Question

If there is an error in measuring its radius by Δr, what is the approximate error in the volume of a sphere?

Answer 1

Answer:

approximate error in the volume of a sphere = 4π r²

Step-by-step explanation:

Volume of a sphere = (4/3) π r³

Where r = Radius

approximate error in the volume of a sphere = Δr

Δr = dV/dr

Δr = d((4/3) π r³ )/dr

=> Δr  = (4/3) π 3r²

=> Δr  = 4π r²

approximate error in the volume of a sphere = 4π r²

Answer 2

Answer:

4πr²Δr

Step-by-step explanation:

Let us assume that the actual radius of the sphere is r and the radius becomes R after making the error of Δr.

So, we can write R=r+Δr ...... (1)

Now, the actual volume of the sphere is v= 4πr³/3 and the erroneous volume V=4πR³/3.

Hence, the error in measuring the volume of the sphere is

Δv=V-v

⇒Δv=4πR³/3 -4πr³/3

⇒Δv=4π/3(R³-r³)

⇒Δv=4π/3{(r+Δr)³-r³} (From equation (1))

⇒Δv=4π/3{(r³+3r²Δr)-r³}  

Here we apply the formula, (a+b)³=a³+3a²b+3ab²+b³, and ignoring the higher-order terms of Δr, since Δr is very small.

⇒Δv=4π/3(3r²Δr)

⇒Δv= 4πr²Δr ... this is the approximate error in the volume of the sphere.

(Answer)