# Calculus - 14

+2 votes

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

asked May 30 1 flag
edited May 31

## 5 Answers

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

Please check this question...

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.
Thankyou brother for your answer.
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?
0 votes
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.
0 votes
it was not mentioned volume
answered May 31 by (490 points)
0 votes
this question is incomplete ,I think it should not be taken for evaluation .How some one can do a question which is not complete great imagination
answered May 31 by (3,450 points)
For this marks will be given for all.
+1 vote

$\\ V=4/3*\pi*R^3\\ \\ \frac{\partial V}{\partial R}=4/3*\pi *3*R^2\\ \\ \partial V=4/3*\pi *3*10*10*\partial R\\ \\ \partial V=4/3*\pi*3*10*10*0.06=24\pi$

Therefore Answer is A

answered Jun 3 by (10,800 points)
Answer: