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A-parabolic-refector-is-formed-by-revolving-the-arc-of-the-parabala-y-2-4ax-from-x-0-to-x-h-about-the-axis-If-the-diameter-of-the-reflector-is-2l-Show-that-the-area-of-the-reflecting-surface-is




Question Number 227055 by Spillover last updated on 28/Dec/25
A parabolic refector is formed by  revolving the arc of the parabala  y^2 =4ax  from x=0    to  x=h  about the axis.If the  diameter  of the reflector is 2l.Show that  the area of the reflecting surface is  ((πl)/(6h^2 )){(l^2 +4h^2 )^(3/2) −l^3 }
$${A}\:{parabolic}\:{refector}\:{is}\:{formed}\:{by} \\ $$$${revolving}\:{the}\:{arc}\:{of}\:{the}\:{parabala} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax}\:\:{from}\:{x}=\mathrm{0}\:\:\:\:{to}\:\:{x}={h} \\ $$$${about}\:{the}\:{axis}.{If}\:{the}\:\:{diameter} \\ $$$${of}\:{the}\:{reflector}\:{is}\:\mathrm{2}{l}.{Show}\:{that} \\ $$$${the}\:{area}\:{of}\:{the}\:{reflecting}\:{surface}\:{is} \\ $$$$\frac{\pi{l}}{\mathrm{6}{h}^{\mathrm{2}} }\left\{\left({l}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{l}^{\mathrm{3}} \right\} \\ $$$$ \\ $$

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