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Here's What I have concluded:
We can find the exact volume of any shape using:
V= [tex]\int\limits_{a}^{b}[/tex]A(x) dx
Where,A(x)is the cross-sectional area at height x
and [a,b] is the height interval
We know that the horizontal cross-sections are hexagonal
∴A=(3√3)/2 a^2
Where a,is the length of a side
Write the side length a,at height x
a= s
∴A=(3√3)/2 s^2
Applying Pythagoras Theorem (s^2=r^2-x^2)
∴A=(3√3)/2(r^2-x^2)
Integrate from x=0,x=h
V= \int\limits_{0}^{h}(3√3)/2(r^2-x^2) dx
V= (3√3)/2 \int\limits_{0}^{h}(r^2 dx-x^2 dx)
=[ (3√3)/2 ( r^2 x-x^3/3)]
=[(3√3)/2 r^2 x-(√3 x^3)/2]
=[(3√3)/2 r^2 h-(√3 h^3)/2]-0
∴V=(-√3 h^3+3√3 hr^2)/2Now the question is:If the safety capacity for the benign use of a camping gas lamp inside a confined space is 20.8 m^3, calculate whether it would be safe to use the lamp within the lightweight ‘pop-up’ tent. (
The height of the tent is 1.89 m)How do i find a value when i have two variables in my volume equation, and I only know height?