Question

The relative error in the determination of the surface area of a sphere is$$\alpha$$. Then the relative error in the determination of volume is:
A- $${\displaystyle \frac{2}{3}}\alpha$$
B- $${\displaystyle \frac{5}{2}}\alpha$$
C- $${\displaystyle \frac{3}{2}}\alpha$$
D- $$\alpha$$

Answer 1

Hint: We are given the relative error in determination of surface area of sphere. Relative error is the ratio of change in area to the original area. We can make use of logarithm to arrive at a meaningful solution. We can Use the expression for the area of sphere to determine the fractional change in the radius of the sphere and then use the expression for the fractional change in the radius of sphere to determine the fractional change or relative error in the volume of the sphere

Complete step by step answer:Use the expression for the surface area and volume of the sphere. And apply logarithm to determine the relative error in the volume.
Write the expression for the surface area $$S$$of the sphere,
$$S=4\pi r^2$$
Here, $$r$$ is the radius of the sphere.

Apply natural logarithm both side
$$\begin{array}{rl}& \ln S=\ln 4\pi r^2\\ & =\ln 4\pi +2\ln r\end{array}$$

Differentiate Above equation with respect to $$r$$
$$\begin{array}{rl}& {\displaystyle \frac{1}{S}}{\displaystyle \frac{dS}{dr}}=0+2{\displaystyle \frac{1}{r}}\\ & \end{array}$$
Rearrange
$${\displaystyle \frac{dr}{r}}={\displaystyle \frac{dS}{2S}}$$

Substitute $$\alpha$$ for $${\displaystyle \frac{dS}{S}}$$
$${\displaystyle \frac{dr}{r}}={\displaystyle \frac{\alpha }{2}}$$ --- (1)

Write down the expression for the volume $$V$$ of the sphere
$$V={\displaystyle \frac{4}{3}}\pi r^3$$

Apply logarithm both sides
$$\begin{array}{rl}& \ln V=\ln {\displaystyle \frac{4}{3}}\pi r^3\\ & =\ln {\displaystyle \frac{4}{3}}\pi +\ln r^3\\ & =\ln {\displaystyle \frac{4}{3}}\pi +3\ln r\end{array}$$

Differentiate above equation with respect to $$r$$
$${\displaystyle \frac{dV}{Vdr}}=0+{\displaystyle \frac{3}{r}}$$

Rearrange
$${\displaystyle \frac{dV}{V}}=3{\displaystyle \frac{dr}{r}}$$

Substitute $${\displaystyle \frac{\alpha }{2}}$$ for $${\displaystyle \frac{dr}{r}}$$
$${\displaystyle \frac{dV}{V}}=3{\displaystyle \frac{\alpha }{2}}$$

Therefore, the fractional change of the volume of the sphere is $$3{\displaystyle \frac{\alpha }{2}}$$ . Hence, the correct choice is C.

Note:when we are concerned with the precision, we use the concept of relative error. Since, it is a ratio it is a unitless quantity. It is mostly used to compare things that are measured in different units. Although it is a measure of precision, the same term can also be used to describe accuracy.