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Homework Help: Percent Uncertainty of a spherical beach ball

  1. Aug 19, 2004 #1
    Question is : What is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 3.86 ± 0.08 m ?

    The answer is 6 % , but Im not getting that, Im pretty sure im on the right path, I found the volume of the ball using V = (4 x 3.14 x r^3) / 3

    Btw, Im pretty sure you need to take into account the ± .08m for 3.86 when you find radius. I got 3.94 and 3.78 Radius's. But I still dont see how I can get 6 %. Heres the percent uncertainty formula. (uncertainty)/(value) x 100

    Maybe Im on the wrong track, please help me through this problem, I am new to Physics and a little rusty on my math, so any pointers are greatly appreciated.

  2. jcsd
  3. Aug 19, 2004 #2
    how much calculus have u done? cuz what i will be posting further requires calc knowledge ...

    V = (4/3)*pi*r^3
    dV/dr = (4/3)*pi*3*r^2 = 4*pi*r^2

    let delta_V (the uncertainty in volume) and delta_r (the uncertainty in radius) ..
    then its easy to show that.
    delta_V = (dV/dr) * delta_r
    so ,
    delta_V = 4*pi*r^2*delta_r

    Now as u said,
    percentage error = (uncertainty)/(value) * 100
    percentage error
    = delta_V / V * 100
    = (4*pi*r^2*delta_r)/((4/3)*pi*r^3) * 100
    = (3/r)*delta_r*100

    put r = 3.86 and delta_r = 0.08
    and u will get 6.22%

    -- AI
  4. Aug 20, 2004 #3
    Here's a method of solving the question without calculus.

    V= (4pi/3)* r^3
    that part is easily calculated to give 240.91 m^3
    For the calculation of error use the following.
    1)Multiplication by a constant does not change the PERCENT ERROR.
    2)The PERCENT ERROR in a value raised to a constant power is power*%error in the value.
    % error in the radius is 2%
    4pi/3 is a constant value and has no effect on the %error of the volume.
    % error in volume = 3*2 = 6%.
    Thus the % error in the volume is 6%.
  5. Aug 20, 2004 #4


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    Science Advisor

    To get a precise value: if r = 3.86 ± 0.08 m , then r could be as large as 3.86+ 0.08 = 3.94 m. Calculate the volume of a sphere with that radius. (256 cubic meters)
    r could be as small as 3.86- 0.08= 3.76. Calculate the volume of that. (222.6 cubic meters)

    The "reference" size is r= 3.86. Calculate the volume of that sphere (241 cubic meters). The true size could be 241-222.6= 18.4 cubic meters too small or 256- 241= 15 cubic meters too small. The larger of those is 18.4 so the "uncertainty" is 18.4 cubic meters. Now divide by the volume, 241 cubic meters, to get the "relative uncertainty" which is actually about 7%.

    A good approximation is to use V= (4/3)πr3 to get dV= 4πr2dr and, then, dV/V= 3 dr/r. With r= 3.86 and dr= 0.08, this is dV/V= 3(0.08/3.86)= 6.2%.

    Notice the calculations done on that approximation are the same as Hypercase gave.
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