Question
Equilateral hexagon is revolving around one of its edges. Find the volume of the solid of revolution.
no idea how to do this can someone please help!!! urgent!
no idea how to do this can someone please help!!! urgent!
Answers
Steve
One way: Theorem of Pappas
If the edge goes from (0,0) to (0,1), then the center is at (√3/2,√3/2).
The area of the hexagon is 3√3/2. The radius of rotation is √3/2, so its path has length π√3.
So, the volume of the solid is π√3 * 3√3/2 = 9π/2
Another way: calculus. Using symmetry, we can rotate a triangle and a rectangle about the y-axis.
Triangle: using discs, the volume is
v = ∫[1,3/2] π(R^2-r^2) dy
where R=(√3(3/2-y) and r=(√3(y-1/2))
v = ∫[1/2,1] π((√3(3/2-y))^2-(√3(y-1/2))^2) dy = 3π/4
Rectangle:
v = ∫[0,1/2] πr^2 dy
where r=√3
v = ∫[0,1/2] 3π dy = 3π/2
That is the volume of the top half of the figure: 9π/4
Double that and you get the first volume: 9π/2
If the edge goes from (0,0) to (0,1), then the center is at (√3/2,√3/2).
The area of the hexagon is 3√3/2. The radius of rotation is √3/2, so its path has length π√3.
So, the volume of the solid is π√3 * 3√3/2 = 9π/2
Another way: calculus. Using symmetry, we can rotate a triangle and a rectangle about the y-axis.
Triangle: using discs, the volume is
v = ∫[1,3/2] π(R^2-r^2) dy
where R=(√3(3/2-y) and r=(√3(y-1/2))
v = ∫[1/2,1] π((√3(3/2-y))^2-(√3(y-1/2))^2) dy = 3π/4
Rectangle:
v = ∫[0,1/2] πr^2 dy
where r=√3
v = ∫[0,1/2] 3π dy = 3π/2
That is the volume of the top half of the figure: 9π/4
Double that and you get the first volume: 9π/2