Use differentials to estimate the amount of material in a closed cylindrical can that is 4040 cm high and 1616 cm in diameter if the metal in the top and bottom is 0.20.2 cm thick, and the metal in the sides is 0.050.05 cm thick. Note, you are approximating the volume of metal which makes up the can (i. e. melt the can into a blob and measure its volume), not the volume it encloses.

The volume of the metal which make up the can = 179.95 cm^3

Step-by-step explanation:

To solve this the can height = 40 cm

The diameter of the can = 16 cm

The thickness of the top of the can = 0.2 cm

The thickness of the sides = 0.05 cm

The outer volume of the cylinder = pi × r^2 × h = pi × (16/2)^2 × 40 = 8042.5 cm^3

However, the cylinder is 0.2 cm thick on top and 0.05 cm thick on the sides hence

Change in volume dV is given by partial differentiation dV/dr = 2×pi×r×h and dV/dh = 2×pi×r^2×dh

Therefore, dV = dr×2×pi×r×h1 + dh×2×pi×r^2 as V is a function of both h and r

Where h1 = the height of cylinder - top of cylinder = 40-0.2×2 = 39.6 cm

Then we have

dV = 0.05×2×pi×8×39.6 + 0.2×2×pi×8^2 = 99.53 cm^3 + 80.42 cm^3 = 179.95 cm^3

Hence the volume of the metal which make up the can = 179.95 cm^3