91Ó°ÊÓ

Since we are revolving around the \(x\)-axis, the height of each disk is given by \(f(x)\) and the radius is \(x\). The volume of each infinitesimally small disk is given by \(\pi(x)f(x)^2dx\). We will integrate this from \(0\) to \(\infty\) to find the total volume.

The integral for the volume is given by: $$V = \int_{0}^{\infty} \pi x\left(\frac{\sqrt{x}}{\sqrt[3]{x^{2}+1}}\right)^2 dx$$

Let's first simplify the integral: $$ V = \pi \int_{0}^{\infty} x\left(\frac{x}{x^2 + 1}\right) dx $$

Now we need to determine whether the integral converges or diverges. We can do this by comparing the integral to the integral of a simpler function that also goes to infinity. In this case, we choose the function \(\frac{x}{x^2}\), because this function is greater than \(\frac{x}{x^2+1}\) on the interval \([1,\infty)\). Let's consider the improper integral of this function: $$ \int_{1}^{\infty} \frac{x}{x^2} dx $$ To evaluate this integral, we find: $$\lim_{b\to\infty}\int_{1}^{b} \frac{x}{x^2} dx = \lim_{b\to\infty}\int_{1}^{b} \frac{1}{x} dx$$ $$= \lim_{b\to\infty}\left[ \ln{(|x|)} \right]_{1}^{b}$$ $$= \lim_{b\to\infty}\left[\ln{(|b|)}-\ln{(|1|)}\right] = \infty$$ Since the improper integral \(\int_{1}^{\infty} \frac{x}{x^2} dx\) diverges, and it is greater than the integral \(\int_{1}^{\infty} \frac{x}{x^2+1} dx\), it implies that the original integral also diverges.

Since the integral for the volume of the solid of revolution diverges, we can say that the volume does not exist.