if 3.8 <= r <= 4.2,
π*3.8^2*12 <= v <= π*4.2^2*12
544.32 <= v <= 665.01
...
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.
4 answers
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?
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.
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?