91Ó°ÊÓ

The integral limits are given as 2 to 5 for \(\rho\), 0 to \(\pi\) for \(\phi\), and 0 to \(2\pi\) for \(\theta\). This suggests that the solid region defined is a portion of a sphere where the radius \(\rho\) varies from 2 to 5, \(\phi\) varies from 0 to \(\pi\) (covering the full spherical spread vertically), and \(\theta\) varies from 0 to \(2\pi\) (covering full rotation around the z-axis). So, the volume described is actually that of a spherical shell (like a hollow ball) with inner radius 2 and outer radius 5.

We now need to evaluate the given integral. As our function is \(\rho^{2} \sin \phi\), our integral becomes: \[\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{2}^{5} \rho^{2} \sin \phi d \rho d \phi d \theta\]First, let's integrate with respect to \(\rho\): It is a simple power rule application and the integral of \(\rho^{2}\) is \(\frac{\rho^{3}}{3}\).Therefore, we have:\[\int_{2}^{5} \rho^{2} d\rho = [\frac{\rho^{3}}{3}]_{2}^{5} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3}\]The original integral simplifies to:\[\int_{0}^{2 \pi} \int_{0}^{\pi} \frac{117}{3} \sin \phi d\phi d\theta\]Next, integrate with respect to \(\phi\). The integral of \(\sin \phi\) is \(-\cos \phi\).Therefore, we have:\[\int_{0}^{\pi} \frac{117}{3} \sin \phi d\phi = -\frac{117}{3} [\cos \phi]_{0}^{\pi} = -\frac{117}{3} (-2) = \frac{234}{3}\]The integral is further simplified to:\[\int_{0}^{2 \pi} \frac{234}{3} d\theta\]Finally, integrate:\[\int_{0}^{2 \pi} \frac{234}{3} d\theta = [\frac{234}{3} \theta]_{0}^{2 \pi} = \frac{468 \pi}{3}\]

The volume of the solid region defined by the given iterated integral is \(\frac{468 \pi}{3}\).