- #1
lazypast
- 85
- 0
Hi
I know calculus can be used to calculate the dimensions of an object and the minimum material which can be used. It is a pressure vessel of a cylindrical shape.
[tex]v= \frac {4 \pi r^3} {3}[/tex]
[tex]A=2 \pi r^2 + 2 \pi rh[/tex]
as the cylinder is hollow the thickness of the walls is found by
[tex]t= \frac {pd} {4 \sigma }[/tex]
sigma = [tex]300x10^6 Nm^{-2}[/tex] (steel i think?)
p = pressure
d = diameter
I have found amount of material can be calculated by surface area multiplied by thickness but would like wo see the minimum amount of material used proved by calculus
I know calculus can be used to calculate the dimensions of an object and the minimum material which can be used. It is a pressure vessel of a cylindrical shape.
[tex]v= \frac {4 \pi r^3} {3}[/tex]
[tex]A=2 \pi r^2 + 2 \pi rh[/tex]
as the cylinder is hollow the thickness of the walls is found by
[tex]t= \frac {pd} {4 \sigma }[/tex]
sigma = [tex]300x10^6 Nm^{-2}[/tex] (steel i think?)
p = pressure
d = diameter
I have found amount of material can be calculated by surface area multiplied by thickness but would like wo see the minimum amount of material used proved by calculus