Answer:
[tex]dV=113.55 \ cm^3[/tex]
Explanation:
The Differential of Multivariable Functions
Given a multivariable function V(r,h), the total differential of V is computed by
[tex]dV=\displaystyle \frac{\partial V}{\partial r} dr+\frac{\partial V}{\partial h} dh[/tex]
The volume of a cylinder of radius r and height h is
[tex]V=\pi\cdot r^2\cdot h[/tex]
Let's compute the partial derivatives
[tex]\frac{\partial V}{\partial r}=2\pi rh[/tex]
[tex]\displaystyle \frac{\partial V}{\partial h}=\pi r^2[/tex]
The total differential is
[tex]dV=(2\pi rh)dr+(\pi r^2)dh[/tex]
The differential dr is approximated to [tex]\Delta r[/tex] and the differential dh is approximated to [tex]\Delta h[/tex]. We can see the increment of radius is the thick of the metal in the sides, and the increment of the height is the thick of the metal in the top and bottom. Thus dr=0.05 cm, dr=0.2 cm
[tex]dV=(2\pi \cdot 3\ cm\cdot 30\ cm)0.2\ cm+(\pi\ (3\ cm)^2)0.05\ cm[/tex]
[tex]dV=113.55 \ cm^3[/tex]
