Integration¶

This chapter doesn’t assume any knowledge about differential geometry. The most versatile way to do integration over manifolds is explained in the differential geometry section.

General Case¶

We want to integrate a function $f$ over a $k$ -manifold in $\R^n$ , parametrized as:

$\mathbf{\varphi}: \R^k \to \R^n\quad \mathbf{\varphi}(t_1, t_2, \dots, t_k) = \mat{\varphi_1(t_1, t_2, \dots, t_k)\cr \varphi_2(t_1, t_2, \dots, t_k)\cr \vdots \cr \varphi_n(t_1, t_2, \dots, t_k)\cr }$

then the integral of $f(x_1, x_2, \dots, x_n)$ over $\varphi$ is:

$\int_M f(x_1, x_2, \dots, x_n)\,\d S = \int_{\R^n} f(\mathbf{\varphi}(t_1, t_2, \dots, t_k))\sqrt{\det{\bf G}}\,\d t_1\d t_2\cdots\d t_k$

where ${\bf G}$ is called a Gram matrix and ${\bf J}$ is a Jacobian:

$({\bf G})_{ij} = ({\bf J}^T{\bf J})_{ij} = J_{ik}J_{kj} = {\partial\varphi_k\over\partial t_i} {\partial\varphi_k\over\partial t_j} ({\bf J})_{ij} = {\partial\varphi_i\over\partial t_j} = \mat{ {\partial\mathbf{\varphi}\over\partial t_1} & {\partial\mathbf{\varphi}\over\partial t_2} & \cdots & {\partial\mathbf{\varphi}\over\partial t_k} \cr \vdots & \vdots & \vdots & \vdots \cr \vdots & \vdots & \vdots & \vdots \cr \vdots & \vdots & \vdots & \vdots \cr }$

The idea behind this comes from the fact that the volume of the $k$ -dimensional parallelepiped spanned by the vectors

${\partial\mathbf{\varphi}\over\partial t_1}, \dots, {\partial\mathbf{\varphi}\over\partial t_k}$

is given by

$V = \sqrt{\det {\bf J}^T{\bf J}}$

where ${\bf J}$ is an $n\times k$ matrix having those vectors as its column vectors.

Example¶

Let’s integrate a function $f(x, y, z)$ over the surface of a sphere in 3D (e.g. $k=2$ and $n=3$ ):

$\mathbf{\varphi}(\theta, \phi) = \mat{ r\sin\theta\cos\phi \cr r\sin\theta\sin\phi \cr r\cos\theta \cr } {\bf J} = \mat{ -r\sin\theta\sin\phi & r\cos\theta\cos\phi \cr r\sin\theta\cos\phi & r\cos\theta\sin\phi \cr 0 & -r\sin\theta \cr } {\bf G} = {\bf J}^T {\bf J} = \mat{ -r\sin\theta\sin\phi & r\sin\theta\cos\phi & 0 \cr r\cos\theta\cos\phi & r\cos\theta\sin\phi & -r\sin\theta \cr } \mat{ -r\sin\theta\sin\phi & r\cos\theta\cos\phi \cr r\sin\theta\cos\phi & r\cos\theta\sin\phi \cr 0 & -r\sin\theta \cr } = \mat { r^2\sin^2\theta & 0 \cr 0 & r^2 \cr } \det{\bf G} = r^4\sin^2\theta \sqrt{\det{\bf G}} = r^2\sin\theta \int_M f(x, y, z) \d S = \int_{\R^n} f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)\, r^2\sin\theta\,\d\theta\,\d\phi = = \int_0^\pi\d\theta \int_0^{2\pi}\d\phi\, f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)\, r^2\sin\theta$

Let’s say we want to calculate the surface area of a sphere, so we set $f(x, y, z) = 1$ and get:

$\int_M \d S = \int_0^\pi\d\theta \int_0^{2\pi}\d\phi\, r^2\sin\theta = 2\pi r^2\int_0^\pi\d\theta \sin\theta = 4\pi r^2$

