Answer:
The estimated amount of metal in the can is 87.96 cubic cm
Step-by-step explanation:
We can find the differential of volume from the volume of a cylinder equation given by
![V= \pi r^2 h](https://tex.z-dn.net/?f=V%3D%20%5Cpi%20r%5E2%20h)
Thus that way we will find the amount of metal that makes up the can.
Finding the differential.
A small change in volume is given by:
![dV =\cfrac{\partial V}{\partial h} dh + \cfrac{\partial V}{\partial r} dr](https://tex.z-dn.net/?f=dV%20%3D%5Ccfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20h%7D%20dh%20%2B%20%5Ccfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20r%7D%20dr)
So finding the partial derivatives we get
![dV =\pi r^2 dh + \pi 2r h dr](https://tex.z-dn.net/?f=dV%20%3D%5Cpi%20r%5E2%20dh%20%2B%20%5Cpi%202r%20h%20dr)
![dV =\pi r^2 dh + 2\pi r h dr](https://tex.z-dn.net/?f=dV%20%3D%5Cpi%20r%5E2%20dh%20%2B%202%5Cpi%20r%20h%20dr)
Evaluating the differential at the given information.
The height of the can is h = 26 cm, the diameter is 10 cm, which means the radius is half of it, that is r = 5 cm.
On the other hand the thickness of the side is 0.05 cm that represents dr = 0.05 cm, and the thickness on both top and bottom is 0.3 cm, thus dh = 0.3 cm +0.3 cm which give us 0.6 cm.
Replacing all those values on the differential we get
![dV =\pi 5^2 (0.6) + 2\pi (5) (26) (0.05)](https://tex.z-dn.net/?f=dV%20%3D%5Cpi%205%5E2%20%280.6%29%20%2B%202%5Cpi%20%285%29%20%2826%29%20%280.05%29)
That give us
![V= 28 \pi \, cm^3](https://tex.z-dn.net/?f=V%3D%2028%20%5Cpi%20%20%5C%2C%20cm%5E3)
Or in decimal value
![\boxed{dV= 87.96 \, cm^3}](https://tex.z-dn.net/?f=%5Cboxed%7BdV%3D%2087.96%20%5C%2C%20cm%5E3%7D)
Thus the volume of metal in the can is 87.96 cubic cm.