91Ó°ÊÓ

First, we express the given equations in cylindrical coordinates, where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The equation for the sphere \(x^2 + y^2 + z^2 = 2\) becomes \(r^2 + z^2 = 2\). The equation for the paraboloid \(z = x^2 + y^2\) becomes \(z = r^2\) in cylindrical coordinates.

The region is bounded above by the sphere and below by the paraboloid. Thus, the bounds for \(z\) are from the paraboloid to the sphere: \(z = r^2\) (lower surface) to \(z = \sqrt{2 - r^2}\) (upper surface).

To find the \(r\) limits, note that the paraboloid intersects the sphere where \(r^2 = z\) and \(z^2 = 2 - r^2\). Solving \(r^2 = \sqrt{2 - r^2}\) \(\Rightarrow\) \(2r^2 = 2\) yields \(r^2 = 1\), so \(r\) ranges from 0 to 1.

Since the problem is rotationally symmetric around the z-axis, \(\theta\) ranges from 0 to \(2\pi\).

The volume is found by integrating in cylindrical coordinates where volume \(dV = r\, dz\, dr\, d\theta\). Setup the integral as: \[ V = \int_{0}^{2\pi} \int_{0}^{1} \int_{r^2}^{\sqrt{2 - r^2}} r\, dz\, dr\, d\theta\]Evaluate the integral step by step, integrating with respect to \(z\) first:\[= \int_{0}^{2\pi} \int_{0}^{1} \left[ r(z) \right]_{r^2}^{\sqrt{2 - r^2}} \, dr \, d\theta= \int_{0}^{2\pi} \int_{0}^{1} r(\sqrt{2-r^2} - r^2) \, dr \, d\theta\]Calculate the integral with respect to \(r\):\[= \int_{0}^{2\pi} \left[ -\frac{1}{3}(2 - r^2)^{3/2} + \frac{r^5}{5} \right]_{0}^{1} \, d\theta= \int_{0}^{2\pi} \left( -\frac{1}{3} + \frac{1}{5} \right) \, d\theta= \int_{0}^{2\pi} \left( -\frac{2}{15} \right) \, d\theta\]Finally, integrate with respect to \(\theta\):\[= -\frac{2}{15} \cdot 2\pi = -\frac{4\pi}{15}\]Since negative volume doesn't make sense, we take the absolute value: \[V = \frac{4\pi}{15}\]

The calculated volume of the region between the spherical surface and the paraboloid is \(\frac{4\pi}{15}\). The negative sign arose from taking the bounded region; however, we interpret volume as a positive quantity.