4) A surface of revolution is generated as follows: Two fixed points (xi, yi) an

Question

4) A surface of revolution is generated as follows: Two fixed points (xi, yi) and (x2, y2) in the x, y plane are joined by a curve y-y(x). (Actually, you will make your life easier if you start out writing this as x -x(y)) The whole curve is now rotated about the x-axis to generate a surface. Show that the curve for which the area of the surface is stationary has the formy = yo cosh- are constants. This is often called the soap bubble problem, since the resulting surface is usually the shape of a soap bubble held by two coaxial rings of radii yi and y2. Hint: when solving the integral with square root, use substitution y = yo cosh u. where xo and yo

Explanation / Answer

4) given two points

(x1,y1) amd (x2,y2)

joined by curve y = y(x)

hence

y1 = y(x1)

y2 = y(x2)

now, at some x1 < x < x2

consider dx

then arc length of the small curve length = dl

dl*sin(theta) = dy

dl*cos(theta) = dx

dy/dx = tan(theta)

area of this curve when rotated about x axis = dA

dA = 2*pi*y*dl

dl = dx/cos(theta) = sqrt(1 + tan^2(theta))dx = sqrt(1 + (dy/dx)^2)dx

dA = 2*pi*y*sqrt(1 + (dy/dx)^2)dx

integrating

A = integral(2*pi*y*sqrt(1 + (dy/dx)^2)dx) from x = x1 to x = 2

if we consider

y = yo*cosh((x - xo)/yo)

dy/dx = sinh((x - xo)/yo)

this will solve the

A = integral(2*pi*y*sqrt(1 + (dy/dx)^2)dx) from x = x1 to x = x2 equation

A = integral(2*pi*y*cosh((x - xo)/yo)dx) from x = x1 to x = x2 equation

A = 2*pi*yo(yo*(sinh(2(x2-xo)/yo) - sinh(2(x1-xo)/yo) + 2(x2 - x1))/4

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