**Proof plan.** We first compute $\omega_n = \mathcal{L}^n(B(0,1))$ by evaluating the Gaussian [integral](/page/Integral) $\int_{\mathbb{R}^n} e^{-|x|^2} \, d\mathcal{L}^n(x)$ in two ways: as a product of one-dimensional Gaussians (giving $\pi^{n/2}$), and via the [Coarea Formula](/theorems/24) in radial coordinates (giving $n\omega_n \cdot \Gamma(n/2)/2$). Equating yields $\omega_n$. The scaling $\mathcal{L}^n(B(0,R)) = \omega_n R^n$ follows from the homogeneity of Lebesgue measure. The surface area formula is then derived from the volume formula using the [Coarea Formula](/theorems/24). **Step 1: Compute the Gaussian integral via Fubini.** By Fubini's theorem and the one-dimensional Gaussian identity $\int_{-\infty}^\infty e^{-t^2} \, d\mathcal{L}^1(t) = \sqrt{\pi}$: \begin{align*} I_n := \int_{\mathbb{R}^n} e^{-|x|^2} \, d\mathcal{L}^n(x) = \prod_{i=1}^n \int_{-\infty}^\infty e^{-x_i^2} \, d\mathcal{L}^1(x_i) = \pi^{n/2}. \end{align*} **Step 2: Compute the Gaussian integral via the Coarea Formula.** Define $u: \mathbb{R}^n \to \mathbb{R}$ by $u(x) := |x|$. Then $\nabla u(x) = x/|x|$ for $x \neq 0$, so $|\nabla u(x)| = 1$ a.e. The level [set](/page/Set) $u^{-1}(r)$ is $\partial B(0, r)$. By the [Coarea Formula](/theorems/24): \begin{align*} I_n = \int_{\mathbb{R}^n} e^{-|x|^2} \cdot 1 \, d\mathcal{L}^n(x) = \int_0^\infty \left(\int_{\partial B(0, r)} e^{-r^2} \, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r) = \int_0^\infty e^{-r^2} \mathcal{H}^{n-1}(\partial B(0, r)) \, d\mathcal{L}^1(r). \end{align*} We do not yet know $\mathcal{H}^{n-1}(\partial B(0, r))$, but we can determine its dependence on $r$. By the scaling property of [Hausdorff measure](/page/Hausdorff%20Measure) — the dilation $x \mapsto rx$ maps $\partial B(0, 1)$ onto $\partial B(0, r)$ with Lipschitz constant $r$, so $\mathcal{H}^{n-1}(\partial B(0, r)) = r^{n-1} \mathcal{H}^{n-1}(\partial B(0, 1))$. Writing $\sigma_n := \mathcal{H}^{n-1}(\partial B(0, 1))$: \begin{align*} I_n = \sigma_n \int_0^\infty e^{-r^2} r^{n-1} \, d\mathcal{L}^1(r). \end{align*} The substitution $s = r^2$, $ds = 2r \, dr$, $r^{n-1} \, dr = \frac{1}{2} s^{(n-2)/2} \, ds$ gives \begin{align*} \int_0^\infty e^{-r^2} r^{n-1} \, d\mathcal{L}^1(r) = \frac{1}{2}\int_0^\infty s^{n/2 - 1} e^{-s} \, d\mathcal{L}^1(s) = \frac{1}{2}\Gamma(n/2). \end{align*} Therefore $I_n = \frac{\sigma_n}{2}\Gamma(n/2)$. **Step 3: Determine $\sigma_n$ and $\omega_n$.** Equating the two expressions for $I_n$: \begin{align*} \pi^{n/2} = \frac{\sigma_n}{2}\Gamma(n/2), \qquad \text{so} \qquad \sigma_n = \frac{2\pi^{n/2}}{\Gamma(n/2)}. \end{align*} For the volume of the unit ball, apply the [Coarea Formula](/theorems/24) to the constant [function](/page/Function) $f = 1$ over $B(0, 1)$: \begin{align*} \omega_n = \mathcal{L}^n(B(0, 1)) = \int_0^1 \mathcal{H}^{n-1}(\partial B(0, r)) \, d\mathcal{L}^1(r) = \sigma_n \int_0^1 r^{n-1} \, d\mathcal{L}^1(r) = \frac{\sigma_n}{n}. \end{align*} Substituting $\sigma_n = 2\pi^{n/2}/\Gamma(n/2)$: \begin{align*} \omega_n = \frac{2\pi^{n/2}}{n\Gamma(n/2)} = \frac{\pi^{n/2}}{\Gamma(n/2 + 1)}, \end{align*} where the last equality uses the functional equation $\Gamma(s + 1) = s\Gamma(s)$ with $s = n/2$. **Step 4: Scaling to radius $R$.** The dilation $x \mapsto Rx$ maps $B(0, 1)$ onto $B(0, R)$. Since Lebesgue measure scales as $\mathcal{L}^n(RB) = R^n \mathcal{L}^n(B)$: \begin{align*} \mathcal{L}^n(B(0, R)) = R^n \omega_n. \end{align*} Similarly, $\mathcal{H}^{n-1}(\partial B(0, R)) = R^{n-1} \sigma_n = n\omega_n R^{n-1}$, completing the proof. **Step 5: Verification for low dimensions.** For $n = 1$: $\omega_1 = \pi^{1/2}/\Gamma(3/2) = \sqrt{\pi}/(\sqrt{\pi}/2) = 2$, so $\mathcal{L}^1(B(0,R)) = 2R$ (the interval $(-R, R)$) and $\mathcal{H}^0(\partial B(0,R)) = 1 \cdot 2 \cdot R^0 = 2$ (two points). For $n = 2$: $\omega_2 = \pi/\Gamma(2) = \pi$, so $\mathcal{L}^2(B(0,R)) = \pi R^2$ and $\mathcal{H}^1(\partial B(0,R)) = 2\pi R$. For $n = 3$: $\omega_3 = \pi^{3/2}/\Gamma(5/2) = \pi^{3/2}/(3\sqrt{\pi}/4) = 4\pi/3$, so $\mathcal{L}^3(B(0,R)) = \frac{4}{3}\pi R^3$ and $\mathcal{H}^2(\partial B(0,R)) = 4\pi R^2$.