Answer:
16.65 cm³
Step-by-step explanation:
Given that :
the diameter of the closed tin can = 12 cm
the height of the closed tin can = 11 cm
The thickness of the closed tin can = 0.04 cm
The main task required to be carried out is to use differentials to estimate the amount of tin in a closed tin can.
We all know that the volume of a tin can = [tex]\pi r ^ 2 h[/tex]
The differential of the volume is :
[tex]dV = \dfrac{\partial V}{\partial r}dr + \dfrac{\partial V}{\partial h} dh \ ----- \ (1)[/tex]
Differentiating V with respect to r; we have:
[tex]\dfrac{\partial V}{\partial r}= 2 \pi rh[/tex]
Differentiating V with respect to h ; we have:
[tex]\dfrac{\partial V}{\partial h} = \pi r^2[/tex]
Hence; the estimate of the amount of tin in a closed tin can is:
= [tex]2 \pi rh \ dr+ \pi r ^2 \ dh[/tex]
= [tex]\pi r(2h \ dr+ r \ dh)[/tex]
= [tex]\pi *6 (2 * 11 * 0.04 + 0.04 * 0.08)[/tex]
where;
dr = increase in radius = 0.04cm (since thickness of the radius = 0.04 cm)
dh = increase in the height = 0.04 + 0.04 = 0.08
= 16.65 cm³
Hence; the amount of tin in a closed tin can is = 16.65 cm³
