Measured length of the rod is $15 \pm 0.4$ cm and the radius of the rod is $6.1 \pm 0.2$ cm. What is the error in calculated volume? (two decimal places)
What i've tried is
$$V=\pi r^2 l$$ $$V=1753.48 \text{ cm}^3$$
$$DV/ 1753.48= 2(0.2/6.1) + (0.4)/15$$ $$DV = 161.74 \text{ cm}^3$$
But I got it wrong.
$V(r,l)=\pi r^2l$, so $$\begin{align} V(r+\Delta r,l+\Delta l)&\approx V(r,l)+\frac{\partial V}{\partial r}\Delta r+\frac{\partial V}{\partial l}\Delta l\\ &\approx V(r,l)+2\pi r l\Delta r+\pi r^2\Delta l \end{align}$$
So error in $V$ is $2\pi r l\Delta r+\pi r^2\Delta l$.
Note this does not factor in statistics/probability. If the given information is meant to convey that $l$ has a Normal distribution centered at $15$ with $\sigma=0.4$, and that $r$ has a Normal distribution centered at $6.1$ with $\sigma=0.2$, then $\Delta V$ should be computed as $$\sqrt{\left(2\pi r l\Delta r\right)^2+\left(\pi r^2\Delta l\right)^2}$$