# If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

## Solution :

Let r be the radius of a sphere and $$\delta$$r be the error in measuring the radius. Then, r = 9 cm and $$\delta$$r = 0.03 cm.

Let V be the volume of the sphere. Then,

V = $${4\over 3}\pi r^3$$  $$\implies$$ $$dV\over dr$$ = $$4\pi r^2$$

$$\implies$$ $$({dV\over dr})_{r = 9}$$ = $$4\pi \times 9^2$$ = $$324 \pi$$

Let $$\delta$$V be the error in V due to error in V due to error $$\delta$$r in r. Then,

$$\delta$$V = $$dV\over dr$$ $$\delta$$r  $$\implies$$  $$\delta$$V =  $$324 \pi\times 0.03$$ = $$9.72 \pi cm^3$$

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