Mass of ellipsoid's surface

Find the mass of ellipsoid's surface $E=\{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\}$ if density $\rho=\frac{r}{4\pi abc}$, where $r=dist(0,T_{(x,y,z)}E)$ and $T_{(x,y,z)}E$ is a surface tangent to $E$.

I know I should integrate. But how? Not to mention I'm not quite sure what the mass of a surface is.

• Hmm... this density looks familiar. Isn't this the surface charge density of a perfect conducting sphere? – achille hui Apr 9 '14 at 15:31

The density looks familiar. In fact, Lord Kelvin has shown that it is the surface charge density of a unit charged perfect conducting ellipsoid. i.e. the integral of the density is $1$.

To evaluate the integral, introduce ellipsoidal polar coordinates $$[0,2\pi] \times [0,\pi] \ni (\theta,\phi)\quad\mapsto\quad\vec{X}(\theta,\phi) = (a\sin\theta\cos\phi, b\sin\theta\sin\phi,c\cos\theta)$$

At a point $\vec{X}(\theta,\phi) = (u,v,w)$ on the ellipsoid, we have

\begin{align} \vec{X}_\theta &= \frac{\partial\vec{X}}{\partial\theta} = (a\cos\theta\cos\phi, b\cos\theta\sin\phi,-c\sin\theta)\\ \vec{X}_\phi &= \frac{\partial\vec{X}}{\partial\phi} = (-a\sin\theta\sin\phi, b\sin\theta\cos\phi,0)\\ \implies X_\theta \times X_\phi &= abc\sin\theta \left(\frac{\sin\theta\cos\phi}{a},\frac{\sin\theta\sin\phi}{b},\frac{\cos\theta}{c}\right)\\ &= abc\sin\theta \left(\frac{u}{a^2},\frac{v}{b^2},\frac{w}{c^2}\right) \end{align} So the normal vector at $(u,v,w)$ is along the direction $\left(\frac{u}{a^2},\frac{v}{b^2},\frac{w}{c^2}\right)$ and hence the equation of tangent plane is given by

$$\frac{u}{a^2}(x-u) + \frac{v}{b^2}(y-v) + \frac{w}{c^2}(z-w) = 0 \quad\iff\quad \frac{ux}{a^2} + \frac{vy}{b^2} + \frac{wz}{c^2} = 1$$ From this we can deduce the distance between the tangent plane and the center is give by $$r = \frac{1}{\sqrt{ \left(\frac{u}{a^2}\right)^2 + \left(\frac{v}{b^2}\right)^2 + \left(\frac{w}{c^2}\right)^2}}$$

Using this, we can express the surface element of the ellipsoid as

$$dS = | \vec{X}_\theta \times \vec{X}_\phi |d\theta d\phi = \frac{abc\sin\theta}{r} d\theta d\phi$$ and hence the surface integral is simply

$$\int_0^{2\pi}\int_0^\pi \frac{r}{4\pi abc} \frac{abc\sin\theta}{r} d\theta d\phi = \frac{1}{4\pi}\int_0^{2\pi}\int_0^\pi \sin\theta d\theta d\phi = 1$$

Update

It just struck me there is no need to introduce any parametrization and evaluate the integral explicitly. If $\hat{n}$ is the outward normal vector at a point $\vec{X}$ on $E$, the tangent plane has the from $\hat{n}\cdot( \vec{x} - \vec{X} )$ and hence $r = \vec{n}\cdot\vec{X}$. Let $V$ be the solid ellipsoid enclosed by the surface $E = \partial V$. We can evaluate the integral using divergence theorem and the formula for the volume of an ellipsoid!

$$\frac{1}{4\pi abc}\int_E \hat{n}\cdot\vec{X} dS = \frac{1}{4\pi abc}\int_V \nabla\cdot\vec{X} dV = \frac{1}{4\pi abc}\int_V 3 dV = \frac{1}{4\pi abc} \left( 4\pi abc \right) = 1$$

The mass of a surface area is the density multiplied by the area. If the density is $P(x, y, f(x, y))$ then the mass is (by definition) $$\int \int_{R} P(x, y, f(x, y))\sqrt{1 + \Big( \frac{\partial f}{\partial x}\Big)^{2} + \Big( \frac{\partial f}{\partial x}}\Big)^{2}dA$$

Let $$(\phi,\theta)\mapsto{\bf z}(\phi,\theta)=(a\cos\theta\cos\phi,b\cos\theta\sin\phi, c\sin\theta)$$ with $0\leq\phi\leq 2\pi$, $\ -{\pi\over 2}\leq\theta\leq{\pi\over2}$ be a parametric representation of $E$ by means of geographical coordinates. The outward normal at ${\bf z}(\phi,\theta)$ is given by $${\bf n}(\phi,\theta)={{\bf z}_\phi\times{\bf z}_\theta\over|{\bf z}_\phi\times{\bf z}_\theta|}\ ,$$ and the scalar surface element by $${\rm d}\omega=|{\bf z}_\phi\times{\bf z}_\theta|\ {\rm d}(\phi,\theta)\ .\tag{1}$$ The points ${\bf x}$ of the tangent plane $T$ at ${\bf z}:={\bf z}(\phi,\theta)$ satisfy the equation $${\bf n}\cdot{\bf x}={\bf n}\cdot{\bf z}\ ,$$ whence $T$ has distance $r(\phi,\theta)={\bf n}\cdot{\bf z}$ from the origin. It follows that the mass of the surface element $(1)$ is given by $${\rm d}m=\rho\>{\rm d}\omega={1\over 4\pi a b c}({\bf z}_\phi\times{\bf z}_\theta)\cdot{\bf z}(\phi,\theta)\ {\rm d}(\phi,\theta)\ ,$$ so that the square roots happily cancel. The total mass in question is then given by the integral $$\int_E {\rm d}m={1\over 4\pi a b c}\int_{-\pi/2}^{\pi/2}\int_0^{2\pi}({\bf z}_\phi\times{\bf z}_\theta)\cdot{\bf z}(\phi,\theta)\ d\phi\ d\theta\ .$$