Speaker A
And then, we still have plus 27/2. And now, we integrate with respect to x, the antiderivative is going to be 27/8 times x cubed divided by 3, which is minus 27/24 x cubed, and then we have plus 27/2 x.
Speaker A
We now need to find big F of 2 minus big F of 0. Substituting 2 for x, we have minus 27/24 times the cube of 2 plus 27/2 times 2 and big F of 0 is equal to 0.
Speaker A
Simplifying, we have 18. So now we know the mass of the lamina is 18.
Speaker A
The next step is to determine the moment about the y-axis, the moment about the x-axis, and then we can determine the center of mass by determining x-bar, which is the moment about the y-axis divided by the mass, and y-bar is equal
Speaker A
to the moment about the x-axis divided by the mass. Let's go and set up the double integrals for the moment about the y-axis and the moment about the x-axis.
Speaker A
So I've already set this up. Notice how for the moment about the y-axis, we have an integrated function of x times the density function, which gives us x times the quantity 3x + 3y.
Speaker A
Distributing, we have 3 x squared plus 3xy. The order of integration is the same, as well as the limits of integration.
Speaker A
Working all this out, the moment about the y-axis is equal to 13.5, and then for the moment about the x-axis, the integrating function is y times the density function, which gives us an integrated function of 3xy plus 3 y squared.
Speaker A
We go through the same process again with the same order of integration, as well as the limits of integration, the moment about the x-axis is equal to 27.
Speaker A
So now that we have the mass and the moment about the y-axis and moment about the x-axis, we can determine the center of mass given by x-bar, y-bar.
Speaker A
The x coordinate of the center of mass is 13.5 divided by 18, which is 0.75, and the y coordinate of the center of mass is 27 divided by 18, which is 1.5.
Speaker A
If we go back to the bounded region R, the point 0.75, 1.5 is approximately here, which is the center of mass.
Speaker A
I hope you found this helpful.
Topics:double integralmass calculationmoment of inertiacenter of masslaminatriangular regiondensity functioncalculusmultivariable integrationMathispower4u
