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I want to find the surface area of a cone (excluding the circular 'base') by doing a volume integral with the appropriate Dirac delta. Schematically

$$ \text{Area}=\int d\text{Vol} \ \delta(...) $$

Let the cone have height $h$ and maximum radius $R=1$. The correct area is $A=\pi \sqrt{1+h^2}$. I put the pointy end at the origin and align the cone with the $z$ axis.

In Cartesian co-ordinates, I have tried

$$ A \stackrel{?}{=}\int dx \ dy \int\limits_0^h dz \ \delta \left( z-h\sqrt{x^2+y^2} \right) =\int\limits_{h\sqrt{x^2+y^2}<h} dx \ dy =\pi \neq A $$

As pointed out in the comments, the correct expression should be

$$\tag{1} A=\int dx \ dy \int\limits_0^h dz \ \left[ \delta \left( z-h\sqrt{x^2+y^2} \right) \sqrt{1+h^2}\right] $$

In spherical co-ordinates, I find by comparing to the correct answer

$$\tag{2} \int d\theta \ d\phi \int\limits_0^L dr \ r^2\sin\theta \left[ \frac{\delta(\theta-\theta_0)}{r} \right]=A $$

Where $L=\sqrt{1+h^2}$ and $\sin \theta_0=1/L$. I am unable to 'derive' (or even intuit) the terms in $[ ...]$ in eq (1) and eq (2). What I know:

  1. The 'composition with a function' property of the delta.
  2. The delta picks up an inverse Jacobian under co-ordinate transforms.
  3. It is necessary to integrate out singular variables in the Jacobian eg. for a point at the origin in spherical co-ordinates.
  4. It is necessary to be careful when using the delta to find areas or lengths.

Question: How to derive (by means other than 'comparing to the correct answer') the expressions in eq. (1) and eq. (2) which include factors appended to the deltas?

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  • $\begingroup$ For (1) you can take $\sqrt{1+h^2} \, \delta(z-\sqrt{x^2+y^2})$ or, if you don't want factors in front of $\delta,$ $$\delta\left(\frac{z-\sqrt{x^2+y^2}}{\sqrt{1+h^2}}\right).$$ $\endgroup$
    – md2perpe
    Sep 15, 2021 at 13:22
  • $\begingroup$ @md2perpe Thank you. Any idea how one would go about deriving this expression? $\endgroup$
    – Sal
    Sep 15, 2021 at 20:45
  • $\begingroup$ The cone area is $$ A = \iint_{x^2+y^2\leq z^2} \sqrt{1+h^2} \, dx \, dy . $$ Insert a unit, $1=\int_{\mathbb{R}} \delta(z-\sqrt{x^2+y^2}) \, dz$ to get $$ A = \iiint_{x^2+y^2\leq 1, z\in\mathbb{R}} \sqrt{1+h^2} \, \delta(z-\sqrt{x^2+y^2}) \, dx \, dy \, dz . $$ Then use that $\delta(ax)=\frac{1}{|a|}\delta(x)$ to rewrite $\sqrt{1+h^2} \, \delta(z-\sqrt{x^2+y^2})$ as $\delta\left(\frac{z-\sqrt{x^2+y^2}}{\sqrt{1+h^2}}\right).$ $\endgroup$
    – md2perpe
    Sep 15, 2021 at 20:53
  • $\begingroup$ @md2perpe What I mean is: "derive without knowing the answer beforehand" $\endgroup$
    – Sal
    Sep 15, 2021 at 20:54

1 Answer 1

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Theorem
Let $\Omega$ be a set with piecewise smooth boundary $\partial\Omega$. Let $\mathbf{n}$ be the outbound normal field on $\partial\Omega$ and $S$ be the area measure on $\partial\Omega.$ Then $\nabla \mathbf{1}_\Omega = -\mathbf{n}S,$ where $\mathbf{1}_\Omega$ is the indicator function on $\Omega$.

Proof
Let $\varphi$ be a test function. Then, $$ \langle \nabla\mathbf{1}_\Omega, \varphi \rangle = -\langle \mathbf{1}_\Omega, \nabla\varphi \rangle = -\int_\Omega \nabla\varphi \, dV = -\oint_{\partial\Omega} \varphi \, \mathbf{n}\,dS = \langle -\mathbf{n}\,dS, \varphi \rangle. $$


The indicator function on the inner of the cone can be written as $u(x,y,z) = H(z-h\rho),$ where $\rho=\sqrt{x^2+y^2}.$ The gradient then is $$ \nabla u = -\frac{hx}{\rho}\delta(z-h\rho) \,\hat{x} -\frac{hy}{\rho}\delta(z-h\rho) \,\hat{y} +\delta(z-h\rho) \,\hat{z} $$ which gives that the area measure is given by the distribution $S=|\nabla u| = \sqrt{1+h^2} \delta(z-h\rho).$


In polar coordinates the indicator function can be written as $u(r,\theta,\phi)=H(\theta_0-\theta)$ which has gradient $$ \nabla u = -\frac{1}{r}\delta(\theta_0-\theta)\,\hat{\theta} $$ giving the area measure distribution $S=|\nabla u| = \frac{1}{r}\delta(\theta_0-\theta).$

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  • $\begingroup$ This is precisely the sort of thing I was looking for. Thank you! $\endgroup$
    – Sal
    Sep 15, 2021 at 22:52

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