# Homework Help: Percent uncertainty in the volume of a spherical beach ball

1. Aug 19, 2004

### Anamoly

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.

Thanks
-Anamoly

2. Aug 19, 2004

### jcsd

If you get in trouble with these questions, one idiot p-proof method is to work out the volume: with a) the measured value, b) the maximum value, c) the minimum value, then by comapring the volumes the uncertainty is obvious.

3. Aug 19, 2004

### Anamoly

a) V=240
b) V=256
c) V=226

(I know the answer, I just want some help how to find it.)

4. Aug 19, 2004

### jcsd

(max. - min.)/2 then just divide and mutiply by 100 to get the answer in percent.

Last edited: Aug 19, 2004
5. Aug 19, 2004

?????
Here is how you do it:

$$uncertanty = \frac{V_{u}- V_{e}}{V_{e}} * 100$$

$$uncertanty = \frac{256 - 240}{240} * 100$$

$$uncertanty = 6.66$$

Last edited: Aug 19, 2004
6. Aug 19, 2004

there we go, sorry for all of the posts, I had a big error, the answer is 3 posts up.

7. Jan 8, 2009

### Raza

$$V=\frac{4}{3} \pi r^3$$

$$dV=\frac{4}{3}(3r^2) \pi$$

$$dV=\frac{4}{3}(44.6988) \pi \times (0.08)=14.97971165$$

$$\frac{14.97971165}{256} \times 100 = 5.8$$

8. Jan 8, 2009

### Staff: Mentor

Do you realize this thread is more than 4 years old...?

9. Jan 8, 2009

### gmax137

Russ - "what happens online, STAYS online..." apparently forever. That's the beauty & the curse of the medium.

10. Jan 9, 2009

### Raza

woah, I am completely sorry.
I was just searching in Google and found this. I didn't realize that it would be that old.
BTW, people who were searching like me now know the answer.

11. Jan 11, 2009

### whulj2006

I agree with that!

12. Feb 8, 2010

### BeerGUT

6 years later and still useful.

13. Sep 14, 2010

### kLPantera

Almost 7 years later and we just did this problem in class! Still useful.