In this note we will try to find a point P on which a thin plate of any given shape balances horizontally. This point is called the center of mass (or center of gravity) of the plate.

We first consider the simpler situation. Suppose that the rod lies along the x-axis with m1 at x1 and m2 at x2 and the center of mass at x. According to the Law of Lever m1(x−x1)=m2(x2−x) or x=m1+m2m1x1+m2x2.
The numbers m1x1 and m2x2 are called the moments of the masses m1 and m2 (with respect to the origin). Above equation says that the center of mass is obtained by adding the moments of masses and dividing by total mass.
In general, if we have a system of n particles with masses m1,m2,…,mn located at the points x1,x2,…,xn on the x-axis, it can be shown similarly that the center of mass of the system is located at x=∑i=1nmi∑i=1nmixi.
Sum of the individual moments M=∑i=1nmixi is called the moment of the system with respect to the origin and m=∑i=1nmi is the total mass of the system.
Now we consider a system of particles with masses m1,m2,…,mn located at the points (x1,y1),(x2,y2),…,(xn,yn) in the xy-plane. By analogy with the one-dimensional case, we define the moment of the system about the y-axis to be My=∑i=1nmixi and the moment of the system about x-axis as Mx=∑i=1nmiyi.
My measures the tendency of the system to rotate about the y-axis and Mx measures the tendency to rotate about the x-axis.
As in the one-dimensional case, the coordinates of the center of mass are given in terms of the moments by the formulas x=mMy and y=mMx where m=∑i=1nmi is the total mass. Since mx=My and my=Mx, the center of mass (x,y) is the point where a single particle of mass m would have the same moments as the system.
Example 1. Find the moments and center of mass of the system of objects that have masses 5, 7, and 8 at the points (2,1), (3,-5), and (-4,4).
My=5⋅2+7⋅3+8⋅(−4)=−1.
Mx=5⋅1+7⋅(−5)+8⋅4=2.
Since m=5+7+8=20 then x=mMy=−201 and y=mMx=202=101.
So, center of mass is located at point (−201,101).
Next we consider a flat plate (called a lamina) with uniform density ρ that occupies a region R under continuous curve y=f(x), above x-axis and between lines x=a and x=b. We
wish to find the center of mass of the plate, which is called the centroid of R. In doing this we use the following simple fact: centroid of a rectangle is its center.
As always we divide the interval [a,b] into n subintervals with endpoints x0,x1,…,xn and equal width Δx. We choose the sample point xi⋆ to be the midpoint xi of the i-th subinterval, that is, xi=21(xi−1+xi).
The centroid of the i-th approximating rectangle Ri is its center Ci=(xi,21f(xi)). Its area is f(xi)Δx, so its mass is ρf(xi)Δx.
The moment of Ri about the y-axis is the product of its mass and the distance from Ci to the y-axis, which is xi. Thus, My(Ri)=(ρf(xi)Δx)xi=ρxif(xi)Δx. Adding these moments, we obtain the moment of the polygonal approximation to R, and then by taking the limit as n→∞ we obtain the moment of R itself about the y-axis: My=limn→∞∑i=1nρxif(xi)Δx=ρ∫abxf(x)dx.
Similarly we compute the moment of Ri about the x-axis as the product of its mass and the distance from Ci to the x-axis: Mx(Ri)=(ρf(xi)Δx)21f(xi)=21ρ(f(xi))2Δx.
Again we add these moments and take the limit to obtain the moment of R about the x-axis:
Mx=limn→∞∑i=1n21ρ(f(xi))2Δx=ρ∫ab21(f(x))2dx.
Just as for systems of particles, the center of mass of the plate is defined so that mx=My and my=Mx. But the mass of the plate is the product of its density and its area: m=ρA=ρ∫abf(x)dx.
Therefore,
x=mMy=ρ∫abf(x)dxρ∫abxf(x)dx=∫abf(x)dx∫abxf(x)dx and y=mMx=ρ∫abf(x)dxρ∫ab21(f(x))2dx=∫abf(x)dx∫ab21(f(x))2dx.
Note, that center of mass doesn't depend on density.
The center of mass of the plate (or the centroid of R) is located at the point (x,y), where
x=A1∫abxf(x)dx=∫abf(x)dx∫abxf(x)dx
y=A1∫ab21(f(x))2dx=∫abf(x)dx∫ab21(f(x))2dx.
Example 2. Find the center of mass of region bounded by y=4−x2 for x≥0.
First let's find area: A=∫02(4−x2)dx=(4x−31x3)∣02=4⋅2−31⋅8=316.
Now, y=A1∫0221(4−x2)2dx=323∫02(16−8x2+x4)dx=
=323(16x−38x3+51x5)∣02=323⋅15256=58.
x=A1∫02x(4−x2)dx=163∫02(4x−x3)dx=
=163(2x2−41x4)∣02=163⋅4=43.
So, center of mass is (43,58).
In a similar way if region R is bounded by curves f(x) and g(x) on interval [a,b] and f(x)≥g(x) for a≤x≤b then
The center of mass of the plate (or the centroid of R) is located at the point (x,y), where
x=A1∫abx(f(x)−g(x))dx=∫ab(f(x)−g(x))dx∫abx(f(x)−g(x))dx
y=A1∫ab21((f(x))2−(g(x))2)dx=∫ab(f(x)−g(x))dx∫ab21((f(x))2−(g(x))2)dx.
Example 3. Find center of mass of the region bounded by y=x2 and y=2−x2.
First we sketch graph and see that bounds of integration are -1 and 1.
Next, since region is symmetric about x-axis then x=0.
A=∫−11(2−x2−x2)dx=(2x−32x3)∣−11=38.
y=A1∫−1121((2−x2)2−(x2)2)dx=163∫−11(4−4x2)dx=
=163(4x−31x3)∣−11=163⋅316.
So, center of mass is (0,1). Alternatively we could notice that region is symmetric about line y=1, so y=1.