# Surface area of cone

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?

• Hello, welcome to Math.SE. For some basic information about writing maths at this site see e.g. here, here, here and here. – Lord_Farin Oct 30 '13 at 18:35

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.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}{\left\vert #1\right\rangle} \newcommand{\ol}{\overline{#1}} \newcommand{\pars}{\left(\, #1 \,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}} \newcommand{\verts}{\left\vert\, #1 \,\right\vert} \newcommand{\wt}{\widetilde{#1}}$ Put the surface of the cone in a plane. It becomes a 'pie/pizza slice' with $$\mbox{radius}\quad R_{\rm pie}\equiv \root{r^{2} + h^{2}} \qquad\mbox{and}\qquad\mbox{angle}\quad \Delta\theta \equiv {2\pi r \over \root{r^{2} + h^{2}}}$$

\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}