Centre of gravity of a lamina

Question

Find the centroid of the region bounded by y=2-2x and y=x

Answer 1

Hi Lio!

 

You do not mention a third bounded side, which is typically either the y-axis (x=0) or the x-axis (y=0).  I will take y=0 to be the third side of the bounded region (the procedure is essentially the same for taking x=0 as the third side).  Also, without information about a distribution function for density, I assume the density in the region is uniform.

 

The centroid coordinates are given by:

 

x_cm = M_x/m

y_cm = M_y/m

 

where

 

m = ∫∫ σ(x,y) dxdy

M_x = ∫∫xσ(x,y)dxdy  (the moment of the lamina about the y-axis)

M_y = ∫∫yσ(x,y)dxdy  (the moment of the lamina about the x-axis)

 

where

σ(x,y) = density distribution function;  in the absence of more information, I am taking this to be a constant (which I can normalize to 1; it can be taken out of the integral ratios for x_cm and y_cm and defined above and cancelled, anyway)

 

With this assumption, m just becomes the double integral of dxdy over the region, which is just the area.  Since this is a line-bounded lamina, I could compute the area geometrically, but I will do the full integration here, for completeness.

 

So, let's set up the double integrals we need to get the coordinates.  First, I need to get the bounds I will use for the x-direction integral, assuming y=0 is the third side of my bounded region.  To do this, I will invert the equations above to get x in terms of y:

 

x=y

x = (2-y)/2

 

The first equation represents the left x-boundary of the lamina, and the lower equation the right x-boundary.  Hence, I will be integrating in the x-direction from y to (2-y)/2.

 

In the y-direction, I will be integrating up to the y-value where the two lines intersect.  That y-value is the one such that:

 

y = (2-y)/2 --> y = 2/3

 

So, the I will integrate y from 0 to 2/3.

 

With all this, I can set up the integrals that define x_cm and y_cm.

 

m = ∫∫dxdy = ∫₀^2/3∫_y^(2-y)/2 dxdy = ∫₀^2/3([(2-y)/2]-y)dy = ∫₀^2/3(1-(3y/2))dy = y-(3y²/4)|₀^2/3 = 1/3

 

M_x= ∫∫xdxdy  = ∫₀^2/3∫_y^(2-y)/2 xdxdy = ∫₀^2/3 ([x²/2]_y^(2-y)/2)  dy = ∫₀^2/3 (1-y-(3y²/4))/2 dy = 1/2[y-(y²/2)-(y³/4)]₀^2/3 = 5/27 (by my calculation)

 

M_y=∫∫ydxdy = ∫₀^2/3∫_y^(2-y)/2 ydxdy = ∫₀^2/3 y(1-(3y/2))dy = ∫₀^2/3 (y-(3y²/2))dy = [(y²/2)-(y³/2)]₀^2/3 = 2/27 (again, by my calculation)

 

That would make the coordinates of the centroid:

 

x_cm = M_x/m = (5/27)/(1/3) = 5/9

y_cm = M_y/m = (2/27)/(1/3) = 2/9

 

I am not guaranteeing the correctness of all the algebra, but, given the configuration and my assumptions, these values seem reasonable.  They are within the bounded region, and in approximately the correct part of the region, based on a quick drawing.

 

The procedure is similar if y=0 is the other boundary, except you have to recompute the limits of integration for that region.

 

Hope this helps!  Let me know if you want to go over any of the details more closely.