What is the volume of the solid whose cross-sections are equilateral triangles perpendicular to the x-axis and with a base on the region bounded by curves y= x, y= -x, and x= 1.
Please explain, because I really need to know how to do this for the rest of my homework.
Please explain, because I really need to know how to do this for the rest of my homework.
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the base length of the triangle at x is f(x) - f(g) = x - (-x) = 2x.
and the area of the equilateral triangle with side length a = 2x is
A = √(3)a^2/4
A = √(3)*(4x^2)/4
A = √(3)*x^2
And the volume of this triangle with width ∆x is
V = √(3)*x^2*∆x
For the whole volume integrate this from x = 0 to 1
V = ∫(√(3)*x^2)dx from 0 to 1
V = √(3)∫(x^2)dx
V = √(3)[x^3/3]
V = √(3)*1/3*1^3
V = √(3)/3
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(as a control, you could also see the solid as a triangular pyramid with base side length 2x = 2 and height 1.
the volume of this pyramid is 1/3*base area*height = 1/3*(√(3)*2^2/4)*1 = √(3)/3 )
OG
and the area of the equilateral triangle with side length a = 2x is
A = √(3)a^2/4
A = √(3)*(4x^2)/4
A = √(3)*x^2
And the volume of this triangle with width ∆x is
V = √(3)*x^2*∆x
For the whole volume integrate this from x = 0 to 1
V = ∫(√(3)*x^2)dx from 0 to 1
V = √(3)∫(x^2)dx
V = √(3)[x^3/3]
V = √(3)*1/3*1^3
V = √(3)/3
-------
(as a control, you could also see the solid as a triangular pyramid with base side length 2x = 2 and height 1.
the volume of this pyramid is 1/3*base area*height = 1/3*(√(3)*2^2/4)*1 = √(3)/3 )
OG