Throughout, ρ refers to the Euclidean distance from a given point to the origin, φ refers to the azimuthal angle in the reference plane from the reference direction to the point, and θ refers to the inclination angle of the point from the zenith direction. The zenith direction is the positive z axis, the xy-plane is the reference plane, and the positive x axis is the reference direction. I'm learning triple integrals in spherical coordinates, and I understand them well (ie. the Jacobian factor, substitution on bounds, order, etc.). I am struggling with understanding the reasoning for using particular bounds for traversing a full sphere. Let me explain with an example: $$\int\limits_0^{2\pi}\int\limits_0^{\pi}\int\limits_0^r\rho^2sin(\theta)d{\rho}d{\theta}d{\varphi}$$ This is how the volume of a sphere of radius r is represented as a triple integral in spherical coordinates. The radius goes from 0 to r, from the center to the surface of the sphere. The inclination angle goes from 0 to π, so at the end of the inner two integrals we have like a semicircular fin off the z axis along the x axis. Then the azimuthal angle ranges from 0 to 2π, rotating the "fin" all the way around the xy-plane, making the sphere. This triple integral gives $\frac{4}{3}{\pi}r^3$ as expected.
Why, however, can we not first go halfway around with the azimuthal angle, and then all the way around the inclination angle? (Or, more specifically for how I'm about to write it, all the way around the inclination angle to make a disc, and then halfway around the azimuthal angle to make a sphere. Same thing though, by Fubini's Theorem.) Like so: $$\int\limits_0^{\pi}\int\limits_0^{2\pi}\int\limits_0^r\rho^2sin(\theta)d{\rho}d{\theta}d{\varphi}$$ Yet this gives zero. So what's wrong? Why must the half rotation be the inclination angle and the full rotation be the azimuthal angle?