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I have a question about the significant figures topic:

Problem: Find the volume of a cylinder having a diameter measured as 2.3 ft and length measured as 8.25 ft.

Solution: V = π(D^2 /4)h = π((2.3)^2)(8.25)/4 = 34.276739 cu ft = 34 cu ft

Question:
In my workbook is explained that the final answer is rounded to two significant figures because the quantity 2.3 has two significant figures.
Why will I have to round to two significant figures if the quantity 8.25 have three significant figures?

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    $\begingroup$ Writing in all capitals doesn't make it more significant. $\endgroup$ – Nick Alexeev Oct 23 '17 at 14:40
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You can only assume the accuracy of the least precise component of the input. so if you have elements with only 2 SF then you can not assert a higher level of precision.

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Selter's answer follows the book's approach. A proper analysis, however, should go a little deeper.
First, we stick with the convention that the uncertainty in any provided value is +/- half the least sig digit, i.e. 2.3 means 2.3 +/- 0.05 . Next, we notice that the diameter enters into the formula squared and the height linearly. The standard approach is to look at the partial derivatives and do variational analysis. Following the mechanisms presented, e.g., in Bevington Chapter 3, we will get ("Var" is variance, or sigma^2)

$ \frac{Var(volume)}{Vol^2} = \frac{\pi}{4} *( 2*\frac{Var(diameter)}{Diameter^2} + \frac{Var(height) }{height^2} ) $

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