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I was trying to solve this problem: A cylindrical roller is exactly 12 inches long and its diameter is measured as 6 ± 0.05 inches. Approximate its volume with an estimate for the error.

I think I could use the differential of y (dy) equation and get f ‘(x), which I assume is the volume in question, however I have no change in volume to equate dy’s equation to. Am I wrong in my assertion? Do I need to do linear approximation for this problem?

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    $\begingroup$ Why don't just compute the volume? $\endgroup$
    – user
    Dec 4, 2023 at 8:21
  • $\begingroup$ Conflict between the original poster's instincts, which seem good to me, and the bizarre nature of the problem. As has already been commented, there is no reason to use calculus here, since the formula is readily available, and you can determine the $~\pm~$ error the same way. On the other hand, the original poster is probably wondering why they would be assigned such a problem in a calculus class (assuming that that is what is happening), when calculus is not needed. What is the point? My only explanation here is that sometimes problem composers do daffy things. $\endgroup$ Dec 4, 2023 at 8:30

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Let's see

$$V=\frac{1}{4} \pi d^2 h$$ $$V=\frac{1}{4} \pi (6\pm0.05)^2(12) $$ $$V=(6\pm0.05)^2 3\pi$$ I think what you want is the volume as is $V \pm dV$, so, that would be something like

$$V=(36\pm 2\times0.05)\times3\pi$$ $$V=(108\pm 0.3)\pi$$ In physics a usually the $\Delta x^2$ term is ignored, as it is beyond the significant figure.

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  • $\begingroup$ How did you do the solutions under V +/- dV? $\endgroup$ Dec 4, 2023 at 9:12

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