Moment of inertia and Center of mass questions

[DWill]

Hi all, I need some help on a couple questions:

1) Find the moment of inertia about the x-axis of a thin plate of constant density 1 (density = 1) bounded by the circle x^2 + y^2 = 4. Then use your result to find Iy and Io for the plate.

Here's how I was thinking to set it up:

Since the region R is a circle, I thought it would be best to take the double integral with order dy dx. So the limits of integration for x would be -2 < x < 2, and for y would be -sqrt(4-x^2) < y < sqrt(4-x^2), right? Since density = 1, wouldn't the setup for the double integral be this (sorry I don't know how to do an integral sign, so a "|" is supposed to represent that):

|| y^2 dy dx, with the limits of integration I said above

I did it this way and eventually I get to the point where I have to take the integral of (sqrt(4-x^2))^3 with respect to x. I haven't figured out how to do this, so I was just wondering if I'm on the right track or not.

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Next question:

2) Find the center of mass of a thin triangular plate bounded by the y-axis and the lines y= x and y = 2 - x if density function = 6x + 3y + 3.

Once again I just want to check if I have set up this problem correctly. The limits of integration should be: 0 < y < 1, and y < x < 2-y ? And the order of integration should be dx dy ? So to find mass M of the plate I did this:

|| (6x + 3y + 3) dx dy, with the limits of integration as above

For My it would simply be: || x(6x+3y+3) dy dx, right?

Btw: all the inequalities should be <= instead of just <

Thanks

 

[Hootenanny]

Hi all, I need some help on a couple questions:

1) Find the moment of inertia about the x-axis of a thin plate of constant density 1 (density = 1) bounded by the circle x^2 + y^2 = 4. Then use your result to find Iy and Io for the plate.

Here's how I was thinking to set it up:

Since the region R is a circle, I thought it would be best to take the double integral with order dy dx. So the limits of integration for x would be -2 < x < 2, and for y would be -sqrt(4-x^2) < y < sqrt(4-x^2), right? Since density = 1, wouldn't the setup for the double integral be this (sorry I don't know how to do an integral sign, so a "|" is supposed to represent that):

|| y^2 dy dx, with the limits of integration I said above

I did it this way and eventually I get to the point where I have to take the integral of (sqrt(4-x^2))^3 with respect to x. I haven't figured out how to do this, so I was just wondering if I'm on the right track or not.

We'll start with question one first. You are indeed on the right track and your working thus far is correct. However, it may be useful to note that since we are dealing with a circular lamina, the integral greatly simplifies with the use of cylindrical coordinates.

 

[Hootenanny]

2) Find the center of mass of a thin triangular plate bounded by the y-axis and the lines y= x and y = 2 - x if density function = 6x + 3y + 3.

Once again I just want to check if I have set up this problem correctly. The limits of integration should be: 0 < y < 1, and y < x < 2-y ? And the order of integration should be dx dy ? So to find mass M of the plate I did this:

|| (6x + 3y + 3) dx dy, with the limits of integration as above

For My it would simply be: || x(6x+3y+3) dy dx, right?

Btw: all the inequalities should be <= instead of just <

Are you sure about your limits of integration? In this case it may be easier to set up the limits if you were to use horizontal sections (as you did above).