The curve traced out by a point at a distance 1 m from the center of a circle of radius 2 m as the circle rolls along the x-axis is called a trochoid and has parametric equations:
x=2θ-sinθ y=2-cosθ
One arch of the trochoid is given by the parameter interval 0≤θ≤2π. Find the area under one arch of the trochoid.
2007-09-04 16:08:10 · 3 answers · asked by dawance88 2 in Science & Mathematics ➔ Mathematics
r^2 = x^2 + y^2 = (2θ-sinθ )^2 + (2-cosθ )^2 = 4θ^2 - 4sinθ + (sinθ)^2 + 4 - 4cosθ + (cosθ)^2 =
4θ^2 + 4sinθ - 4cosθ + 5.
Integrate this expression with respect to θ from 0 to 2π and you should have your area.
2007-09-04 16:37:20 · answer #1 · answered by Mathsorcerer 7 · 1⤊ 0⤋
first substitute to cos^3(x) via cos^2(x) cos(x) and to (one million-sin^2(x))cosx Now you have quintessential of sin^2(x) [one million-sin^2(x)]cos(x) dx substitute the variable: u = sin(x) and du = cos(x)dx Now you have quintessential of u^2[ one million -u^2]du ==> quintessential of (u^2 - u^4) du which will effect u^3/3 - u^5/5 + C and now substitute returned u via sin(x) and get the main suitable answer: sin^3(x) /3 - sin^5(x) /5 + C ok
2016-10-09 23:39:20 · answer #2 · answered by ? 4 · 0⤊ 0⤋
ehem. WHAT?????
2007-09-04 16:14:07 · answer #3 · answered by Ashlynn W 2 · 0⤊ 1⤋