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Math Forum / Mathematics / Recreational Math / July 2009Parabolic Mirror
A parabolic mirror is formed by revolving the part of the parabola y^2 = 4ax from x = 0 to x = h about the axis of the parabola. If the width of the mirror is 2k, show that the area of its surface is [(pi*k)/6h^2]*[(k^2+4h^2)^(3/2)-k^3] Hi Bill, The surface area of your paraboloid of revolution is a fairly straightforward integration, provided you remember to integrate w.r.to the inclined surface ds, rather than the horizontal projection dx. We have y^2=4ax .: dy/dx=2a/y and (dy/dx)^2=a/x At x=h y=k .:a=k^2/4h And ds=sqrt(dx^2+dy^2) = dx.sqrt(1+(dy/dx)^2) Now dA=2Pi.y.ds = 2Pi.y.dx.sqrt(1+(dy/dx)^2) =2Pi.2.sqrt(a.x).dx.sqrt(1+a/x) =4Pi.sqrt(a).sqrt(a+x).dx :A=4Pi.sqrt(a).Integral 0->h sqrt(a+x).dx =4Pi.sqrt(a).2/3((a+h)^(3/2) - a^(3/2)) Substitute a=k^2/4h :A=Pi.k/(6.h^2).(k^2+4.h^2)^(3/2) - k^3) QED. Regards, Peter Scales. Hi Peter, Thanks for the response, which is fine. Rehads Bill |
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