Riemanns Sum Problem: Find the exact volume

[MathsKid007]

View attachment 8261

This is what i have so far
We can find the exact volume of any shape using:
V= \(\displaystyle int[a,b] A(x) dx\)
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
\(\displaystyle ∴A=(3√3)/2 a^2\)
Where a,is the length of a side
Write the side length a,at height x
a= s
\(\displaystyle ∴A=(3√3)/2 s^2\)
\(\displaystyle V= int[0,h](3√3)/2 x^2 dx\)
\(\displaystyle V= (3√3)/2 int[0,h]x^2 dx\)
\(\displaystyle = (3√3)/2*x^3/3\)
\(\displaystyle =[(√3 x^3)/2] [0,h]\)
\(\displaystyle V=(√3 h^3)/2\)
Is this correct?

 

[Opalg]

This is what i have so far
We can find the exact volume of any shape using:
V= \(\displaystyle int[a,b] A(x) dx\)
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
\(\displaystyle ∴A=(3√3)/2 a^2\)
Where a,is the length of a side
Write the side length a,at height x
a= s
\(\displaystyle ∴A=(3√3)/2 s^2\)
\(\displaystyle V= int[0,h](3√3)/2 x^2 dx\)
\(\displaystyle V= (3√3)/2 int[0,h]x^2 dx\)
\(\displaystyle = (3√3)/2*x^3/3\)
\(\displaystyle =[(√3 x^3)/2] [0,h]\)
\(\displaystyle V=(√3 h^3)/2\)
Is this correct?

Hi MathsKid, and welcome to MHB!

I agree with your solution up to the point $A = \frac{3\sqrt3}2s^2$. But you have gone wrong in the next line, when you form the integral. You need to apply Pythagoras to write $s^2 = r^2 - x^2$, so that $A = \frac{3\sqrt3}2(r^2 - x^2)$. Then you can integrate that from $x=0$ to $x=h$.