91Ó°ÊÓ

We are given the following surfaces: 1. \(x^2 + y^2 = 1 + z^2\) 2. \(x^2 + y^2 = 2 - z^2\)

To find the region enclosed by these surfaces, we can set the equations equal to each other and solve for \(z\): \((1 + z^2) = (2 - z^2)\) \(2z^2 = 1\) \(z^2 = \frac{1}{2}\) \(z = ±\frac{1}{\sqrt{2}}\) The region lies between these two \(z\) values, starting at the origin (because we are given that the region contains the origin). So, the region of integration is \(-\frac{1}{\sqrt{2}} ≤ z ≤ \frac{1}{\sqrt{2}}\). We can find the corresponding bounds for \(x\) and \(y\) as follows: For the lower surface, \(x^2 + y^2 = 1 + z^2\) Since \(|z| ≤ \frac{1}{\sqrt{2}}\), we know that \(0 \le x^2 + y^2 \le 2\).

We will use cylindrical coordinates, which are of the form \((r, \theta, z)\). Recall that in cylindrical coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). Thus, the triple integral that represents the volume is: \(\int\int\int_{V} r \,dz\,dr\,d\theta\)

Using the bounds we found in Step 2, we can determine the bounds for the integral: 1. For the \(z\) coordinate: \(-\frac{1}{\sqrt{2}} ≤ z ≤ \frac{1}{\sqrt{2}}\) 2. For the \(r\) coordinate: \(0 \le r ≤ \sqrt{2 - z^2}\) 3. For the \(\theta\) coordinate: \(0 \le \theta ≤ 2\pi\)

Now we can plug in the bounds and evaluate the triple integral: \(\int_{- \frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{0}^{2\pi} \int_{0}^{\sqrt{2 - z^2}} r \,dr\,d\theta\,dz\) First, evaluate the inner integral: \(\int_{- \frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{0}^{2\pi} \frac{1}{2} (2 - z^2) \,d\theta\,dz\) Next, evaluate the middle integral: \(\int_{- \frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \pi (2 - z^2) \,dz\) Finally, evaluate the outer integral: \\ \(\pi \int_{- \frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (2 - z^2) \,dz = \pi [\frac{4}{3\sqrt{2}}]\) \\ So, the volume of the region enclosed by the surfaces \(x^2 + y^2 = 1 + z^2\) and \(x^2 + y^2 = 2 - z^2\) and containing the origin is \(\frac{4\pi}{3\sqrt{2}}\).