Physlib.SpaceAndTime.Space.Integrals.Basic

For any dimension $d$, the volume measure on the $d$-dimensional space $\text{Space } d$ is equal to the additive Haar measure normalized with respect to the standard orthonormal basis of the space.

In the $(d+1)$-dimensional Euclidean space $\text{Space}(d+1)$, for any point $x$ and any strictly positive radius $r > 0$, the volume (Lebesgue measure) of the closed ball $\bar{B}(x, r) = \{y \in \text{Space}(d+1) \mid \text{dist}(x, y) \leq r\}$ is non-zero.

For any natural number $d$, let $x$ be a point in the $(d+1)$-dimensional real Euclidean space $\text{Space}(d+1)$ and $r$ be a real number. The volume (standard Lebesgue measure) of the closed ball centered at $x$ with radius $r$ is finite, i.e., $\text{volume}(\{y \in \text{Space}(d+1) \mid \text{dist}(x, y) \leq r\}) \neq \infty$.

In the 3D Euclidean space $\text{Space } 3$, the volume (Lebesgue measure) of the open unit ball centered at the origin, defined as $\{x \in \text{Space } 3 \mid \|x\| < 1\}$, is equal to $\frac{4}{3}\pi$.

In the 2-dimensional Euclidean space $\text{Space } 2$, the volume (Lebesgue measure) of the open unit ball centered at the origin with radius $1$, defined as $\{x \in \text{Space } 2 \mid \|x\| < 1\}$, is equal to $\pi$.

In the 2-dimensional Euclidean space $\text{Space } 2$, the real-valued volume (Lebesgue measure) of the open unit ball centered at the origin with radius $1$ is equal to $\pi$: $$\text{volume}_{\mathbb{R}}(\{x \in \text{Space } 2 \mid \|x\| < 1\}) = \pi$$ where the unit ball is defined as the set of points $x$ such that their norm $\|x\|$ is less than $1$, and $\text{volume}_{\mathbb{R}}$ denotes the volume expressed as a real number.

In the 3-dimensional Euclidean space $\text{Space } 3$, the real-valued volume (Lebesgue measure) of the open unit ball centered at the origin with radius $1$, defined as $\{x \in \text{Space } 3 \mid \|x\| < 1\}$, is equal to $\frac{4}{3}\pi$.

Let $f: \text{Space}(1) \to \mathbb{R}$ be a function. The integral of $f$ over the one-dimensional space $\text{Space}(1)$ with respect to the volume measure is equal to the integral of $f$ composed with the inverse of the linear isometric equivalence $\text{oneEquiv} : \text{Space}(1) \cong \mathbb{R}$ over the real numbers $\mathbb{R}$ with respect to the standard Lebesgue measure: $$\int_{\text{Space}(1)} f(x) \, dV = \int_{\mathbb{R}} f(\text{oneEquiv}^{-1}(x)) \, dx$$ where $\text{oneEquiv}^{-1}$ is the map that sends a real number $x$ to the corresponding vector in $\text{Space}(1)$.

For any natural number $d$, let $V = \text{Space}(d+1)$ be the $(d+1)$-dimensional real inner product space and $F$ be a normed real vector space. For a function $f: V \to F$, the integral of $f$ over $V$ with respect to the standard volume measure can be expressed in spherical coordinates as: $$\int_{V} f(x) \, dV(x) = \int_{S^d \times (0, \infty)} f(r \omega) \, d(\sigma \otimes \mu_d)(\omega, r)$$ where $S^d \subset V$ is the unit sphere, $\sigma$ is the surface measure on $S^d$, and $\mu_d$ is the radial measure on the interval $(0, \infty)$ defined by $d\mu_d(r) = r^d \, dr$, where $d$ is the dimension of the space minus one.

Let $\lambda$ be the standard Lebesgue measure (volume) on the real numbers $\mathbb{R}$. The measure on the interval of positive real numbers $(0, \infty)$ obtained by the pullback of $\lambda$ under the inclusion map $\iota: (0, \infty) \hookrightarrow \mathbb{R}$ is $s$-finite.

For any natural number $d$ and any measurable function $f: \text{Space}(d+1) \to [0, \infty]$, the Lebesgue integral of $f$ with respect to the volume measure on the $(d+1)$-dimensional Euclidean space $\text{Space}(d+1)$ is given by: $$\int_{\text{Space}(d+1)} f(x) \, d\text{vol} = \int_{S^d \times (0, \infty)} f(r\omega) \, d(\sigma \times \mu_d)$$ where $S^d$ is the unit sphere in $\text{Space}(d+1)$, $\omega \in S^d$ is a unit vector, $r \in (0, \infty)$ is the radius, $\sigma$ is the spherical measure, and $\mu_d$ is the radial measure on $(0, \infty)$ defined by $r^d dr$.

For any natural number $d$ and any measurable function $f: \text{Space}(d+1) \to [0, \infty]$, the Lebesgue integral of $f$ with respect to the volume measure on the $(d+1)$-dimensional real vector space $\text{Space}(d+1)$ is given by: $$\int_{\text{Space}(d+1)} f(x) \, d\text{vol} = \int_{S^d \times (0, \infty)} f(r\omega) \cdot r^d \, d(\sigma \times \lambda)$$ where $S^d$ is the unit sphere in $\text{Space}(d+1)$, $\sigma$ is the spherical measure on $S^d$, and $\lambda$ is the standard Lebesgue measure on the positive real line $(0, \infty)$. In this expression, $r \in (0, \infty)$ represents the radius and $\omega \in S^d$ is a unit vector.