# Calculus - 14

The radius of sphere is measured as 10cm with an error 0.06cm. Find approximate error in the volume.

(A) $24\pi$

(B) $72\pi$

(C) $36\pi$

(D) $12\pi$

EDIT: Marks for all for now

edited May 31

Sir, here u didn't mention...whether we have to find approximation error for area or value or some other....

For volume..it is 24 pie
answered May 31 by (640 points)
I dont know about the question but can you explain it to me if the question is asked for the volume as you have mentioned some answer 24pi.
Volume = 4/3 pi R^3
R =10
r = 0.06
Delta v = dv/dr * r
Solve it
4/3 * pi * 3*10*10*0.06
24 pi

But bro here they didn't mention whether area or volume or some other...But method is this
updated the question.
This is some generalized method to solve such type of question or this is your approach?
U can use it for almost every problem....
Okay. Got it. Thanks
Marks will be given to all in this question or not?
Here not considered for volume or for area so I think question should be wrong
answered May 31 by (17,270 points)
Yes bro...it is wrong.
it was not mentioned volume
answered May 31 by (490 points)
$\\ V=4/3*\pi*R^3\\ \\ \frac{\partial V}{\partial R}=4/3*\pi *3*R^2\\ \\ \partial V=4/3*\pi *3*10*10*\partial R\\ \\ \partial V=4/3*\pi*3*10*10*0.06=24\pi$