Center of gravity of a portion of cylinder

Attachments

1. Figure # 1
2. Figure # 2 reproduced from http://www.lmnoeng.com/Volume/InclinedCyl.htmhttp://www.lmnoeng.com/Volume/InclinedCyl.htm

Description
The blue is the mass inside a cylinder. In steady condition, I would want to know where its Center of gravity point shall be from either of the ends. Technically its like one mentioned in figure # 2 .It can be regarded as a portion of cylinder

Trials :
1) Divided them as a rectangle and a triangle and tried to combine the CG . (answer did not match with 3D models)
Workout :

Combined Center of gravity = area of Rectangle*Center of gravity of Rectangle + Area of triangle* Center of gravity of triangle/(Area of Rectangle+ Area of Triangle)
In this case, (taking all from left end : CG=(500*50+250*33.33)/(500+250)

No clue why it doesn’t match with the 3D modeling software’s answers. Per software, Center of gravity = 41.89 from left


Any ideas ?

Attachment 69029

Attachment 69030

Attached Images

	figure 1.jpg (32.5 KB)
	figure 2.jpg (12.7 KB)