If the radius of a sphere is measured as 9 te | vedclass

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Question

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Given $r = 9 	ext{ cm}$ and $\Delta r = 0.03 	ext{ cm}$

Vedclass · Accepted Answer

$9.72$

Vedclass · Answer

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Given $r = 9 	ext{ cm}$ and $\Delta r = 0.03 	ext{ cm}$. The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$. The approximate error in volume $dV$ is given by $dV = \left( \frac{dV}{dr} ight) \Delta r$. Substituting the values,$dV = (4 \pi r^2) \Delta r = 4 \pi (9)^2 (0.03)$. $dV = 4 \pi (81) (0.03) = 324 \pi (0.03) = 9.72 \pi 	ext{ cm}^3$. Thus,the approximate error in calculating the volume is $9.72 \pi 	ext{ cm}^3$.

If the radius of a sphere is measured as $9 \text{ cm}$ with an error of $0.03 \text{ cm}$,then find the approximate error in calculating its volume. (in $\pi \text{ cm}^3$)

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