0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hmm... this density looks familiar. Isn't this the surface charge density of a perfect conducting sphere? $\endgroup$ – achille hui Apr 9 '14 at 15:31
1
$\begingroup$

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!

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

$\endgroup$
1
$\begingroup$

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) \begin{equation} \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 \end{equation}

$\endgroup$
0
$\begingroup$

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 $$\int_E {\rm d}m={1\over 4\pi a b c}\int_{-\pi/2}^{\pi/2}\int_0^{2\pi}({\bf z}_\phi\times{\bf z}_\theta)\cdot{\bf z}(\phi,\theta)\ d\phi\ d\theta\ .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.