Special Cases¶

k = n¶

$\det{\bf G} = \det {\bf J}^R{\bf J} = (\det {\bf J})^2 \d S = |\det {\bf J}|\,\d t_1\,\d t_2\cdots\d t_k$

k = 1¶

$\det{\bf G} = \det \left( \left({\d\varphi_1\over\d t}\right)^2+ \left({\d\varphi_2\over\d t}\right)^2+ \cdots \right) = \left|{\d\mathbf{\varphi}\over\d t}\right|^2 \d S = \left|{\d\mathbf{\varphi}\over\d t}\right|\,\d t$

k = n - 1¶

$\det{\bf G} = \det {\bf J}^R{\bf J} = =\det(\cdots)^2 + \det(\cdots)^2+\cdots+\det(\dots)^2 = =\left|\det\mat{ {\partial\mathbf{\varphi}\over\partial t_1} & {\partial\mathbf{\varphi}\over\partial t_2} & \cdots & {\partial\mathbf{\varphi}\over\partial t_k} & {\bf e}_1 \cr \vdots & \vdots & \vdots & \vdots & {\bf e}_2 \cr \vdots & \vdots & \vdots & \vdots & \vdots \cr \vdots & \vdots & \vdots & \vdots & {\bf e}_n \cr }\right|^2 \equiv |\omega_\varphi|^2 \d S = |\omega_\varphi|\,\d t_1\,\d t_2\cdots\d t_k$

$\omega_\varphi$ is a generalization of a vector cross product. The $\det(\cdots)$ symbol means a determinant of a matrix with one row removed (first term in the sum has first row removed, second term has second row removed, etc.).

k = 2, n = 3¶

$\det{\bf G} = \left|{\partial\mathbf{\varphi}\over\partial t_1}\times {\partial\mathbf{\varphi}\over\partial t_2}\right|^2 \d S = \left|{\partial\mathbf{\varphi}\over\partial t_1}\times {\partial\mathbf{\varphi}\over\partial t_2}\right|\,\d t_1\,\d t_2$

y = f(x, z)¶

$\det{\bf G} = 1 +\left({\partial f\over\partial x}\right)^2 +\left({\partial f\over\partial z}\right)^2 \d S = \sqrt{1 +\left({\partial f\over\partial x}\right)^2 +\left({\partial f\over\partial z}\right)^2 }\,\d x\,\d z$

in general for $x_j = f(x_1, x_2, \dots, x_n)$ we get:

$\det{\bf G} = 1 +\left({\partial f\over\partial x_1}\right)^2 +\left({\partial f\over\partial x_2}\right)^2 +\cdots \d S = \sqrt{1 +\left({\partial f\over\partial x_1}\right)^2 +\left({\partial f\over\partial x_2}\right)^2 +\cdots }\,\d x_1\,\d x_2\cdots\d x_n$

The “ $x_j$ ” term is missing in the sums above.

Implicit Surface¶

For a surface given explicitly by

$F(x_1, x_2, ..., x_n) = 0$

we get:

$\d S = |\nabla F| \left|{\partial F\over\partial x_n}\right|\,\d x_1\cdots\d x_{n-1}$

Orthogonal Coordinates¶

If the coordinate vectors are orthogonal to each other:

${\partial\mathbf{\varphi}\over\partial t_i} \cdot {\partial\mathbf{\varphi}\over\partial t_i} = 0 \quad\text{for $i\neq j$}$

we get:

$\d S = \left|{\partial\mathbf{\varphi}\over\partial t_1}\right| \left|{\partial\mathbf{\varphi}\over\partial t_2}\right| \cdots \left|{\partial\mathbf{\varphi}\over\partial t_k}\right| \d t_1\cdots\d t_k$

Motivation¶

Let the $k$ -dimensional parallelepiped P be spanned by the vectors

${\partial\mathbf{\varphi}\over t_1}, \dots, {\partial\mathbf{\varphi}\over t_k}$

and let $\mathbf{J}$ is $n\times k$ matrix having these vectors as its column vectors. Then the area of P is

$V = \sqrt{\det {\bf J}^T{\bf J}}$

so the definition of the integral over a manifold is just approximating the surface by infinitesimal parallelepipeds and integrating over them.

Example¶

Let’s calculate the total distance traveled by a body in 1D, whose position is given by $s(t)$ :

$l = \int_\gamma \d s = \int_{t_1}^{t_2}\left|{\d s\over \d t}\right| \d t = = \int_{t_1}^{t'}\left|{\d s\over \d t}\right| \d t + \int_{t'}^{t''}\left|{\d s\over \d t}\right| \d t + \cdots + \int_{t''''^{\cdots}}^{t_2}\left|{\d s\over \d t}\right| \d t = =|s(t')-s(t_1)|+|s(t'')-s(t')|+\cdots+|s(t_2)-s(t''''^{\cdots})|$

where $t'$ , $t''$ , ... are all the points at which $\left|{\d s\over \d t}\right|=0$ , so each of the integrals in the above sum has either positive or negative integrand.

Integration¶