Percent Uncertainty in Spherical Beach Ball Volume

[George3]

Homework Statement

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 0.84 + or - 0.06m?
Answer must have 2 sig figs.

Homework Equations

The Attempt at a Solution

I found the volume of the sphere with .84 radius which would be 3.0 m^3 and then I don't know what to do next.

 

[kuruman]

If y is related to x by a power law,

y = axⁿ, then the fractional uncertainty is

$$\frac{\Delta y}{y}=n\frac{\Delta x}{x}$$

 

[Fightfish]

Consider what happens to the uncertainty in a variable when you cube the variable (should be a standard treatment in the manipulation of uncertainties in variables)

 

[George3]

Consider what happens to the uncertainty in a variable when you cube the variable (should be a standard treatment in the manipulation of uncertainties in variables)

So are you saying I should cube the .06 and then divide that by the 3.0m^3?

 

[TVP45]

Could you show your equations and attempts, please?

 

[George3]

(.84)^3(pi)(4/3) = 2.5 m^3

.06^3 = 2.2x10^-4

2.2x10^-4 / 2.5 = 8.6x10^5

(8.6x10^5) x 100% = .0086%

 

[kuruman]

Not how it works. Here, n = 3. Therefore, the fractional uncertainty is

$$\frac{\Delta V}{V} = 3\times\frac{0.06}{0.84}$$

Multiply by 100 and you have percent uncertainty.