thanks for any help.
I'm trying to find the surface area of a cone via integration. I know that the parametric equation of a cone is $$x=u\cos(p) \\ y=u\sin(p) \\ z=u$$
So as a vector, $\vec{R} = \langle u\cos(p), u \sin(p), u \rangle$.
Since the area equals the double integral of $ds$, and $d\vec{s} = \dfrac{d\vec{R}}{du} du \times \dfrac{d\vec{R}}{dp} dp$, I work out that:
$$\vec{ds} = u\,du \, dp \, \langle -\cos(p),-\sin(p),1\rangle \\ ds = n \, d\vec{s} \\ ds = u \, du \, dp $$
I would expect I'd get the correct answer if I integrated this between the limits 0 to $2\pi$ and 0 to $h$, however I get $h^2\pi$ which is incorrect. Could someone point me to where I'm going wrong?
$\endgroup$ 12 Answers
$\begingroup$You actually had just about everything right, except that you skipped an important step: your normal vector to the surface $ \ \vec{ds} \ $ is correct, but you need to integrate its length over the surface of the cone nappe in order to obtain the surface area.
I'll generalize the problem a little, since the choice of proportions for the cone hides one of the factors in the surface area result. For a cone nappe with a height $ \ h \ $ and a "base radius" $ \ r \ $ , we can use similar triangles to find the parametrization (using your notation)
$$ x \ = \ \left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ y \ = \ \left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ z \ = \ u \ \ , $$
with the domain $ \ 0 \ \le \ u \ \le \ h \ , \ 0 \ \le \ p \ < \ 2 \pi \ $ . An "upward" normal vector is then given by
$$ \vec{R_u} \ \times \ \vec{R_p} \ \ " = " \ \ \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \left( \frac{r}{h} \right) \cos \ p&\left( \frac{r}{h} \right) \sin \ p\quad&1\\ -\left( \frac{r}{h} \right) u \ \sin \ p&\left( \frac{r}{h} \right) u \ \cos \ p\quad&0\end{array}\right| $$
$$ = \ \langle \ -\left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ -\left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ \left( \frac{r}{h} \right)^2 u \ \rangle \ \ . $$
So, up to this point, your procedure is fine. What is needed now is the "norm" of this vector:
$$ \| \ \vec{R_u} \ \times \ \vec{R_p} \ \| \ \ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ \cos^2 \ p \ + \ \left( \frac{r}{h} \right)^2 u^2 \ \sin^2 \ p \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ \ . $$
$$ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ . $$
It is the "magnitude" of the infinitesimal patches associated with the normal vectors that we wish to integrate over the domain of the parameters. Thus,
$$ S \ \ = \ \ \int_0^{2 \pi} \int_0^h \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ du \ dp $$
$$ = \ \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \int_0^{2 \pi} dp \ \int_0^h \ u \ \ du $$
$$ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ 2 \pi \ \cdot \ \left(\frac{1}{2}u^2 \right) \vert_0^h \ \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ \pi \ h^2 $$
$$ = \ \pi \ h \ \cdot \ \left(\frac{r}{h} \right) \ \cdot \ h \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ = \ \pi \ r \ \sqrt{ r^2 \ + \ h^2 } \ \ , $$
or $ \ \pi \ $ times the "base radius" times the "slant height" of the cone nappe, as the surface area is frequently expressed. In your use of the "standard cone", for which $ \ r \ = \ h \ $ , this formula gives us $ \ S \ = \ \pi \ \sqrt{2} \ h^2 \ $ , as you will find for your calculations, with the restoration of the omitted step.
$\endgroup$ $\begingroup$$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Put the surface of the cone in a plane. It becomes a 'pie/pizza slice' with $$ \mbox{radius}\quad R_{\rm pie}\equiv \root{r^{2} + h^{2}} \qquad\mbox{and}\qquad\mbox{angle}\quad \Delta\theta \equiv {2\pi r \over \root{r^{2} + h^{2}}} $$
$\endgroup$\begin{align} \color{#f44}{\large\mbox{Surface}} =\half\,R_{\rm pie}^{2}\,\Delta\theta =\half\,\pars{\root{r^{2} + h^{2}}}^{2}{2\pi r \over \root{r^{2} + h^{2}}} =\color{#f44}{{\large\pi r\root{r^{2} + h^{2}}}} \end{align}
