31. A set of plastic spheres are to be made with a diameter of 1cm. If the manufacturing process is accurate to 1mm, what is the propagated error in volume of the spheres?
The volume \(V\) of a sphere is given by the formula: \[V = \frac{4}{3}\pi r^3\]
The differential \(dV\) of the volume \(V\) with respect to the radius \(r\) is given by the derivative of the volume formula, which is: \[dV = 4\pi r^2 dr\]
We were gvien that the diameter was 1 cm which gives us a radius, \(r\) of 0.5 cm and the error in the diameter is 0.1 cm which gives us an error of the radius, \(dr\), of 0.05 cm. Substituting those values into the formula and calculating it using R, gives us 0.157 cubic centimeters:
# Define the radius and the error in the radius
r <- 0.5 # cm
dr <- 0.05 # cm
# Compute the propagated error in the volume
dV <- 4 * pi * r^2 * dr
# Print the propagated error
print(dV)## [1] 0.1570796