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I needed help with this question:

The diameter and altitude of a can in the shape of a right circular cylinder are measured 4cm and 6cm respectively. The possible error in each measurement is 0.1cm.Find approximately the maximum possible error in values computer for the volume.

My try:

$V=24π$

$$ V= \pi r^2 h $$

$$ dV= \frac {dV}{dr} dr + \frac {dV}{dh} dh$$ $$ = (2\pi rh) dr + (\pi r^2)dh$$

$$ \frac {dV}{V} = 2(dr/r)+ (dh/h) = 2(0.1)+ (0.1) = 0.3 $$

$dV=0.3X 24π=7.2π cc$

But the answer is $1.6πcc$ . What I did wrong?

John Alexiou
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Aladdin
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1 Answers1

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You made $2$ basic mistakes of assuming the measurement errors are relative to the measurement values themselves instead of being absolute, plus you are given a possible diameter measurement error that you treated as being of the radius instead.

For the appropriate way to solve this problem, you correctly got to

$$dV = (2\pi rh) dr + (\pi r^2)dh \tag{1}\label{eq1}$$

You're given the diameter is $4$, so the radius is $r = 2$, and the height is $h = 6$. Also, $dh = 0.1$ at the most, and the largest error in the diameter is $0.1$, so the maximum error in the radius is half of this at $0.05$, i.e., $dr = 0.05$. Substitute these values into \eqref{eq1} to get the approximately largest error in the volume measurement is

$$dV = 24\pi(0.05) + 4\pi(0.1) = 1.6\pi\text{ cc} \tag{2}\label{eq2}$$

John Omielan
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