How Does Surface Area Error Affect Volume Error in a Sphere?

[Procrastinate]

There is a 0.8% error in the surface area of a sphere. Find the resulting % error in the volume of the sphere.

I am having trouble solving this and I was wondering if someone could give me a hint?

I currently have reached a dead end with:

2 delta r
-------- = %SA
r

However, I don't have any known r values to work with...

 

[Cyosis]

S= 4 \pi r^2 , \; V=\frac{4}{3} \pi r^2. Solve r in terms of S and plug it into the formula for the volume.

 

[Mark44]

You will also need differentials dr, dS, and dV, since you want dS/S. Recall that dS = dS/dr * dr.

 

[Procrastinate]

My final answer was 2.4%; was that right?

 

[HallsofIvy]

There is an engineer's rule of thumb, derived from the differential of Mark44, that "if measurements add, their errors add, if measurements multiply, their percentage errors add. Since here, you have a percentage error of .008 in r, and V= (4/3)\pi r^3 involves multiplying r three times, yes, the percentage error in V is 3(.008)= 0.024 or 2.4%.

 

[Redbelly98]

Recall that 0.8% is the error in S, not r. So 2.4% is incorrect.

Procrastinate, try finding the % error in r first, what do you get for that?