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**1. Homework Statement**

Give the X, Y, and Z coordinates of the center of gravty of a half cubical block of uniform density. The diagram below shows how the block was cut.

The diagram depicts a cube with sides of length 1 cut diagonally in half. the six vertices are as follows:

(0,0,0) (1,0,0) (0,0,1)

(0,1,0) (1,1,0) (0,1,1)

**2. Homework Equations**

X = (w1x1 + w2x2 + ... wnxn) / (w1 + w2 + ... wn)

Y = (w1y1 + w2y2 + ... wnyn) / (w1 + w2 + ... wn)

Z = (w1z1 + w2z2 + ... wnzn) / (w1 + w2 + ... wn)

Where each set of points is the center of the shape (x1,y1,z1) and the w represent the weight (w1).

**3. The Attempt at a Solution**

Center Points of the five shapes:

1) (1/3, 1, 1/3); W1 = .5

2) (.5, .5, 0); W2 = 1

3) (0, .5, .5); W3 = 1

4) (1/3, 0, 1/3); W4 = .5

5) (.5, .5, .5); W5 = sqrt(2)

X= [ (1/3) (.5) + (.5) (1) + (0) (1) + (1/3) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))

X= .349

Y= [ (1) (.5) + (.5) (1) + (.5) (1) + (0) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))

Y= .5

Z= [ (1/3) (.5) + (0) (1) + (.5) (1) + (1/3) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))

Z= .349

The X,Y,Z coordinates of the center of gravity of the half cubical: (.349, .5, .349)

My only concern is the i took the surface area of the different shapes to be the weight of the different parts. I'm not sure if this is the correct way to go about the problem. (I do know for 2D shapes that the area of the shape can be used as its weight when dealing in the Cartesian plane if the density is uniform since the area is proportional to the weight).