Center of Mass
Math 211  Prof. Leise

Email completed labs to tleise@amherst.edu by noon the Wednesday after Thanksgiving.

We can use double integrals to calculate the mass and center of mass of a thin sheet of material, often called a "lamina" in calc texts.  Once we have triple integrals, we can directly extend these ideas to general three-dimensional objects.

Consider a lamina, parallel to the xy-plane, whose density might vary with x and y, but doesn't change in the z-direction.

Recall that mass equals density times volume. The density function (rho) ρ(x,y) gives the mass density per unit volume at each point (x,y) (we're assuming the density doesn't change in the z-direction). We use an integral to find the mass for a lamina:  divide the object into tiny pieces with dimensions dx × dy × thickness of the lamina, then the mass of each tiny piece is ρ(x,y)×dx×dy×thickness.  Adding up the masses of all the tiny pieces yields:

mass = thickness times ∫∫ρ(x,y)dxdy (integrating over the region of the xy-plane covered by the lamina)

To integrate in Mathematica, you give the outer integral’s bounds first, then the inner integral’s bounds.   For example, to integrate the function f(x,y)=x+y on the triangle bounded by x=1, y=0 and y=x:

Integrate[x+y,{x,0,1},{y,0,x}]

Exercise 1.  Consider a small rectangular sheet of metal (perhaps an aluminum alloy) whose thickness is 1mm and whose density varies slightly: ρ(x,y)=13.4(1+0.01x-0.02y) grams per cubic centimeter for 0≤x≤10cm and 0≤y≤5cm.  Find the total mass of this sheet.

Exercise 2.  Consider a piece of material whose thickness is 0.01 meters and which is bounded by y=0, x=1, and y=2x in the first quadrant of the xy-plane.  Suppose the density is described by ρ(x,y)=6x+6y+6 kilograms per cubic meter.  Find the total mass of this lamina.

Next we want to find the center of mass ((OverBar[x]),(OverBar[y])).  Imagine taking a piece of cardboard and trying to balance it on your fingertip.  That balance point marks the center of mass.  The idea is that we need to balance the weight on each side of your fingertip, recalling that the lever principle:  the force equals the mass times the distance from the pivot point.  The pivot point here will be the center of mass ((OverBar[x]),(OverBar[y])).  

If we consider balancing in the x-direction, we want ∫∫(x-(OverBar[x]))ρ(x,y)dxdy=0, which says that the mass at the points with x>(OverBar[x]) (where x-(OverBar[x])>0) equals the mass at points with x<(OverBar[x]) (where x-(OverBar[x])<0).  We can do the same reasoning in the y-direction to get ∫∫(y-(OverBar[y]))ρ(x,y)dxdy=0.
  
Expanding the integrals yields ∫∫x ρ(x,y)dxdy - (OverBar[x])∫∫ρ(x,y)dxdy = 0 and ∫∫y ρ(x,y)dxdy - (OverBar[y])∫∫ρ(x,y)dxdy = 0.

Solving for the center of mass coordinates yields   (OverBar[x])=(1/(∫∫ρ(x,y)dxdy))∫∫x ρ(x,y)dxdy  and   (OverBar[y])=(1/(∫∫ρ(x,y)dxdy))∫∫y ρ(x,y)dxdy.

Exercise 3.  Find the center of mass for the rectangular sheet of metal in Exercise 1.

Exercise 4.  Find the center of mass for the triangular lamina in Exercise 2.

Exercise 5. Find the center of mass of a lamina bounded by y=x^2 and y=Sqrt[x] in the first quadrant of the xy-plane if the thickness is W and the density is ρ(x,y)=a*x for a>0.  Does the center of mass depend on either the thickness W of the lamina or the value of a in the density function?  Explain your answer.