Parameterize the surface of a cone, with base radius $a$ and height $h$, and so express its surface area as a multiple integral. Hence show its surface area equals $\pi a \sqrt{a^2+h^2}$.
Re-evaluate this surface area by considering how a sector of a disc may be folded to make a cone.
Solution.
The cone has parameterization $\mathbf r=\left(r\cosθ,r\sinθ,\frac{rh}a\right),0≤r≤a,0≤θ≤2π$. $|\mathbf r_r∧\mathbf r_θ|=\left|\left(\cosθ,\sinθ,\frac{h}a\right)∧\left(-r\sinθ,r\cosθ,0\right)\right|=r\sqrt{1+\frac{h^2}{a^2}}$.$$\int_0^a\int_0^{2π}r\sqrt{1+\frac{h^2}{a^2}}\mathrm{~d}θ\mathrm{~d}r=πa\sqrt{a^2+h^2}$$The cone can be unfolded into a sector of radius $\sqrt{a^2+h^2}$ and arc length $2πa$, so the central angle $θ=\frac{2πa}{\sqrt{a^2+h^2}}$, area=$\frac12θ(a^2+h^2)=πa\sqrt{a^2+h^2}$.
Show that the solid angle at the apex of a cone with semiangle $\alpha$ is $2 \pi(1-\cos \alpha)$.
If a sphere has radius $R$ and its centre at distance $D$ from an observer, with $D \gg R$, show that the sphere occupies, as a fraction $\frac{1}{2}\left(1-\frac{\sqrt{D^{2}-R^{2}}}{D}\right) \approx \frac{R^{2}}{4 D^{2}}$ of the observer's view. Use this to explain how the sun (at radius $7 \times 10^{5} \mathrm{~km}$ and distance $1.5 \times 10^{8} \mathrm{~km}$) and moon (at radius $1.8 \times 10^{3} \mathrm{~km}$ and distance $3.8 \times 10^5 \mathrm{~km}$) occupy roughly the same amount of the sky.
Solution.
Let the vertex be at the origin and base be the intersection of the unit sphere and the plane $z=\cosα$.\begin{align*}&\int_0^{\sinα}\int_0^{2π}\frac{\cosα}{(r^2+\cos^2α)^{3/2}}r~\mathrm dθ~\mathrm dr\\=&2π\cosα\int_0^{\sinα}\frac{r\mathrm dr}{(r^2+\cos^2α)^{3/2}}\\=&2π(1-\cosα)\end{align*}Substituting $α=\sin^{-1}\frac RD$, we find the fraction that the sphere occupies is $\frac{2π\left(1-\sqrt{1-\frac{R^2}{D^2}}\right)}{4π}≈\frac{R^2}{4D^2}$, which is of the same order of magnitude for the sun and the moon.
Evaluate $\iint_{\Sigma} \mathbf{F} \cdot \mathrm{d} \mathbf{S}$ where $\mathbf{F}=\left((x-1) x^2 y,(y-1)^2 x y, z^2-1\right)$ and $\Sigma$ is the surface of the unit cube $[0,1]^3$.
Solution. The integral is 0 for the surfaces $x=0$, $x=1$, $y=0$, $y=1$ and $z=1$.
For $z=0$, the normal pointing outward the body, so it is in direction of $-\bf k$: $\int_0^1\int_0^1(-1)(-1)~\mathrm dx~\mathrm dy=1$, so the integral is 1.
Two points are chosen at random on the surface of the sphere $\Sigma$ with equation $x^2+y^2+z^2=a^2$. Explain why the mean distance $\mu$ between the points equals the integral
$\mu=\frac{1}{4 \pi a^{2}} \iint_{\Sigma} \sqrt{x^2+y^2+(z-a)^2} \mathrm{~d} S$ and hence determine $\mu$.
Solution.
The surface area of Σ is $4πa^2$, so the probability density function is $\frac1{4πa^2}$.
Without loss of generality, suppose that one of the points lies at $(0,0,a)$, and the other point be $(x,y,z)$, where $x^2+y^2+z^2=a^2$. The distance between the two points is $\sqrt{x^{2}+y^{2}+(z-a)^2}=\sqrt{2a^2-2az}$.
