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My question is why the volume of 3-sphere that lies between a rotation angle of $\theta$ and $\theta+d\theta$ is $$2\pi \sin^2(\theta/2) d\theta$$

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Since you talk about a "rotation angle", I presume that you're using the correspondence between unit quaternions, the 3-sphere and rotations in three-dimensional Euclidean space. (That should preferably have been provided as context in the question.)

The rotation angle $\theta$ of a rotation corresponds to an angle $\theta/2$ in the rotation's representation as a unit quaternion, which can be regarded as a vector on the 3-sphere (i.e. a 4-dimensional unit vector) forming an angle $\theta/2$ with the "real" axis. It's a bit confusing to call the volume element you're interested in "the volume of 3-sphere that lies between a rotation angle of $\theta$ and $\theta + d\theta$", since those are rotation angles, whereas in talking about the 3-sphere one would usually call this the volume of 3-sphere that lies between an angle of $\phi$ and $\phi+d\phi$, where $\phi=\theta/2$.

To calculate that volume, consider how you would calculate the surface of 2-sphere that lies between $\phi$ and $\phi+d\phi$. This is $d\phi$ times the circumference of the 1-sphere (circle) at $\phi$, which has radius $\sin\phi$ and hence circumference $2\pi\sin\phi$. Thus, the surface element of the 2-sphere between $\phi$ and $\phi+d\phi$ is $2\pi\sin\phi d\phi$.

Analogously, the volume of 3-sphere that lies between $\phi$ and $\phi+d\phi$ is $d\phi$ times the surface of the 2-sphere at $\phi$, which has radius $\sin\phi$ and hence surface $4\pi\sin^2\phi$. Thus, the volume element of the 3-sphere between $\phi$ and $\phi+d\phi$ is $4\pi\sin^2\phi d\phi$. Then you get your result by substituting $\phi=\theta/2$.

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