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I am studying maths as a hobby and am doing a chapter on calculus and small changes and errors. I am trying to understand the following problem. I cannot get the answer in the text book, which is 6%. I have not tried to calculate the area change as I obviously am making a fundamental error.

If a 2% error is made in measuring the diameter of a sphere, find approximately the resulting percentage errors in the volume and surface area.

I have said:

Let V = volume and D = diameter

$V = \frac{4 \pi r^3}{3} = \frac{\pi D^3}{6}$

Now if $\delta V, \delta D$ represent small changes in V and D respectively:

$\frac{\delta V}{\delta D} \approx \frac{dV}{dD}$ and $\frac{dV}{dD} = \frac{\pi D^2}{2}$ so

$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100}.\frac{\pi D^2}{2} = \frac{\pi D^2}{100}$

Percentage error = $\frac{\delta V}{V} = \frac{\pi D^2}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^2}{100}.\frac{6}{\pi D^3} = \frac{6}{100D}$ but the answer is actually 6%.

Where have I gone wrong.

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$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100} \times D \times \frac{\pi D^2}{2} = \frac{\pi D^3}{100}$

Please note $\delta D = \frac{2}{100} \times D $ (New $D$ minus old $D$).

$\frac{\delta V}{V} = \frac{\pi D^3}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^3}{100}.\frac{6}{\pi D^3} = \frac{6}{100}$

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To answer the question you can either use calculus via Math Lover's method, or use knowledge of scale factors. Diameter is a length, so the length scale factor is either $0.98$ or $1.02$.

So the area scale factor is $0.98^2$ or $1.02^2$ which is $ \sim 0.96$ and $1.04 $ respectively: either way an error of 4%.

The volume scale factor is $0.98^3$ or $1.02^3$ which is $ \sim 0.94$ and $1.06 $ respectively: either way an error of $ \sim 6$%.

Here I am using the well-known result:

For small positive $x, \quad (1+x)^n \sim 1+nx$, which comes from the Binomial expansion of $(1+x)^n$.

You could argue that this answer is unsatisfactory, because proving that area scale factor = (length scale factor)$^2$ and volume scale factor = (length scale factor)$^3$ probably requires calculus...

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