## 1.8 Another Example

Let’s compute the volume of a sphere of radius 4.58 inches. This time, we will use 3-place logarithms (i.e., logarithms to three decimal places).

The formula for the volume of a sphere is $V = \frac43 \pi r^3$ So, using the rules we have just learned, $\log V = \log 4 - \log 3 + \log \pi + \log 4.58^3 = \log 4 - \log 3 + \log \pi + 3\times \log 4.58 \\ = 0.602 - 0.477 + 0.497 + 3\times( 0.661 ) \\ = 2.605 = 0.605 + 2$ The number with logarithm 0.605 is 4.027. The “2” tells us to multiply by 100. So, this gives the final value of the volume, $$V$$ = 402.7 cubic inches. (Again, check with computer: 402.4250839 cubic inches. We have about a 0.07% error in our answer, using 3-place logarithms.) Of course, if our knowledge of the radius is only to three digits as given above, then no more than the first three digits in the answer should be of any significance. In real life, of course, one must always perform an appropriate error analysis of a result, but that is beyond the scope of our present tutorial.