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.
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
Is this right? If not can you please give me the solution? I spent a while on this.
v = πr^2h
dv = 2πrh dr + πr^2 dh
h is assumed exact, so dh=0.
dv = 2πrh dr
dr = 0.2, so
dv = 2π(4)(12)(.2) = 19.2π
and that's ±
Your estimate assumes that the height was also measured with an error. The problem does not say that. If that is stipulated, then your answer is correct.
To find the resulting possible error in the volume of the cylinder, we need to calculate the maximum and minimum volumes that can result from the given measurement with its possible error.
Let's start by calculating the maximum volume by considering the maximum possible radius and height:
Maximum radius = given radius + possible error = 4 inches + 0.2 inches = 4.2 inches
Maximum height = given height = 12 inches
Using these values, we can calculate the maximum volume:
Vmax = π * (Maximum radius)^2 * Maximum height
= π * (4.2 in)^2 * 12 in
≈ 3.14 * 17.64 in^2 * 12 in
≈ 3.14 * 211.68 in^3
≈ 665.856 in^3
Next, let's calculate the minimum volume by considering the minimum possible radius and height:
Minimum radius = given radius - possible error = 4 inches - 0.2 inches = 3.8 inches
Minimum height = given height = 12 inches
Using these values, we can calculate the minimum volume:
Vmin = π * (Minimum radius)^2 * Minimum height
= π * (3.8 in)^2 * 12 in
≈ 3.14 * 14.44 in^2 * 12 in
≈ 3.14 * 173.28 in^3
≈ 544.5312 in^3
Therefore, the resulting possible error in the volume of the cylinder is the difference between Vmax and Vmin:
Error in volume = Vmax - Vmin
= 665.856 in^3 - 544.5312 in^3
≈ 121.3248 in^3
Hence, the resulting possible error in the volume of the cylinder is approximately 121.3248 in^3.