19. Center of mass Find the center of mass of a thin plate of ...

Question

19. Center of mass Find the center of mass of a thin plate of constant density covering the region in the first and fourth quadrants enclosed by the curves y=1/(1+x²) and y=-1/(1+x²) and by the lines x=0 and x=1.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the center of mass of a thin plate of constant density covering the region in the 1st and 4th quadrants. Enclosed by the curves, y is equal to e to the power of x and y is equal to e to the power of x, and by the vertical lines x is equal to 0 and x is equal to the natural log of 2. OK, so it appears for this particular prong we're asked to take all the information that is provided to us and we're asked to determine what the center of mass is for this specific thin plate, and specifically this thin plate has constant density. And that's important to keep in mind. So ultimately we are asked to determine what the center of mass is for this thin plate based on the information that's provided to us. So now that we know what we're ultimately trying to solve for. Our first step that we need to take is we need to take a moment to come up with a brief plan for how we're going to solve this problem, and to begin, we need to note that in order to find the center of mass, which the center of mass is denoted as parentheses of x, commay bar in parentheses, in order to find. The center of mass of a thin plate with constant density, we will need to calculate the total area and the moments about the axes, and the coordinates will be defined as x bar will be equal to m subscript Y, MY for short, divided by a, and y bar will be equal to m subscript x MX divided by a. And we need to note that our first major step to solve this problem is we will need to find the total area which is denoted as A. So the region enclosed between Y upper, so Y upper will be equal to e to the power of x and y lower. Will be equal to, so once again, Y upper, and then we need to note that y lower will be equal to e to the power of x, and the height of our plate will be H X, and HX will be equal to EX minus parentheses of e X, which will be equal to 2 multiplied by E the power of x. And we need to find the area by integrating from X is equal to 0 to X is equal to the natural log of 2. So we need to note that the area A will be equal to the integral evaluated from 0 to the natural log of 2 of 2 multiplied by e to the power of x. DX, which will be equal to square brackets of, so square brackets of -2 multiplied by e to the power of x, and this will be evaluated once again from 0 to the natural log of 2. So that will mean that A is equal to -2 multiplied by e to the power of the natural log of 2 minus 2 multiplied by e to the power of 0, which when we plug this expression into our calculator, we will find that it will be simply equal to 1. So that means without a doubt that the total area is simply 1. Our second step now is we need to find the moment about the X axis, which will be M X and Y bar. So for a region bounded by two functions, the moment about the x axis will be written as m x is equal to the integral evaluated from a to b of 1/2 multiplied by square brackets of parentheses of y upper to the 2 minus y lower. To the power of 2. DX fantastic. So now we need to substitute in our functions, so we will find that. M X will be equal to drum roll. The integral evaluated from 0 to the natural log of 2 of 1/2 multiplied by square brackets of parentheses of e to the power of x 2 minus e to the power of x to the power of 2. DX fantastic. And we will find that we could write that MX will be equal to the natural law, so it's gonna be MX. Which will be equal to the integral evaluate from 0 to the natural log of 2. Of 1/2 multiplied by square brackets of e to the power of 2x minus e to the power of 2x. DX, which will be equal to the integral evaluated from 0 to the natural log of 2 of 0 DX, which is just simply equal to 0. So now, at this point, we need to calculate Y bar, and Y bar, as we remember, should be equal to MX divided by A. So we need to note that 0 divided by 1 will be simply equal to 0. Fantastic. So that means, without a doubt, the vertical center of mass will be 0. Our third step now is we need to find the moment about the Y axis, which is MY. And the moment about the y axis is the integral of, and that's we can note, it's the integral of X multiplied by HXDX. So that means we can go ahead and write that MX, when, we solve for MX, so now we need to solve for MY. So MY. Will be equal to the integral evaluated from 0 to the natural log 2 of x multiplied by parentheses of 2 multiplied by e to the power of x dx, which will be equal to 2 multiplied by the integral evaluated from 0 to the natural log of 2 of x multiplied by e to the power of x dx. Fantastic. So now in order to solve this specific integral at this stage, we will need to recall and use the integration by parts method. So we will state that u is equal to x, which will mean that duu by dx will be equal to 1, which will mean that duu will be equal to dx. Fantastic. So moving right along here, we also need to let dV be equal to e to the power of x dx, which will mean that the integral of dV will be equal to the integral of e to the power of x dx, which will mean that V will be simply equal to e to the power of x. So that will mean that my will be equal to 2 multiplied by parentheses of square brackets of x multiplied by e to the power of x, and this will be evaluated from 0 to the natural log of 2. Minus the integral evaluated from 0 to the natural log of 2 of parentheses of e to the power of x dx, so that will mean that my will be equal to 2 multiplied by square brackets of x multiplied by e to the power of x minus e to the power of x. evaluate from 0 to the natural log of 2. So we now need to evaluate our limits. So we need to take my is equal to 2 multiplied by parentheses of square brackets of the natural, so negative log of 2 multiplied by e to the power of the natural log of 2 minus e to the power of the natural log of 2. Minus square brackets of 0 minus the 0. Fantastic. Don't forget your square bracket and your parenthesis. Fantastic. So we will find that we can go ahead and write that MY will be equal to 2 multiplied by parentheses of 1/2 multiplied by the natural log of 2 minus 1/2 + 1. Which will be equal to when we plug this expression to our calculator and write our answer in its most simplified form will be simply equal to 1 minus the natural log of 2. Fantastic. Our 4th and final step is we now need to calculate the center of mass coordinate x bar. And as we should recall, x is equal to my divided by a, which is equal to 1 minus the natural log of 2 divided by 1, which will be equal to 1 minus the natural log of 2. Which will mean that our final answer for the center and mass of this thin plate, which is parentheses xy bar once again in parentheses, will be equal to parentheses of 1 minus the natural log of 2.0, and that's it. That is our final answer. Hooray, we did it. We solve for our final answer. So going back to the top to recap what we've learned, we can safely say without a doubt that our final answer. For the center of mass for this thin plate that has constant density will be once again parentheses of 1 minus the natural log of 2.0, and that's it. That's our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

19. Center of mass Find the center of mass of a thin plate of constant density covering the region in the first and fourth quadrants enclosed by the curves y=1/(1+x²) and y=-1/(1+x²) and by the lines x=0 and x=1.

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