In Calculus III class today we learned how to evaluate triple integrals in spherical coordinates. One of the example questions we worked on was to use a triple integral to find the volume of a shape.
Use spherical coordinates to find the volume of the region $Q$ bounded by the cylinders $r = 1$ and $r = 2$, and the cones $\varphi = \frac{\pi}{3}$ and $\varphi = \frac{\pi}{6}$.
The image provided by the book is.
While working ahead I found $dV = \rho^2 \sin\varphi d\rho d\varphi d\theta$, $\rho_1 = \csc\varphi$, and $\rho_2 = 2\csc\varphi$ and used this information to evaluate the integral:
$$\iiint_QdV = \int_0^{2\pi} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \int_{\csc\varphi}^{2\csc\varphi} \rho^2 \sin\varphi d\rho d\varphi d\theta = \frac{28\pi}{3\sqrt3}$$
The teacher finally caught up and agreed on everything I wrote, but she wrote the middle integral as $\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}$. This makes no sense to me because $\frac{\pi}{6} < \frac{\pi}{3}$ and when increasing the value of $\varphi$ $\frac{\pi}{6}$ is encountered first followed later by $\frac{\pi}{3}$.
Her argument for doing it this way is that $\frac{\pi}{3}$ is encountered first when traveling from the xy-plane in the positive z direction, which was what we considered when finding the limits of integration in rectangular coordinates. Doing it her way the answer is $-\frac{28\pi}{3\sqrt3}$ which is negative, and therefore doesn't represent volume. I don't think that her way is correct. Especially since it directly contradicts the questions we worked over before this one.
Which way is the right way? I want to be sure I understand this correctly.