Answer to Question #204651 in Electrical Engineering for Kabir

Question #204651
  • Find the area of the surface of the solid formed by the revolution the curve, x=a(θ - sinθ) and Y=a(1- cosθ) about x-axis (y=0).
1
Expert's answer
2021-06-09T11:15:02-0400

"A = 2\\pi\\int_{\\alpha}^{\\beta}x(\u03b8) \\sqrt{[x\u2032(\u03b8)]2+[y\u2032)]2d\u03b8}"

"Parameter\\space \u03b8 \\space varies\\space from\\space 0 \\space to \\space2\u03c0"

Derivatives:

"x\n\u2032\n(\n\u03b8\n)\n=\n(\n\u03b8\n\u2212\nsin\n\u03b8\n)\n\u2032\n=\n1\n\u2212\ncos\n\u03b8\n,"

"y\n\u2032\n(\n\u03b8\n)\n=\n(\n1\n\u2212\ncos\n\u03b8\n)\n\u2032\n=\nsin\n\u03b8\n,"

"[x\u2032(\u03b8)]^2+[y\u2032(\u03b8)]^2=(1\u2212cos\u03b8)^2+sin^2\u03b8=1\u22122cos\u03b8+cos^2 \\theta+sin^2 \\theta"

"=2-2cos \\theta=4 sin ^2 \\frac{\\theta}{2}"

"A=2\\pi\\int_0^{2x} [(\\theta-sin \\theta).2sin \\frac{\\theta}{2}]d\\theta=4 \\pi[\\int_0^{2x} \\theta sin \\frac{\\theta }{2} d \\theta- \\int_0^{2x} sin \\theta sin \\frac{\\theta}{2} d\\theta]"

"I_2=\\int_0^{2x} sin \\theta sin \\frac{\\theta}{2}d \\theta=\\frac{4}{3} sin^3 \\frac{\\theta}{2} |_0^{2x}=0"

The area of the surface is ;

"=4x[I_1-I_2]"


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