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Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(excluding the base).

User Melv
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1 Answer

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We can parameterize this part of a cone by


\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\frac hau\right\rangle

with
0\le u\le a and
0\le v\le2\pi. Then


\mathrm dS=\|\mathbf s_u*\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+(h^2)/(a^2)}u\,\mathrm du\,\mathrm dv

The area of this surface (call it
\mathcal S) is then


\displaystyle\iint_(\mathcal S)\mathrm dS=\sqrt{1+(h^2)/(a^2)}\int_(v=0)^(v=2\pi)\int_(u=0)^(u=a)u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+(h^2)/(a^2)}\pi=a√(a^2+h^2)\pi
User Wolfgang Jeltsch
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