Drew Youngren dcy2@columbia.edu
where $\vec r: D \to \Omega \subset \RR^3$ is a parametrization of surface $\Omega$.
If $f \equiv 1$, we get surface area.
Find the moment of inertia of a uniformly-dense spherical shell about its central axis.
Solution. \[ \begin{align*} \vec r_u \times \vec r_v &= \begin{vmatrix} \vec i & \vec j & \vec k \\ R\cos u \cos v & R\cos u \sin v & -R\sin u \\ -R\sin u \sin v & R\sin u \cos v & 0 \\ \end{vmatrix} \\ &= \bv{ R^2 \sin^2 u \cos v \\ R^2 \sin^2 u \sin v \\ R^2 \sin u \cos u} = R \sin (u)\, \vec r(u,v)\end{align*} \] has magnitude $R^2 \sin u$.
Solution cont'd. \[I_z = \iint_{\mathcal S} (x^2 + y^2)\,\mu\,dS \] \[ = \mu\int_0^{2\pi}\int_0^\pi (R\sin u)^2 R^2 \sin u \,du\,dv \] \[ = \frac83 \pi \mu R^4 = \frac23 M R^2 \]
A surface $\Omega$ is oriented if there is a continuous choice of unit normal vector $\vec N$ at each point of the surface.
In other words, an orientation is choice of positive "side" of a surface. By convention we encode this in the parametrization $\vec r(u,v)$. \[ \vec N = \frac{\vec r_u \times \vec r_v}{|\vec r_u \times \vec r_v | } \]
A graph $\langle u, v, f(u,v) \rangle$ could be oriented upward. \[ \vec N \sim \langle -f_x, -f_y, 1 \rangle\]
A surface of revolution $\langle u, g(u) \sin v, g(u)\cos v \rangle $ could be oriented outward. \[ \vec N \sim g(u) \langle -g'(u), \sin v, \cos v \rangle\]
The Möbius strip is, famously, non-orientable.
We will not deal with such surfaces in this class, but it should be known such exist.
Find a parametrization of the unit sphere oriented inward.
Solution. \[ \vec r(u,v) = \bv{ \sin(v) \cos(u) \\ \sin(v) \sin(u) \\ \cos(v) \\ }\] so $v$ plays the role of $\phi$ and $u$ that of $\theta$ is one (of many) ways to do this.
A flux integral of a vector field $\vec F$ through a surface $\Omega$ (somethimes just the "surface integral of a vector field") with orientation $\vec N$ corresponding to parametrization $\vec r: D \to \RR^3$ is
\[\iint_\Omega \vec F\cdot d\vec S = \iint_\Omega \vec F \cdot \vec N\,dS \]
Click
and find Story → Flux Integrals for a
visualization.
\[ \iint\limits_\Omega \vec F \cdot \vec N\,dS = \iint\limits_D \vec F(\vec r(u,v))\cdot \frac{\vec r_u\times \vec r_v}{|\vec r_u \times \vec r_v|} |\vec r_u \times \vec r_v|\,dA \]
\[ = \iint\limits_D \vec F(\vec r(u,v))\cdot \vec r_u\times \vec r_v\,dA\]
That integrand $\vec F(\vec r(u,v))\cdot \vec r_u\times \vec r_v$ is a triple-product, or, rather, a volume of a parallelopiped, two sides defined by the surface, and one by the vector field.
Let $\vec F = \vec i$ (a constant) and $\Omega$ the unit square in the $yz$-plane. Compute the flux $\iint_\Omega \vec F\cdot d\vec S$.
Solution.$\int_0^1 \int_0^1 \vec i \cdot \vec i\, du\,dv = 1$
Repeat for the part of the surface $x = 3y(1 - y)$ with $0 \leq y,z \leq 1$.
Solution.$\int_0^1 \int_0^1 \vec i \cdot (\vec i + (6u - 3)\vec j)\, du\,dv = 1$
Find the flux of the vector field $z\,\vec k$ through the piece of the cone $z=\sqrt{x^2 + y^2}$ below $z=2$, oriented upward.
Solution. \[ \vec r(u,v) = \langle u \cos v, u \sin v, u \rangle \qquad 0 \leq u \leq 2, 0 \leq v \leq 2 \pi \]
\[\vec r_u \times \vec r_v = \langle -u \cos v, -u \sin v, u \rangle \]
\[ \iint_{\mathcal C} \vec F\cdot d\vec S = \int_0^{2\pi} \int_{0}^{2} u^2\,du\,dv = \frac{16\pi}{3} \]
If $\vec F(x,y) = \langle P(x,y),Q(x,y) \rangle $, then
\[\operatorname{curl} \vec F (x,y) = Q_x - P_y = \text{ ``circulation density''}\]It's a little complicated
Consider a "vector" of operators
Let's ignore that $\nabla$ is not technically a vector and use it with vector operations anyway:
Let $f(x,y,z)$ be a scalar field. Then "vector times scalar is vector" becomes
\[ \nabla f = \frac{\partial f}{\partial x} \,\vec i + \frac{\partial f}{\partial y} \,\vec j + \frac{\partial f}{\partial z} \,\vec k \]
aka the gradient
Let $ \vec F(x,y,z) = P(x,y,z)\,\vec i + Q(x,y,z)\,\vec j + R(x,y,z)\,\vec k$ be a vector field. Then "vector dot vector is scalar" becomes
\[ \nabla \cdot \vec F = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \]
aka the divergence
Let $ \vec F(x,y,z) = P(x,y,z)\,\vec i + Q(x,y,z)\,\vec j + R(x,y,z)\,\vec k$ be a vector field. Then "vector cross vector is vector" becomes
\[ \nabla \times \vec F = \begin{vmatrix} \vec i & \vec j & \vec k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ P & Q & R \end{vmatrix} \]
\[ = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \,\vec i + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\,\vec j + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\,\vec k \]
aka the curl