The radius of a 12 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.2 inch. What is the resulting possible error in the volume of the cylinder? Include units in your answer.
if 3.8 <= r <= 4.2,
π*3.8^2*12 <= v <= π*4.2^2*12
544.32 <= v <= 665.01
...
r = 4 in.
h = 12 in.
V = pi*r^2 * h = 3.14*4^2 * 12 = 602.88
m^3.
r = 4.2 in.
h = 12.2 in.
V = 3.14*4.2^2 * 12.2 = 675.75 in^3
675.75-602.88 =72.87 in^3?
is that right?
Hmmm
3.14*4.2^2 * 12.2 = 665.75
other than that, your logic is sound.
Also, note that the error using 4.0-0.2 in for the low-side value differs by a smaller amount.
So my updated answer would be that the possible error is the volume of the cylinder is 63.87 in^3?
To find the resulting possible error in the volume of the cylinder, we need to calculate the maximum and minimum possible values for the volume.
The formula for the volume of a cylinder is V = πr^2h, where r is the radius and h is the height of the cylinder.
Given that the measured radius is 4 inches with a possible error of ±0.2 inches, we have two scenarios:
1. Maximum radius scenario: r = 4 + 0.2 = 4.2 inches
2. Minimum radius scenario: r = 4 - 0.2 = 3.8 inches
Now, let's assume the height of the cylinder is H inches (not given in the question). The volume of the cylinder in these two scenarios would be:
1. Maximum volume scenario:
V_max = π(4.2)^2H = 55.42H cubic inches
2. Minimum volume scenario:
V_min = π(3.8)^2H = 45.52H cubic inches
To find the resulting possible error in the volume, we subtract the minimum volume from the maximum volume:
Possible error in volume = V_max - V_min
= (55.42H - 45.52H) cubic inches
= 9.9H cubic inches
So, the resulting possible error in the volume of the cylinder is 9.9H cubic inches. The units depend on the height (H) of the cylinder, which was not given in the question.