Write the function y = kx^2 to describe the parabola. Find k by using the information about the height and radius of the parabolic section. Rewrite the function with this new value substituted in for k.
Example:
Find the center of mass of a uniform bowl cut in a parabolic section. The bowl's height is 0.1 m and its radius is 0.1 m.
(0.1, 0.1) is a point on the bowl. Plug in 0.1 for x and 0.1 for y to solve for k.
0.1 = k(0.1)^2
0.1 = k*.01
k = 10
y = 10x^2
Change y(x) to x(y) by rearranging the equation until x is by itself on the left side. This is because you are integrating over y, in the vertical direction, so you need to know the horizontal dimensions of each slice in terms of x. This is the same as dA, the derivative with respect to area.
Example:
y = 10x^2
0.1y = x^2
x = + and -sqrt(0.1y)
Because the equation splits into two identical parts, rewrite it as:
x = 2*sqrt(0.1y)
dA = 2*sqrt(0.1y) dy
Set up the integral for the y-coordinate. Because you took slices of area with a uniform density, the dm can be rewritten as D*dA, where D is the density, and dA = 2*sqrt(0.1y) dy.
Example:
ycm = (1 / M) S y dm
ycm = (1 / M) 2D*S y * sqrt(0.1y) dy
The limits of integration are 0 and 0.1 (the height of the section).
Rewrite M, the mass, as an integral, using the same information as for the previous integral, but leaving out the extra *y.
Example:
M = 2D*S sqrt(0.1y) dy
The limits of integration are 0 and 0.1 (the height of the section).
Write a ratio of the two integrals to take into account the 1 / M. Solve by integrating.
Example:
ycm = 2D*S y * sqrt(0.1y) dy / 2D*S sqrt(0.1y) dy
sqrt(0.1) is a constant and can be brought outside the integral, so it cancels, just like the 2 and the D.
y * sqrt(y) = y^1 * y^0.5 = y^1.5
ycm = S y^1.5 dy / S y^0.5 dy
ycm = 0.4y^2.5 / (2/3)y^1.5 = 0.6y
Evaluate from 0 to 0.1:
ycm = 0.06 - 0 = 0.06