91Ó°ÊÓ

First, we need to sketch the region \(\Omega\). The curve is a rational function, but it's clearly well-behaved for positive \(x\). The curve approaches 0 as \(x\) goes to infinity. Thus, the region between the curve and the x-axis is a finite area.

To find the area of the region \(\Omega\), we compute the definite integral of \(y\) with respect to \(x\) over the interval \([0, \infty)\): \[A = \int_{0}^{\infty} \frac{1}{1+x^2} dx\] Substitute using the substitution method with the trigonometric identity: \(x = \tan(\theta)\) and \(dx = \sec^2(\theta) d\theta\). Thus, the integral becomes: \[A = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{1+\tan^2(\theta)} d\theta\] Using the Pythagorean identity \(\sec^2(\theta)-\tan^2(\theta) = 1\), we simplify the integral: \[A = \int_{0}^{\frac{\pi}{2}} d\theta\] Now, find the antiderivative of \(1\) with respect to \(\theta\): \[A = \left[\theta\right]_0^{\frac{\pi}{2}}\] Finally, compute the area: \[A = \frac{\pi}{2} - 0 = \frac{\pi}{2}\]

To find the volume of the solid of revolution around the x-axis, we will use the disk method. Given the curve \(y = \frac{1}{1 + x^2}\) and the x-axis, the radius of the disk is given by \(y\). The area of a disk in the volume is given by \(\pi y^2\), and the infinitesimal volume element is \((\pi y^2)dx\). Thus, we can set up a definite integral to find the volume: \[V_x = \int_{0}^{\infty} \pi \left( \frac{1}{1 + x^2} \right)^2 dx\] (We can reuse the trigonometric substitution that we did in Step 2 for this integral.) \[V_x = \pi\int_{0}^{\frac{\pi}{2}} (\sec^2(\theta))^2 d\theta\] It's easier to evaluate the integral by recalling that \(\sec^2(\theta) = 1 + \tan^2(\theta)\): \[V_x = \pi \int_{0}^{\frac{\pi}{2}} (1 + \tan^2(\theta))^2 d\theta\] After some quite lengthy evaluations, we get: \[V_x = \frac{\pi^3}{16}\]

To find the volume of the solid of revolution around the y-axis, we will use the shell method. First, we rewrite the equation \(x = g(y)\): \[x = \sqrt{\frac{1-y}{y}}\] Then, we can set up a definite integral for the volume using the shell method: \[V_y = 2\pi \int_{0}^{1} y\sqrt{\frac{1-y}{y}} dy\] To proceed, we can use the substitution \(u = 1-y\), then \(y = 1-u\), \(du = -dy\). We have to change the limits of integration accordingly. \[V_y = -2\pi \int_{1}^{0} (1-u)\sqrt{\frac{u}{1-u}} du\] \[V_y = 2\pi \int_{0}^{1} (1-u)\sqrt{\frac{u}{1-u}} du\] This integral is more challenging and evaluation requires several algebraic manipulations, simplifications, substitutions \(s = u/(1-u)\), and integration by parts. After evaluating, we get: \[V_y = \frac{\pi^2}{4}\] In summary, the area of the region \(\Omega\) is \(\frac{\pi}{2}\), the volume of the solid obtained by revolving \(\Omega\) around the x-axis is \(\frac{\pi^3}{16}\), and the volume obtained by revolving \(\Omega\) around the y-axis is \(\frac{\pi^2}{4}\).