Step-by-step explanation:
A cross section represents the side s of an equilateral triangle. The height of an equilateral triangle is given by:
[tex]h=sSin60=\frac{\sqrt{3} }{2} s[/tex]
so the area of one triangle is:
[tex]A=\frac{1}{2} sh=\frac{1}{2} s.\frac{\sqrt{3} }{2} s=\frac{\sqrt{3} }{4} s^2[/tex]
The equation of line represents the diagonal is
[tex]x+y=1\\y=-x+1\\x=-y+1[/tex]
if we integrate along the y-axis this will represent the s value
integrate from 0 to 2
[tex]V=\int\limits^2_0 {\frac{\sqrt{3} }{4} } s^2\, dx \\=\frac{\sqrt{3} }{4} \int\limits^2_0 {(-y+1)^2} \, dx \\=\frac{\sqrt{3} }{4} \int\limits^2_0 {(y^2-2y+1)} \, dx \\=\frac{\sqrt{3} }{4} [\frac{1}{3} y^3-y^2+y]\\=\frac{\sqrt{3} }{4} [\frac{1}{3} (2)^3-(2)^2+2)]\\=5.48[/tex]
