If there is an error of 0.1% in the measurement of the radius of a sphere,
Question:

If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

Solution:

Let x be the radius and y be the volume of the sphere.

$y=\frac{4}{3} \pi x^{3}$

Let $\Delta x$ be the error in the radius and $\Delta y$ be the error in the volume.

Then, $\frac{\Delta x}{x} \times 100=0.1$

$\Rightarrow \frac{d x}{x}=\frac{1}{1000}$

Now, $y=\frac{4}{3} \pi x^{3}$

$\Rightarrow \frac{d y}{d x}=4 \pi \mathrm{x}^{2}$

$\Rightarrow \mathrm{dy}=4 \pi \mathrm{x}^{2} \mathrm{dx}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\frac{4 \pi \mathrm{x}^{2} \mathrm{dx}}{\frac{4}{3} \pi \mathrm{x}^{3}}=\frac{3}{\mathrm{x}} \mathrm{dx}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\frac{3}{1000}$

$\Rightarrow \frac{\Delta \mathrm{y}}{\mathrm{y}} \times 100=0.3$

Hence, the percentage error in the calculation of the volume of the sphere is 0.3.

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