## Motivation

In the study of Partial Differential Equations, particularly when dealing with the Laplace or [Wave equations](/page/Wave%20Equation), we often encounter [functions](/page/Function) that depend solely on the distance from the origin. These **radial functions** possess symmetries that drastically simplify integration over balls in $\mathbb{R}^n$. To rigorously compute integrals like $\int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x)$, we must employ tools from Geometric Measure Theory, specifically the Coarea and Area formulas. These formulas generalize standard calculus techniques like polar coordinates to arbitrary dimensions and geometries.

## Formal Definition

We define the open ball and the concept of a radial function.

[definition:Radial Function] Let $n \in \mathbb{N}$ with $n \ge 2$. The **open ball** of radius $R > 0$ centered at the origin is the set $B(0,R) \subseteq \mathbb{R}^n$ defined by: \begin{align*} B(0,R) &:= \{ x \in \mathbb{R}^n : |x| < R \}. \end{align*} A function $g: B(0,R) \to \mathbb{R}$ is called **radial** if there exists a profile function $f: [0, R) \to \mathbb{R}$ such that: \begin{align*} g(x) &= f(|x|) \quad \text{for all } x \in B(0,R). \end{align*} We assume $f$ is Borel measurable and integrable with respect to the relevant measures. [/definition]

## Examples

We demonstrate two distinct rigorous methods to reduce the $n$-dimensional integral of a radial function to a 1-dimensional integral.

[example:Method Via Coarea] **Method 1: Level Set Analysis.** We utilize the [Coarea Formula (Classical)](/theorems/24) to disintegrate the domain $B(0,R)$ into level [sets](/page/Set) of the distance function. Define the Lipschitz mapping $u: \mathbb{R}^n \to \mathbb{R}$ by: \begin{align*} u(x) &:= |x|. \end{align*} The gradient is $\nabla u(x) = \frac{x}{|x|}$ for $x \neq 0$, implying $|\nabla u(x)| = 1$ almost everywhere. Applying the Coarea formula to the function $g(x) = f(|x|)$ over the domain $B(0,R)$: \begin{align*} \int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x) &= \int_{B(0,R)} f(u(x)) |\nabla u(x)| \, d\mathcal{L}^n(x) \\ &= \int_{-\infty}^{\infty} \left( \int_{u^{-1}(r) \cap B(0,R)} f(r) \, d\mathcal{H}^{n-1}(y) \right) dr. \end{align*} The level set $u^{-1}(r)$ is the sphere $\partial B(0,r)$. For $r \in (0, R)$, this intersection is the full sphere; otherwise it is empty. Thus: \begin{align*} \int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x) &= \int_{0}^{R} f(r) \mathcal{H}^{n-1}(\partial B(0,r)) \, dr. \end{align*} [/example]

