[proofplan] We first compute $d\omega$ directly and show that the terms from differentiating the numerator cancel exactly with the terms from differentiating the radial factor $(x^2+y^2+z^2)^{-3/2}$. To prove non-exactness, we restrict $\omega$ to the unit sphere $S^2 \subset \mathbb{R}^3_0$ and identify the restriction with the positively oriented area form on $S^2$. Its integral is $4\pi$, whereas [Stokes' theorem](/theorems/1530) on the closed oriented surface $S^2$ would force the integral of an exact $2$-form to vanish. [/proofplan]

[step:Differentiate the solid angle form and prove it is closed]Let \begin{align*} r: \mathbb{R}^3_0 &\to (0,\infty) \\ (x,y,z) &\mapsto (x^2+y^2+z^2)^{1/2} \end{align*} and define the smooth function \begin{align*} f: \mathbb{R}^3_0 &\to (0,\infty) \\ (x,y,z) &\mapsto r(x,y,z)^{-3}. \end{align*} Also define the smooth $2$-form $\alpha \in \Omega^2(\mathbb{R}^3_0)$ by \begin{align*} \alpha := x\,dy \wedge dz + y\,dz \wedge dx + z\,dx \wedge dy. \end{align*} Then $\omega = f\alpha$. Since \begin{align*} df = -3r^{-5}(x\,dx+y\,dy+z\,dz), \end{align*} and \begin{align*} d\alpha &= dx \wedge dy \wedge dz + dy \wedge dz \wedge dx + dz \wedge dx \wedge dy \\ &= 3\,dx \wedge dy \wedge dz, \end{align*} the graded Leibniz rule gives \begin{align*} d\omega &= d(f\alpha) \\ &= df \wedge \alpha + f\,d\alpha. \end{align*} We compute the first term. All wedge products with repeated factors vanish, so \begin{align*} df \wedge \alpha &= -3r^{-5}(x\,dx+y\,dy+z\,dz) \wedge (x\,dy \wedge dz + y\,dz \wedge dx + z\,dx \wedge dy) \\ &= -3r^{-5}(x^2+y^2+z^2)\,dx \wedge dy \wedge dz \\ &= -3r^{-3}\,dx \wedge dy \wedge dz. \end{align*} The second term is \begin{align*} f\,d\alpha = 3r^{-3}\,dx \wedge dy \wedge dz. \end{align*} Therefore \begin{align*} d\omega = -3r^{-3}\,dx \wedge dy \wedge dz + 3r^{-3}\,dx \wedge dy \wedge dz = 0. \end{align*} Thus $\omega$ is closed.[/step]

[guided]The form $\omega$ is a product of a scalar radial factor and a polynomial-looking $2$-form. We isolate those two pieces because the [exterior derivative](/theorems/1525) obeys a product rule. Define \begin{align*} r: \mathbb{R}^3_0 &\to (0,\infty) \\ (x,y,z) &\mapsto (x^2+y^2+z^2)^{1/2} \end{align*} and \begin{align*} f: \mathbb{R}^3_0 &\to (0,\infty) \\ (x,y,z) &\mapsto r(x,y,z)^{-3}. \end{align*} Define the smooth $2$-form $\alpha \in \Omega^2(\mathbb{R}^3_0)$ by \begin{align*} \alpha := x\,dy \wedge dz + y\,dz \wedge dx + z\,dx \wedge dy. \end{align*} Then $\omega=f\alpha$. We first differentiate the scalar factor. Since $f=(x^2+y^2+z^2)^{-3/2}$, the chain rule gives \begin{align*} df = -3r^{-5}(x\,dx+y\,dy+z\,dz). \end{align*} Next we differentiate $\alpha$. The coordinate $1$-forms $dx,dy,dz$ are closed, so only the coefficient functions contribute: \begin{align*} d\alpha &= d(x\,dy \wedge dz) + d(y\,dz \wedge dx) + d(z\,dx \wedge dy) \\ &= dx \wedge dy \wedge dz + dy \wedge dz \wedge dx + dz \wedge dx \wedge dy. \end{align*} The permutations $(dy,dz,dx)$ and $(dz,dx,dy)$ are cyclic permutations of $(dx,dy,dz)$, hence have positive sign. Therefore \begin{align*} d\alpha = 3\,dx \wedge dy \wedge dz. \end{align*} Now apply the graded Leibniz rule to $\omega=f\alpha$: \begin{align*} d\omega = d(f\alpha) = df \wedge \alpha + f\,d\alpha. \end{align*} The purpose of the computation is to see that these two terms cancel. In $df\wedge\alpha$, every wedge product containing a repeated coordinate $1$-form is zero. Hence the only surviving products are \begin{align*} x\,dx \wedge x\,dy \wedge dz,\qquad y\,dy \wedge y\,dz \wedge dx,\qquad z\,dz \wedge z\,dx \wedge dy. \end{align*} Each of these is a positive multiple of $dx \wedge dy \wedge dz$, so \begin{align*} df \wedge \alpha &= -3r^{-5}(x^2+y^2+z^2)\,dx \wedge dy \wedge dz \\ &= -3r^{-3}\,dx \wedge dy \wedge dz. \end{align*} On the other hand, \begin{align*} f\,d\alpha = 3r^{-3}\,dx \wedge dy \wedge dz. \end{align*} Adding the two contributions gives \begin{align*} d\omega = -3r^{-3}\,dx \wedge dy \wedge dz + 3r^{-3}\,dx \wedge dy \wedge dz = 0. \end{align*} Thus $\omega$ is closed.[/guided]

