How to understand integral of wedge product?

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Taken from Bott and Tu (page 70):

Let $\sigma$ be the positive generator of $H^{n-1}(S^{n-1})$ and $\psi = \pi^* \sigma$ the corresponding generator of $H^{n-1}(\mathbf R^n - 0)$. [...] Let $d\rho = \rho'(r)\,dr$ be a bump form on $\mathbf R$ with total integral $1$. Then $(d\rho) \wedge \psi$ is a compactly supported form on $\mathbf R^n$ with total integral $1$.

If it's not already obvious, $\pi$ is the projection $\mathbf R^n - 0 \to S^{n-1}$ and $\rho$ is a function of the radius $r$ in $\mathbf R^n$.

Question: How am I to understand the last sentence intuitively?

So far, the picture in my head is for $n = 3$. I imagine $\sigma \in H^2(S^2)$ to be a function that eats a $2$-dimensional subset of $S^2$ and spits out its area. Then $\psi \in H^2(\mathbf R^3)$ is a function that eats a $2$-manifold in $\mathbf R^3$, projects it onto the unit sphere, and spits out its area. I'm not sure how these help me understand what $d\rho \wedge \psi$ is doing, or why it should integrate to $1$.