Final answer:
The integral of zdv over the region E bounded by two spheres in the first octant is evaluated using spherical coordinates. The radial distance varies from 5 to 6 and the angular coordinates vary from 0 to \(\pi/2\). After separation and integration, the final value simplifies to \(\frac{3}{2}(36\sqrt{36} - 25\sqrt{25})\).
Step-by-step explanation:
To evaluate the integral \(\int\int\int_E zdv\) in spherical coordinates where E lies between the spheres \(x^2+ y^2 + z^2 = 25\) and \(x^2 + y^2 + z^2 = 36\) in the first octant, we need to set the limits for the spherical coordinates: \(\rho, \theta, \phi\).
The region E is bounded by spheres of radii 5 and 6, so the radial coordinate \(\rho\) varies from 5 to 6. The first octant corresponds to angle limits of \(\theta\) from 0 to \(\pi/2\) and \(\phi\) from 0 to \(\pi/2\). The volume element in spherical coordinates is \(\rho^2 \sin(\phi) d\rho d\theta d\phi\). And since z = \rho \cos(\phi), the integral becomes:
\(\int_{0}^{\pi/2}\int_{0}^{\pi/2}\int_{5}^{6} (\rho^3 \cos(\phi) \sin(\phi)) d\rho d\theta d\phi\).
To solve this integral, we can separate it into the product of three integrals, one for each variable, and integrate each one independently.