The volume V of a solid right circular cylinder is given by V = πr2h where r is the radius of the cylinder and h is its height. A soda can has inner radius r = 1.5 inches, height h = 9 inches, wall thickness 0.02 inches, and top and bottom thickness 0.05 inches. Use linearization to compute the volume, in in3, of metal in the walls and top and bottom of the can. Give your answer to 2 decimal places.

Respuesta :

Answer:

[tex]Volume = 2.40 inch^3[/tex]

Step-by-step explanation:

Given data:

inner radius of can 1.5 inch

height of can 9 inch

thickness of can dr = 0.02 inch

top and bottom thickness 0.05 inch

so dh = 0.05+ 0.05 = 0.10 inch

we know that

[tex]volume =\pi r^2 h[/tex]

By using total differentiation method we have

[tex]dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh[/tex]

     [tex]= \frac{\partial (\pi r^2 h) }{\partial r} dr + \frac{\partial (\pi r^2 h)}{\partial h} dh[/tex]

    [tex] =\pi h\frac{\partial (r^2)}{\partial r} dr +  \pi r^2\frac{\partial h}{\partial h} dh[/tex]

     [tex]= \pi h(2r) dr + \pi r^2 dh[/tex]

puttinfg all value to get required value of volume

[tex]dV = = \pi h(2r) dr + \pi r^2 dh[/tex]

[tex]= \pi (9)(2\times 1.5)(0.02)+ \pi (1.5)^2 (0.10)[/tex]

        [tex]Volume = 2.40 inch^3[/tex]

   

Volume of walls can be taken as outer - inner volume. The volume of the metal in the walls and top and bottom of the can is 2.41 inch³

How to find the volume of a container?

Take the volume of whole container(including walls), then subtract inner volume from it.

Volume of walls of container = Volume of container with walls' volume - inner volume of the container.

Thus, for the given  case, as shown in the image attached, we get:

Total volume of metal used = volume of walls of can+ volume of top and bottom of the can

  • Volume of walls of can = volume of can - volume of inner can
  • Volume of can = [tex]\pi r^2 h = \pi (1.5 + 0.02)^2 \times 9 \approx 65.325 \: \rm inch^3[/tex]radius of total can = inner radius + thickness of wall)
  • Volume of inner can = [tex]\pi r^2 h = \pi (1.5 )^2 \times 9 \approx 63.62 \: \rm inch^3[/tex]

Volume of walls of can = 65.325 - 63.62 ≈ 1.705 inch³

Volume of top and bottom are same. They're like very tiny heighted cylinder with same radius r = 1.5 inches (they aren't 1.5 + 0.02 since they've to fit in)

Thus,

Volume of top + bottom = 2 times volume of top = [tex]2 \times \pi (1.5)^2 \times 0.05 \approx 0.707 \: \rm inch^3[/tex]

Total volume of metal used ≈ 1.705 + 0.707 ≈ 2.41 inch³

Thus,  The volume of the metal in the walls and top and bottom of the can is 2.41 inch³

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