[step:Identify the restriction of $\omega$ to the unit sphere with the oriented area form]Let \begin{align*} i:S^2 &\to \mathbb{R}^3_0 \\ p &\mapsto p \end{align*} be the inclusion map, where $S^2=\{p\in\mathbb{R}^3: |p|=1\}$ is oriented by the outward unit normal. Let $\sigma \in \Omega^2(S^2)$ denote the oriented area form characterized by \begin{align*} \sigma_p(u,v)=\det(p,u,v) \end{align*} for every $p\in S^2$ and every $u,v\in T_pS^2$, where $\det(p,u,v)$ is computed using the standard orientation of $\mathbb{R}^3$. We claim that $i^*\omega=\sigma$. Fix $p=(x,y,z)\in S^2$ and tangent vectors $u,v\in T_pS^2$. Since $|p|=1$, the denominator in $\omega_p$ is equal to $1$. Evaluating the numerator on $(u,v)$ gives \begin{align*} (i^*\omega)_p(u,v) &= \omega_p(u,v) \\ &= x\,(dy\wedge dz)_p(u,v) + y\,(dz\wedge dx)_p(u,v) + z\,(dx\wedge dy)_p(u,v) \\ &= \det(p,u,v) \\ &= \sigma_p(u,v). \end{align*} Therefore $i^*\omega=\sigma$.[/step]

[guided]We now restrict the form to the unit sphere. The inclusion map is \begin{align*} i:S^2 &\to \mathbb{R}^3_0 \\ p &\mapsto p. \end{align*} The sphere $S^2=\{p\in\mathbb{R}^3: |p|=1\}$ is oriented by the outward unit normal. Define the oriented area form $\sigma \in \Omega^2(S^2)$ by \begin{align*} \sigma_p(u,v)=\det(p,u,v), \end{align*} for $p\in S^2$ and $u,v\in T_pS^2$. This formula says that the ordered pair $(u,v)$ is positively oriented in the tangent plane exactly when the ordered triple $(p,u,v)$ is positively oriented in $\mathbb{R}^3$. We prove $i^*\omega=\sigma$ by evaluating both forms on arbitrary tangent vectors. Fix $p=(x,y,z)\in S^2$ and $u,v\in T_pS^2$. Since $p$ lies on the unit sphere, \begin{align*} (x^2+y^2+z^2)^{3/2}=1. \end{align*} Thus the denominator in $\omega_p$ disappears on $S^2$. By the definition of pullback, \begin{align*} (i^*\omega)_p(u,v) = \omega_p(di_p(u),di_p(v)). \end{align*} The differential $di_p:T_pS^2\to T_p\mathbb{R}^3\cong\mathbb{R}^3$ is the inclusion of the tangent plane, so we may write this as $\omega_p(u,v)$. Expanding the $2$-form gives \begin{align*} (i^*\omega)_p(u,v) &= x\,(dy\wedge dz)_p(u,v) + y\,(dz\wedge dx)_p(u,v) + z\,(dx\wedge dy)_p(u,v). \end{align*} This is precisely the [cofactor expansion](/theorems/398) of the determinant of the matrix with columns $p,u,v$, hence \begin{align*} (i^*\omega)_p(u,v)=\det(p,u,v)=\sigma_p(u,v). \end{align*} Since $p,u,v$ were arbitrary, $i^*\omega=\sigma$ on all of $S^2$.[/guided]

