# Percentage change in volume of sphere calculation when there is an error in calculating diameter.

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.

## 2 Answers

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

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

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

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...