RMS error in volume of a cylinder

[OONeo01]

Homework Statement

The length and radius of a perfect cylinder are measured with an RMS error of 1%. The RMS error of the inferred volume of the cylinder is.. ?

Homework Equations

V=∏r²h
Hence dV/V=2dr/r+dh/h

The Attempt at a Solution

I took dr/r and dh/h as 1%. So got the final answer as 2x1%+1%=3%
But the question came with 4 options(one of which is correct):
1.7%, 3.3%, 0.5% and 1%

So my answer is wrong ! What am I overlooking ? Is there any special precautions to be taken about RMS error ?

 

[mfb]

The uncertainties of the two measurements should be independent, so I would add those contributions in quadrature and not linear. But then the result is 2.2% :D.
1.7% would be the answer if all three dimensions had been measured independently.

 

[OONeo01]

1.7% would be the answer if all three dimensions had been measured independently.

Could you please explain that, or give some link where it has been explained ? I didn't quiet understand what you meant by that.. X-)

 

[haruspex]

The two errors might add up, but also they might partly cancel. So the RMS error in the volume will be less than simply 3 x one dimension of error. Instead, you use a root-sum-squares way of adding them up. But mfb is right that the answer should not be 1.7 either. That would be the answer for volume of a rectangular block with independent 1% errors in each of the three dimensions: √(1²+1²+1²) = √3.
In the present case, there are only two measurements. An error of x% in the radius will produce an error of 2x% in the cross-sectional area, so we have √(1²+2²) = √5 ≈ 2.2, which is not in the list.

 

[OONeo01]

Ok. So √3 which matches with one of the given options doesn't really make much sense.

Thanks a lot for the help guys. What was more important to me was how to find the RMS error(of any measurement, not just the volume of a cylinder) rather than just matching answers with options. I guess I learned that. :-)