I'm trying to determine the uncertainty in volume , $ V = b \cdot h \cdot l$ , of a rectangular block with the following measurements,
$$ b \quad=\quad 2 \pm 0.1 $$ $$ h \quad=\quad 2 \pm 0.2 $$ $$ l \quad=\quad 2 \pm 0.3 $$
I thought that the uncertainty is given by
$$ \delta_{volume} \quad=\quad {V}_{max} - {V} $$
And therefore
$$ V \quad=\quad b \cdot h \cdot l \quad=\quad 2 \cdot 2 \cdot 2 \quad=\quad 8 $$ $$ V_{max} \quad=\quad 2.1 \cdot 2.2 \cdot 2.3 \quad=\quad 10.626 $$
Subtracting those we get $$ \longrightarrow \quad \delta_{volume} \quad=\quad {V}_{max} - {V} \quad=\quad 10.626 - 8 \quad=\quad 2.626 $$
But the answer is apparently wrong, the answer should be $\frac{4}{10} \sqrt{14} $ . I don't even know where the square root comes from! Any tips?
Given a size $A(x, y, z)$ calculated with more basic sizes $x, y, z$, which have uncertainties $\Delta x, \Delta y, \Delta z$ respectively, we know that the uncertainty in $A$ is: $$\Delta A = \sqrt{(\frac{\partial A}{\partial x}\cdot \Delta x)^2 + (\frac{\partial A}{\partial y}\cdot \Delta y)^2 + (\frac{\partial A}{\partial z}\cdot \Delta z)^2}$$ Here, you have $V(b, h, l)$, with known uncertainties $0.1, 0.2, 0.3$, so: $$\Delta V = \sqrt{(bh\Delta l)^2 +(hl\Delta b)^2+(bl \Delta h)^2}$$ Plugging in your values gives: $$\Delta V= \sqrt{(4\cdot 0.3)^2 +(4\cdot 0.1)^2+(4\cdot 0.2)^2}=0.4\sqrt{3^2+1^2+2^2}=0.4\sqrt{14}$$
Which is exactly the answer. Note: this formula can be extended to any amount of "basic" variables.