# Surface area of cone using Dirac delta and volume integral

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}

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?

• 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 $\delta,$ $$\delta\left(\frac{z-\sqrt{x^2+y^2}}{\sqrt{1+h^2}}\right).$$ Sep 15, 2021 at 13:22
• @md2perpe Thank you. Any idea how one would go about deriving this expression?
– Sal
Sep 15, 2021 at 20:45
• 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 $\delta\left(\frac{z-\sqrt{x^2+y^2}}{\sqrt{1+h^2}}\right).$ Sep 15, 2021 at 20:53
• @md2perpe What I mean is: "derive without knowing the answer beforehand"
– Sal
Sep 15, 2021 at 20:54

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 $$S=|\nabla u| = \frac{1}{r}\delta(\theta_0-\theta).$$

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