[example:Method Via Area] **Method 2: Parametrization (Change of Variables).** We derive the result using the [Change of Variables (general)](/theorems/22). **Step 1: Constructing the Chart (Almost Everywhere).** A global chart for the sphere $S^{n-1}$ does not exist; however, we can parametrize "almost all" of the sphere, which is sufficient for Lebesgue integration. Let $N \subset S^{n-1}$ be a [closed set](/page/Closed%20Set) of measure zero (e.g., the "cut" removing the date line). Let $U \subset \mathbb{R}^{n-1}$ be an [open set](/page/Open%20Set) and let $\omega: U \to S^{n-1} \setminus N$ be a smooth diffeomorphism (a local coordinate chart). We define the mapping $\Psi: (0, R) \times U \to B(0,R) \setminus \Sigma$ by: \begin{align*} \Psi(r, \theta) &:= r \omega(\theta), \end{align*} where $\Sigma$ is the set of measure zero corresponding to the radial projection of $N$. $\Psi$ is a diffeomorphism between open sets. **Step 2: The Jacobian Computation.** We compute the differential $D\Psi$. The tangent space of the domain is $\mathbb{R} \times \mathbb{R}^{n-1}$. \begin{align*} \frac{\partial \Psi}{\partial r} &= \omega(\theta) \in \mathbb{R}^n \\ \frac{\partial \Psi}{\partial \theta_j} &= r \frac{\partial \omega}{\partial \theta_j} \in \mathbb{R}^n \quad \text{for } j=1, \dots, n-1. \end{align*} Since $\omega(\theta) \in S^{n-1}$, it is orthogonal to its tangent vectors $\frac{\partial \omega}{\partial \theta_j}$. Thus, the matrix $D\Psi$ has orthogonal columns (if the chart $\omega$ is orthogonal) or block structure. The volume element transforms as: \begin{align*} |\det D\Psi(r, \theta)| &= \left| \det \left( \omega(\theta), r\frac{\partial \omega}{\partial \theta_1}, \dots, r\frac{\partial \omega}{\partial \theta_{n-1}} \right) \right|. \end{align*} Using linearity of the determinant in columns, we factor out $r$ from the $n-1$ angular columns: \begin{align*} |\det D\Psi(r, \theta)| &= r^{n-1} \left| \det \left( \omega(\theta), \frac{\partial \omega}{\partial \theta_1}, \dots, \frac{\partial \omega}{\partial \theta_{n-1}} \right) \right|. \end{align*} The remaining determinant is exactly the surface area element of the unit sphere in the chart $\omega$, denoted $J_\omega(\theta)$. Thus: \begin{align*} |\det D\Psi(r, \theta)| &= r^{n-1} J_\omega(\theta). \end{align*} **Step 3: Integral Transformation and Fubini's Theorem.** By the Change of Variables theorem: \begin{align*} \int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x) &= \int_{(0,R) \times U} f(|\Psi(r,\theta)|) |\det D\Psi(r,\theta)| \, d\mathcal{L}^n(r,\theta). \end{align*} Note that $|\Psi(r,\theta)| = r$. Substituting the Jacobian: \begin{align*} \dots &= \int_{(0,R) \times U} f(r) r^{n-1} J_\omega(\theta) \, d\mathcal{L}^n(r,\theta). \end{align*} The domain $(0,R) \times U$ is a product space. We apply **Fubini's Theorem** to convert the single integral over the product space into iterated integrals: \begin{align*} \dots &= \int_0^R \left( \int_{U} f(r) r^{n-1} J_\omega(\theta) \, d\theta \right) dr. \end{align*} Since $f(r)r^{n-1}$ does not depend on $\theta$, we factor it out: \begin{align*} \dots &= \int_0^R f(r) r^{n-1} \left( \int_{U} J_\omega(\theta) \, d\theta \right) dr. \end{align*} The inner integral $\int_U J_\omega(\theta) \, d\theta$ is precisely the definition of the surface area $\mathcal{H}^{n-1}(S^{n-1} \setminus N)$. Since $N$ has measure zero, this equals $\mathcal{H}^{n-1}(S^{n-1})$. \begin{align*} \int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x) &= \mathcal{H}^{n-1}(S^{n-1}) \int_0^R f(r) r^{n-1} \, dr. \end{align*} [/example]

## Key Results

The examples above converge to the fundamental integration formula for radial functions, linking the volume of the ball to the surface area of the sphere.

[theorem:Polar Coordinates Formula] **Hypothesis:** Let $f: [0, R) \to \mathbb{R}$ be such that $x \mapsto f(|x|)$ is in $L^1(B(0,R))$. **Conclusion:** \begin{align*} \int_{B(0,R)} f(|x|) \, d\mathcal{L}^n(x) &= n \alpha(n) \int_0^R f(r) r^{n-1} \, dr, \end{align*} where $\alpha(n) = \mathcal{L}^n(B(0,1))$ is the volume of the unit ball. This result relies on the identity $\mathcal{H}^{n-1}(\partial B(0,r)) = n \alpha(n) r^{n-1}$. [/theorem]

**Note on Rigor:** It is not necessarily true that radial symmetry implies [integrability](/page/Integral). *Counter-example:* Let $f(r) = r^{-n}$ on $B(0,1)$. While radially symmetric, the integral $\int_0^1 r^{-n} r^{n-1} \, dr = \int_0^1 r^{-1} \, dr$ diverges, so $f(|x|) \notin L^1(B(0,1))$.

## Problems

[problem] Compute the value of the integral $I = \int_{B(0,1)} |x|^2 \, d\mathcal{L}^3(x)$ using the polar coordinates formula. [/problem]

[solution] **Step 1:** Identify the parameters. Here $n=3$, $R=1$, and the radial profile is $f(r) = r^2$. The volume of the unit ball in $\mathbb{R}^3$ is $\alpha(3) = \frac{4}{3}\pi$. **Step 2:** Apply the Polar Coordinates Formula. Using [theorem:Polar Coordinates Formula]: \begin{align*} I &= 3 \alpha(3) \int_0^1 f(r) r^{3-1} \, dr \\ &= 3 \left( \frac{4}{3}\pi \right) \int_0^1 r^2 \cdot r^2 \, dr. \end{align*} **Step 3:** Evaluate the 1D integral. \begin{align*} I &= 4\pi \int_0^1 r^4 \, dr \\ &= 4\pi \left[ \frac{r^5}{5} \right]_0^1 \\ &= \frac{4\pi}{5}. \end{align*} [/solution]

## References

1. L.C. Evans, *Partial Differential Equations* (1998). 2. H. Federer, *Geometric Measure Theory* (1969).