I have a simple question with a possibly trivial mistake in my answer but I'm unable to figure it out.
The question:
A sphere fits inside a cube. The length of the cube and the diameter of the sphere are $10.0 \pm 0.2 \; \text{cm}$. What is the ratio $\frac{\text{percentage uncertainty of the volume of the sphere}}{\text{percentage uncertainty of the volume of the cube}}$?
My answer:
I took the following $2$ approaches. The first one yields the correct answer while the second one doesn't. I would appreciate it if someone can explain the mistake in the second approach.
First, let me define some variables:
- $V_s$ is the volume of the sphere.
- $V_c$ is the volume of the cube.
- $r$ is the radius of the sphere.
- $u$ is the diameter of the sphere.
- $l$ is the length of the cube.
- $a$ is the percentage uncertainty of the volume of the sphere.
- $b$ is the percentage uncertainty of the volume of the cube.
- $\Delta x$ refers to the uncertainty in $x$.
Approach $1$:
$$V_s = \frac{4\pi}{3}\cdot r^3$$ $$\Rightarrow dV_s = \frac{4\pi}{3}\cdot 3r^2 \; dr = 4\pi r^2 \; dr$$ $$\Rightarrow dV_s = \frac{\pi\cdot u^2}{2} \; du$$ $$\Rightarrow \Delta V_s \approx \frac{\pi\cdot u^2}{2} \; \Delta u \quad (1)$$ $$\Rightarrow \Delta V_s \approx \frac{\pi\cdot 10^2}{2} \cdot 0.2 = 10\pi$$ $$V_s = \frac{\pi\cdot u^3}{6} = \frac{1000\pi}{6}$$ $$\Rightarrow a = \frac{100\Delta V_s}{V_s} = 10\pi \cdot \frac{6}{1000\pi} \cdot 100 = 6 \%$$ $$V_c = l^3$$ $$\Rightarrow b = \frac{100\Delta V_c}{V_c} = 100 \cdot \frac{3\Delta l}{l} = \frac{100 \times 3 \times 0.2}{10} = 6 \%$$ $$\therefore \frac{a}{b} = \frac{6 \%}{6 \%} = \fbox{1}$$
Before moving to my second approach, I have the following question:
Considering that the calculation in $(1)$ was approximate, would the value of $\frac{a}{b}$ be a precise value or only an estimate? If it is a precise value, why so?
Approach $2$: $$u = l$$ $$V_s = \frac{\pi}{6}\cdot l^3$$ $$V_c = l^3$$ $$\frac{a}{b} = \frac{\frac{\Delta V_s}{V_s}}{\frac{\Delta V_c}{V_c}} = \frac{\frac{\pi}{6}(\frac{3\Delta l}{l})}{\frac{3\Delta l}{l}} = \boxed{\frac{\pi}{6}}$$
However, $1 \neq \frac{\pi}{6}$, so what is the mistake? I believe that saying $\frac{\Delta V_s}{V_s} = \frac{\pi}{6}(\frac{3\Delta l}{l})$ is wrong, but (if it is), I'm not sure why.
$$\Delta V_s=\frac{\pi}{2}l^2\Delta l,\,V_s=\frac{\pi}{6}\cdot l^3\\\frac{\Delta V_s}{V_s}=\frac{\pi}{2}\frac{6}{\pi}\cdot\frac{l^2}{l^3}\Delta l=\frac{3\Delta l}{l}\therefore\frac{a}{b}=1$$
I am a bit dubious about the whole process though; the question never specified that the cube and sphere fit snugly with one another - it could be the case that the cube has length of $10.2$ units and the sphere has diameter of $9.8$ units... more room for movement and uncertainty there (possibly).
So the danger of calculating uncertainty using calculus (built for limitingly small $\delta$) and extrapolating that to real-valued, "large" $\Delta$, is that the linear approximations of calculus are perfect as $\delta\to0$, but are only approximate for any $\Delta\gt0$. So, take for example the cube. You, using calculus, calculated the uncertainty of its volume as $6\%$. However, if you calculate the forward uncertainty $\frac{10.2^3-10^3}{10^3}$ you find it is equal to $6.1208\%$ - which is close, but not exact due to the error of the calculus linear approximation increasing for larger $\Delta$. The range of the volume of the cube is $9.8^3\le V_c\le 10.2^3$; to find the uncertainty, take $10.2^3-9.8^3$ as the size of the range, and divide it by $10^3$ (the base value), and you get an uncertainty of $12.0016\%$. Notice how $9.8^3$ is quite a bit smaller, relatively speaking, than $10.2^3$ - linearly approximating a cubic won't be too precise! Now let's look at the sphere uncertainty: $\frac{\pi}{6}\left(10.2^3-9.8^3\right)$ divided by $\frac{\pi}{6}10^3$ will in fact give us the exact same uncertainty as the volume of the cube. So the answer is still $1$, but approaches $1$ and $2$ weren't the most rigorous - what would be better is the realisation that both $V_s$ and $V_c$ are functions of $l^3$, and differ only by a constant multiple, so their uncertainties, if using the same length $l$ with the same uncertainty in $l$, will be the same. If my definition of uncertainty doesn't meet with yours, do say, but I think my definition and calculation of uncertainty is the most natural.