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

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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

Mathematics
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you just take the +0.2 to the cylinder height 0.2+4, find the volume and find the volume again for 4-0.2 and find the volume again and get the difference
\[Error=\pi r^2(h+0.2) - \pi r^2(h-0.2)\]
i don't know it's calculus and i'm pretty sure it some how involves the derivative of the volume just not too sure in what way

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oh another of those sneaky questions...
indeed
Thisis prolly what it's asking the volume different between this|dw:1438420033333:dw|
yes it wants the difference between the volume when r = 4.2 and volume when r = 4
pi*4.2^2*12 - pi*4^2*12 = 665 - 603.2
\(V = \pi \ r^2 \ h\) \(dV = 2 \pi r \ dr \ h\) \(dr = ±0.2\)
oh I see then that will give +/- 60.32 cu ins
that might be how they want it done [depends on what is being studied by the OP, i suppose]; but there clearly is nothing wrong in plugging in to the volume equation itself as that will give the exact answer.....
The volume of a right circular cylinder is proportional to the square of it's radius \[V \alpha r^2\] This means any change in the radius will a change the volume by the square of the radius Suppose for the radius of 4 inches, we had a volume V1 before \[V_{1}=16k\] Now when we change the radius by plus minus 2, we get \[V_{2}=(4 \pm 0.2)^2k=(16+0.04 \pm 1.6)k = (16 \pm 1.6)k\] We can ignore the 0.04 as it is quite small Dividng V2 by V1 we get \[\frac{V_{2}}{V_{1}}=\frac{16 \pm 1.6}{16}\] Therefore \[V_{2}=(1 \pm 0.1) V_{1}\]\[V_{2}=V_{1} \pm 0.1V_{1}\]
good result
i fell asleep, sorry. but thank you for your help
its an open ended question and i need to do it using calculus, does anyone know how to do that that way?
  • phi
See Irish boy's post. \[ dV = 2 \pi r \ dr \ h \] or if we replace the infinitesimals we have \[ \Delta V= 2 \pi r h\ \Delta r \] replace the \(\Delta r\) with the uncertainty \(\pm 0.2\) and calculate the uncertainty in V
  • phi
**perhaps it is better to say \[ \Delta V \approx 2 \pi r h\ \Delta r \]

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