# Divergence Theorem Breaks Down for Surface Area of Cone

I am looking at an alternative method for finding the surface area of a closed 3-dimensional figure using the divergence theorem. The method seems to work for the surface area of a sphere but does not compute correctly for the surface of a right cone. There might be a logical problem, since I am comparing the surface area of a scalar field and the flux of a vector field. The method is encapsulated in the following statement:

$$\boxed{A(S) =\iint \limits_S 1~ \mathrm d S = \iint \limits_S \hat {\mathbf n}. \hat {\mathbf n} ~\mathrm d S = \iint \limits_S \hat {\mathbf n}. \mathrm d \mathbf S = \iiint \limits_V \mathrm{div} ( \hat {\mathbf n} )~ \mathrm d V }$$

For the surface area of a sphere we have: $$x^2+y^2+z^2 = r^2 \\ \mathbf n = ( 2x, 2y , 2z) \\ \| \mathbf n \| = \sqrt{ (2x)^2 +(2y)^2 +(2z)^2 } = 2 \sqrt{ x^2+y^2+z^2} \\ \hat {\mathbf n} = \frac{\mathbf n}{\| \mathbf n \|} = \left( \frac{x}{\sqrt{ x^2+y^2+z^2}}, \frac{y}{\sqrt{ x^2+y^2+z^2}} ,\frac{z}{\sqrt{ x^2+y^2+z^2}} \right) \\ \mathrm{div} ( \hat {\mathbf n}) =\frac{2}{\sqrt{x^2+y^2+z^2}} \\ A(S) = \iint \limits_S \hat {\mathbf n}. \hat {\mathbf n} \mathrm d S =\iiint \limits_V \frac{2}{\sqrt{x^2+y^2+z^2}} { \rm d} V=\int_{0}^{2\,\pi}\!\int_{0}^{r}\!\int_{0}^{\pi}\!2\,\rho\,\sin \left( \phi \right) \,{\rm d}\phi\,{\rm d}\rho\,{\rm d}\theta=4 \pi {r}^{2}$$ Everything looks right above. But the method breaks down (or my calculations) when I look at the surface of a cone. Here is my work. Note that I am looking at the surface of an inverted right cone above the $z=0$ plane with radius and height $1$, with a closed disc on the top : $$x^2+y^2-z^2=0 \\ \mathbf n = ( 2x, 2y , -2z) \\ \| \mathbf n \| = 2\sqrt{ x^2 +y^2 +z^2 } \\ \hat {\mathbf n} = \frac{\mathbf n}{\| \mathbf n \|} = \left( \frac{x}{\sqrt{ x^2+y^2+z^2}}, \frac{y}{\sqrt{ x^2+y^2+z^2}} ,\frac{-z}{\sqrt{ x^2+y^2+z^2}} \right) \\ \mathrm{div} ( \hat {\mathbf n}) =\frac{2z^2}{(x^2+y^2+z^2)^{3/2}} \\ \, \\ A(S) = \iint \limits_S \hat {\mathbf n}. \hat {\mathbf n} \mathrm d S =\iiint \limits_V \frac{2z^2}{(x^2+y^2+z^2)^{3/2}} {\rm d} V \\ =\int_{0}^{2\,\pi}\!\int_{0}^{1}\!\int_{r}^{1}\!{\frac {{2z}^{2}}{ \left( {r}^{2}+{z}^{2} \right) ^{3/2}}}\,r \, {\rm d}z\,{\rm d}r\,{\rm d} \theta=2\,\pi-\sqrt {2}\pi$$ This is clearly not the correct answer. The surface area of a right cone with radius and height $1$ should be $\sqrt 2 \pi + \pi$ . I suspect the problem is due to the consistencey of the outward/inward orientation of the normal vectors. In the case of a sphere there is a consistent outward normal for $\hat {\mathbf n}$ everywhere on the surface.

In the case of the cone, the lateral area has normals oriented outwards while points on the disc cap have normals oriented inwards. I'm thinking the normals should be be consistent with respect to their inwards or outwards orientation over the entire closed surface, viz. they are all inward or all outward but not both. Also it would be nice if we could coin a new word for this inward/outward sense of a normal.

Is this the problem, and is there a way to fix it?

• Another possibiity--- does the surface need to be smooth? Also did you try with reversed direction of the normals over the flat part? – coffeemath Dec 7 '17 at 12:11
• Using $\hat {\mathbf n}=\left( \frac{-x}{\sqrt{ x^2+y^2+z^2}}, \frac{-y}{\sqrt{ x^2+y^2+z^2}} ,\frac{z}{\sqrt{ x^2+y^2+z^2}} \right)$ produces $=\int_{0}^{2\,\pi}\!\int_{0}^{1}\!\int_{r}^{1}\!{\frac {{-2z}^{2}}{ \left( {r}^{2}+{z}^{2} \right) ^{3/2}}}\,r \, {\rm d}z\,{\rm d}r\,{\rm d} \theta=-2\,\pi+\sqrt {2}\pi$ – john Dec 7 '17 at 22:58
• Also regarding your comment about the surface being smooth it does not appear it has to be smooth in this paraboloid example. – john Dec 7 '17 at 23:16
• @coffeemath but the cone is smooth outside of a single point, which should not contribute to the integral no? – qbert Dec 8 '17 at 21:02

The $\hat {\mathbf n}$ that you use for top disc is not the normal for the disc. Instead of $\hat {\mathbf n} = (0, 0, 1)$ you use $$\hat {\mathbf n} = \left( \frac{x}{\sqrt{x^2+y^2+1}}, \frac{y}{\sqrt{x^2+y^2+1}}, \frac{1}{\sqrt{x^2+y^2+1}} \right)$$
• Wouldn't the unit normal to the top disc be $\hat {\mathbf n} = (0, 0, 1)$? Or are you saying that the $\hat {\mathbf n}$ defined in the problem above reduces to $\left( \frac{x}{\sqrt{x^2+y^2+1}}, \frac{y}{\sqrt{x^2+y^2+1}}, \frac{\color{red}{-1}}{\sqrt{x^2+y^2+1}} \right)$. Also note that there is a $-1$ since $z=1$ on the top disc. Can we incorporate this into a single $\hat {\mathbf n}$ for $\iiint \limits_V \mathrm{div} ( \hat {\mathbf n} )~ \mathrm d V$ to get the correct surface area. – john Dec 7 '17 at 22:48
• The correct unit normal on the top disc is $\hat{\mathbf n} = (0, 0, 1)$ but you have used $\left( \frac{x}{\sqrt{x^2+y^2+1}}, \frac{y}{\sqrt{x^2+y^2+1}}, \frac{\color{red}{-1}}{\sqrt{x^2+y^2+1}} \right)$ when using the divergence theorem. Since the normal is discontinuous where the disc meets the lateral surface, we can't use the correct normal for all of the surface. – md2perpe Dec 8 '17 at 6:14
• This paraboloid example has a situation as the cone - the normal is not well defined when the disc meets the lateral surface but the divergence theorem still works. Also we could define the normal on the boundary of the disc to be $(0,0,1)$ so it is well defined and thus continuous. – john Dec 8 '17 at 9:44
• But in that example the normal isn't used as the vector field. The surface is also split into two peaces. Your calculations don't give you the area of the disc $\int_{\text{disc}} dS$, but $\int_{\text{disc}} \frac{-1}{\sqrt{x^2+y^2+1}} dS$. – md2perpe Dec 8 '17 at 10:20