$\mu=\frac{1}{4 \pi a^{2}}\iint_{\Sigma} \sqrt{2a^2-2az} \mathrm{~d} S$
$\hphantom{\mu}=\frac{1}{4 \pi}\int_{θ=0}^π\int_{ϕ=0}^{2π}\sqrt{2a^2-2a^2\cosθ}~\sinθ\mathrm{~d}ϕ\mathrm{~d}θ$
$\hphantom{\mu}=\frac{1}{2}\int_{θ=0}^π\sqrt{2a^2-2a^2\cosθ}~\sinθ\mathrm{~d}θ$
$\hphantom{\mu}=\frac{4a}3$.
(i) Calculate the surface integrals $\iint_{\Sigma} f \mathrm{~d} S$ and $\iint_{\Sigma} f \mathrm{~d} \mathbf{S}$ where $f(x, y, z)=\left(x^2+y^2+z^2\right)^2$ and $\Sigma=\left\{(x, y, z) \in \mathbb{R}^3: x^2+y^2=z^2, y \geqslant 0,0 \leqslant z \leqslant 2\right\}$.
(ii) Parametrise the various parts of the boundary $\partial \Sigma$ and determine $\int_{\partial \Sigma} f \mathrm{~d} s$ and $\int_{\partial \Sigma} f \mathrm{~d} \mathbf{r}$.
Solution.
(i)The cone $x^2+y^2=z^2$ has parameterization $\mathbf r=(r\cosθ,r\sinθ,r)$, $\left(x^2+y^2+z^2\right)^2=4r^4$, $\mathbf r_r∧\mathbf r_θ=(-r \cos\theta,-r \sin\theta,r)$, $|\mathbf r_r∧\mathbf r_θ|=\sqrt2r$.
$\iint_{\Sigma} f \mathrm{~d} \mathbf{S}=\int_0^π\int_0^24r^4(-r\cosθ,-r\sinθ,r)~\mathrm dr~\mathrm dθ=\left(0,-\frac{256}{3},\frac{128 \pi }{3}\right)$
$\iint_{\Sigma} f \mathrm{~d} S=\int_0^π\int_0^24r^4\sqrt2r~\mathrm dr~\mathrm dθ=\frac{128 \sqrt{2} \pi }3$
(ii)$\partial \Sigma$ can be parameterized as $(r,0,r)$ ($r$ from 0 to 2) and then $(\cosθ,\sinθ,2)$ (θ from 0 to π) and then $(r,0,-r)$ ($r$ from -2 to 0).
$\int_{\partial \Sigma}f\mathrm{~d}s=\int_0^24r^4\sqrt2\mathrm dr+\int_0^π64\mathrm dθ+\int_{-2}^04r^4\sqrt2\mathrm dr=\frac{256 \sqrt{2}}{5}+128π$.
$\int_{\partial \Sigma} f \mathrm{~d} \mathbf{r}=\int_0^24r^4(1,0,1)\mathrm dr+\int_0^π64(-\sinθ,\cosθ,0)\mathrm dθ+\int_{-2}^04r^4(1,0,-1)\mathrm dr=\left(-\frac{1024}{5},0,0\right)$.
A spherical shell $\Sigma$ with equation $x^{2}+y^{2}+z^{2}=1$ has density $\rho(x, y, z) \geqslant 0$. Show that its moment of inertia about an axis through the points $(\pm a, \pm b, \pm c)$ on the shell equals $I(a, b, c)=\iint_{\Sigma}\left(1-(a x+b y+c z)^{2}\right) \rho(x, y, z) \mathrm{d} S$.
Find this value when $\rho$ is constant. Conversely, if $I(a, b, c)$ is constant for all $(a, b, c) \in \Sigma$, need $\rho(x, y, z)$ be constant?Error! Click to view log.
Solution.
By a rotation of coordinates, $(a,b,c)\mapsto(0,0,1)$.
Since $r=\sinθ$ and $\mathrm dz=\mathrm dθ$, we have $\mathrm dA=2πr\mathrm dz=2π\sinθ\mathrm dθ$.
$\mathrm dI=r^2\mathrm dm=r^2ρ\mathrm dA=2πρ\sin^3θ\mathrm dθ⇒I=2πρ\int_0^π\sin^3θ\mathrm dθ$
$\rho\iint_{\Sigma}\left(1-z^2\right) \mathrm{d} S=\rho\int_0^{2 \pi}\int_0^{\pi } \left(1-\cos^2\theta\right)\sin \theta \mathrm d\theta\mathrm dϕ=2πρ\int_0^π\sin^3θ\mathrm dθ$
The converse is not true, for example, take a part from the shell and put it in the reflected position about the center, but the moment of inertia is unchanged for all axes through center.