[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.
[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]
[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 $i^*\omega=\sigma$.
[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]
[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.
[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]
[/step]
