Question

The radius of a sphere shrinks from $10$cm to $9.8$cm. Find approximate error in its volume.

Answer 1

$\Delta V=\frac{dV}{dr}\Delta r$

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dr}=4\pi r^2$

$=4\times \pi \times \left(10{)}^2\right(9.8-10)=-80\pi c{m}^3$

approximate decrease in volume is $80\pi c{m}^3$

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