Answer: mass (m) = 4 kg
center of mass coordinate: (15.75,4.5)
Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.
The region D is shown in the attachment.
From the image of the triangle, lamina is limited at x-axis: 0≤x≤2
At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):
<u>Points (0,0) and (2,1):</u>
y = 
y = 
<u>Points (2,1) and (0,3):</u>
y = 
y = -x + 3
Now, find total mass, which is given by the formula:

Calculating for the limits above:

where a = -x+3




m = 2(-4+6)
m = 4
<u>Mass of the lamina that occupies region D is 4.</u>
<u />
Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:


and
are moments of the lamina about x-axis and y-axis, respectively.
Calculating moments:
For moment about x-axis:








Now to find the x-coordinate:
x = 
x = 
x = 15.75
For moment about the y-axis:







63
To find y-coordinate:
y = 
y = 
y = 4.5
<u>Center mass coordinates for the lamina are (15.75,4.5)</u>