Question

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

Answer 1

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

Answer 2

\[Error=\pi r^2(h+0.2) - \pi r^2(h-0.2)\]

Answer 3

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

Answer 4

oh another of those sneaky questions...

Answer 5

indeed

Answer 6

Thisis prolly what it's asking the volume different between this

Answer 7

yes it wants the difference between the volume when r = 4.2 and volume when r = 4

Answer 8

pi*4.2^2*12 - pi*4^2*12 = 665 - 603.2

Answer 9

\(V = \pi \ r^2 \ h\)
\(dV = 2 \pi r \ dr \ h\)
\(dr = ±0.2\)

Answer 10

oh I see
then that will give +/- 60.32 cu ins

Answer 11

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.....

Answer 12

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
\[V_{2}=(1 \pm 0.1) V_{1}\]\[V_{2}=V_{1} \pm 0.1V_{1}\]

Answer 13

good result

Answer 14

i fell asleep, sorry. but thank you for your help

Answer 15

its an open ended question and i need to do it using calculus, does anyone know how to do that that way?

Answer 16

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

Answer 17

**perhaps it is better to say
\[ \Delta V \approx 2 \pi r h\ \Delta r \]