How to Calculate the Correct Percentage Uncertainty in Metal Pipe Volume?

[harvey1999]

A manufacturer needs to determine the volume of metal used to produce a metal pipe of the following dimensions:
Length, L 40 + 1 m
External diameter, D 12.0 + 0.2 cm
Internal diameter, d 10.0 + 0.2 cm
Estimate the greatest possible percentage error of the volume.

the correct ans is 22.5%.

but no matter how many times I've tried, my ans is always 34%. Help pls thank you! :)

 

[Qwertywerty]

Could you show us your working?

 

[harvey1999]

Could you show us your working?

V=(pi)(R^2)(L) - (pi)(r^2)(L)

delta(R)=(1/2)(0.2)=0.1cm
delta(r)=(1/2)(0.2)=0.1cm

delta(V) = delta(pi*R^2*L) + delta(pi*r^2*L)

Let x=(pi)(R^2)(L)

delta(x)/x = 2(0.1/6) + 1/40= 7/120

delta(x)=7/120 * (pi*R^2*L)=7/120 * 14 400 pi = 840pi

Let y=pi*r^2*L

delta(y)/y = 2(0.1/5) + 1/40=13/200

delta(y)=13/200 * pi*r^2*L=13/200 * 10 000pi= 650pi

delta(V) = delta(x) + delta(y)
=1490pi
delta(V)/V *100% = 1490pi/4400pi *100 = 34%

fyi 4400pi is the value of V
thank you

 

[Drakkith]

Hi harvey,

I've moved your thread to the homework forums, but in the future please post all homework questions in the appropriate homework forum and use the template provided.

 

[haruspex]

You have counted the 1/40 twice over. There may be other errors.
Safest way is just to compute the maximum possible volume.

 

[harvey1999]

You have counted the 1/40 twice over. There may be other errors.
Safest way is just to compute the maximum possible volume.

i'm sorry, what do you mean by counting 1/40 twice? is my method not the correct approach anyway?

 

[haruspex]

i'm sorry, what do you mean by counting 1/40 twice? is my method not the correct approach anyway?

You counted it here:

delta(x)/x = 2(0.1/6) + 1/40= 7/120

and here:

delta(y)/y = 2(0.1/5) + 1/40=13/200

Your problem starts with this line:

delta(V) = delta(pi*R^2*L) + delta(pi*r^2*L)

Let's back up one step:
$\Delta V = \Delta (\pi*R^2*L-\pi*r^2*L)$
You cannot split that as $\Delta (\pi*R^2*L)+\Delta(\pi*r^2*L)$ because the two terms are not independent.
If the actual length is L-ΔL on the inner radius it will be L-ΔL on the outer radius. Your analysis exaggerates the range by allowing a volume like
$\pi((R-\Delta R)^2(L-\Delta L)-(r+\Delta r)^2(L+\Delta L))$

 

[harvey1999]

You counted it here:

and here:

Your problem starts with this line:

Let's back up one step:
$\Delta V = \Delta (\pi*R^2*L-\pi*r^2*L)$
You cannot split that as $\Delta (\pi*R^2*L)+\Delta(\pi*r^2*L)$ because the two terms are not independent.
If the actual length is L-ΔL on the inner radius it will be L-ΔL on the outer radius. Your analysis exaggerates the range by allowing a volume like
$\pi((R-\Delta R)^2(L-\Delta L)-(r+\Delta r)^2(L+\Delta L))$

however, it is true that if R=A+B,
then delta(R) = delta(A) + delta(B).
i don't get what you mean by independent though. I'm sorry but i just want to fully understand this, thank you!