If the radius of a sphere is measured as 7 m with an error of 0.02m, find the approximate error in calculating its volume.
Given the radius of a sphere is measured as 7 m with an error of 0.02 m.
Let x be the radius of the sphere and Δx be the error in measuring the value of x.
Hence, we have x = 7 and Δx = 0.02
The volume of a sphere of radius x is given by
![]()
On differentiating V with respect to x, we get
![]()
![]()
We know![]()
![]()
![]()
When x = 7, we have
.
![]()
![]()
Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as
![]()
Here,
and Δx = 0.02
⇒ ΔV = (196π)(0.02)
∴ ΔV = 3.92π
Thus, the approximate error in calculating the volume of the sphere is 3.92π m3.
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.