A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is

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KollegeApply · Accepted Answer

80000 $\pi$ mm3 Let x be the radius of the sphere and y be its volume. x = 100, x + ∆ x = 98 \Rightarrow ∆ x = - 2 y = $\frac{4}{3}$$\pi$ x^3 \Rightarrow dy/dx = 4$\pi$ x^2 \Rightarrow \left( dy/dx \right)_{x = 100} = 40000$\pi$ \therefore ∆ y = dy = dy/dxdx = 40000$\pi$ $\times$ \left( - 2 \right) = - 80000$\pi$ \text { Hence, the decrease in the volume of the sphere is } 80000$\pi$ \text{mm}^ 3.

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