[step:Use the nonzero spherical integral to rule out exactness]Let $\mathcal{H}^2$ denote the $2$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $S^2$, and let $\mu_{S^2}$ denote the oriented surface measure determined by the outward orientation and the area form $\sigma$. Thus a top-degree form $\tau\in\Omega^2(S^2)$ is integrated by writing $\tau=h\sigma$ for the unique smooth function $h:S^2\to\mathbb{R}$ and setting \begin{align*} \int_{S^2} \tau\,d\mu_{S^2} := \int_{S^2} h\,d\mathcal{H}^2. \end{align*} Since $i^*\omega=\sigma$ and $\sigma$ is the oriented area form on the unit sphere, the representing function is $h=1$, so \begin{align*} \int_{S^2} i^*\omega\,d\mu_{S^2} = \int_{S^2} 1\,d\mathcal{H}^2 = \mathcal{H}^2(S^2) = 4\pi. \end{align*} Suppose, for contradiction, that there exists a smooth $1$-form $\eta\in\Omega^1(\mathbb{R}^3_0)$ such that $\omega=d\eta$. Pullback commutes with exterior differentiation, so \begin{align*} i^*\omega = i^*(d\eta) = d(i^*\eta). \end{align*} The form $i^*\eta$ is a smooth $1$-form on the compact oriented manifold $S^2$, and $\partial S^2=\varnothing$. By Stokes' theorem for compact oriented manifolds with boundary (citing a result not yet in the wiki: Stokes' theorem), \begin{align*} \int_{S^2} i^*\omega\,d\mu_{S^2} = \int_{S^2} d(i^*\eta)\,d\mu_{S^2} = \int_{\partial S^2} i^*\eta\,d\mu_{\partial S^2} = 0, \end{align*} where $\mu_{\partial S^2}$ denotes the boundary orientation measure, and the last equality holds because $\partial S^2=\varnothing$. This contradicts $\int_{S^2} i^*\omega\,d\mu_{S^2}=4\pi$. Therefore no smooth $1$-form $\eta\in\Omega^1(\mathbb{R}^3_0)$ satisfies $\omega=d\eta$, and $\omega$ is not exact.[/step]

[guided]The obstruction to exactness is detected by integrating over the unit sphere. Let $\mathcal{H}^2$ denote the $2$-dimensional Hausdorff measure on $S^2$, and let $\mu_{S^2}$ denote the oriented surface measure determined by the outward orientation and the area form $\sigma$. Concretely, if $\tau\in\Omega^2(S^2)$ is written as $\tau=h\sigma$ for the unique smooth function $h:S^2\to\mathbb{R}$, then \begin{align*} \int_{S^2} \tau\,d\mu_{S^2} := \int_{S^2} h\,d\mathcal{H}^2. \end{align*} From the previous step, the pullback of $\omega$ to $S^2$ is the oriented area form $\sigma$. Therefore $i^*\omega=1\cdot\sigma$, and hence \begin{align*} \int_{S^2} i^*\omega\,d\mu_{S^2} = \int_{S^2} 1\,d\mathcal{H}^2 = \mathcal{H}^2(S^2) = 4\pi. \end{align*} Now assume, toward a contradiction, that $\omega$ is exact on $\mathbb{R}^3_0$. This means there is a smooth $1$-form $\eta\in\Omega^1(\mathbb{R}^3_0)$ such that \begin{align*} \omega=d\eta. \end{align*} Pulling back along the inclusion $i:S^2\to\mathbb{R}^3_0$ gives \begin{align*} i^*\omega = i^*(d\eta). \end{align*} Pullback commutes with exterior differentiation, so \begin{align*} i^*(d\eta)=d(i^*\eta). \end{align*} Thus the restricted $2$-form $i^*\omega$ would be exact on $S^2$: \begin{align*} i^*\omega=d(i^*\eta). \end{align*} We now apply Stokes' theorem for compact oriented manifolds with boundary (citing a result not yet in the wiki: Stokes' theorem). Its hypotheses are satisfied: $S^2$ is a compact oriented smooth manifold, $i^*\eta$ is a smooth $1$-form on $S^2$, and the boundary of the closed manifold $S^2$ is empty. Therefore \begin{align*} \int_{S^2} d(i^*\eta)\,d\mu_{S^2} = \int_{\partial S^2} i^*\eta\,d\mu_{\partial S^2} = 0, \end{align*} where $\mu_{\partial S^2}$ denotes the boundary orientation measure, and the last equality holds because $\partial S^2=\varnothing$. Combining this with $i^*\omega=d(i^*\eta)$ gives \begin{align*} \int_{S^2} i^*\omega\,d\mu_{S^2}=0. \end{align*} This contradicts the already computed value \begin{align*} \int_{S^2} i^*\omega\,d\mu_{S^2}=4\pi. \end{align*} The contradiction proves that $\omega$ is not exact on $\mathbb{R}^3_0$.[/guided]