Question

Use differentials to estimate the amount of material in a closed cylindrical can that is 40 cm high and 16 cm in diameter, with the metal in the top and bottom approximately 0.1 cm thick, and the metal in the sides approximately 0.05 cm thick. Note that you are measuring the volume of the metal which makes up the can, not the volume it encloses.

a) The differential for the volume is: dV = drt dr
b) The values of dr and dh are: dr = 0.2 cm and dh = 0.05 cm
c) The approximate volume of the material is 576/5Ï€ cm^3.

Answer 1

To estimate the amount of material in the cylindrical can, calculate the volume of the can and subtract the volume of the empty space inside. The volume of the can is 8069.12cm³. The approximate volume of the material in the can is 345.3359875cm³.

Step-by-step explanation:

To estimate the amount of material in the cylindrical can, we need to calculate the volume of the can and then subtract the volume of the empty space inside. The volume of the can is given by the formula V = πr²h, where r is the radius and h is the height. In this case, the can is 16 cm in diameter, so the radius is 8 cm, and the height is 40 cm. Plugging these values into the formula, we have V = 3.142 × (8 cm)² × 40 cm = 8069.12 cm³.

The volume of the empty space inside the can can be calculated by subtracting the volume of the inner cylinder from the volume of the outer cylinder. The inner cylinder has a radius of (8 cm - 0.05 cm) = 7.95 cm and a height of (40 cm - 0.1 cm) = 39.9 cm. The outer cylinder has a radius of 8 cm and a height of 40 cm. So the volume of the inner cylinder is V = 3.142 × (7.95 cm)² × 39.9 cm = 7723.7840125 cm³ and the volume of the outer cylinder is V = 3.142 × (8 cm)² × 40 cm = 8069.12 cm³. Subtracting the volume of the inner cylinder from the volume of the outer cylinder, we get 8069.12 cm³ - 7723.7840125 cm³ = 345.3359875 cm³.

Therefore, the approximate volume of the material in the can is 345.3359875 cm³.