I find that the Torricelli's trumpet has very interesting properties.
It's surface:
$$\int_{1}^{\infty}\frac{2\pi}{x}dx=2\pi\ln{x}|_{1}^{\infty}\rightarrow\infty.$$
is infinity.
Its volume:
$$\int_{1}^{\infty}\frac{\pi}{x^{2}}dx=-\frac{\pi}{x}|_{1}^{\infty}=\pi.$$
is finite.
I extend it so its radius will be:
$$y=\frac{1}{\sqrt{x^2+c}},c\ is\ a\ constant.$$
Its surface will be:
$$\int_{1}^{\infty}\frac{2\pi}{\sqrt{x^2+c}}dx=2\pi\ln \left|\frac{x}{\sqrt{c}}+\sqrt{\frac{c+x^2}{c}}\right||_1^\infty\rightarrow\ diverges.$$
Then its volume:
$$\int_{1}^{\infty}\frac{\pi}{x^2+c}dx=\frac{\pi ^2}{2\sqrt{c}}-\frac{\pi \arctan \left(\sqrt{\frac{1}{c}}\right)}{\sqrt{c}}\rightarrow\ converges.$$
I tried to extend further for the radius:
$$y=\frac{1}{\sqrt[3]{x^3+c}}.\\$$
Its surface: $$\int_{1}^{\infty}\ \frac{2\pi}{\sqrt[3]{x^3+c}} dx.\\$$
Its volume: $$ \int_{1}^{\infty}\ \frac{\pi}{(x^3+c)^{2/3}} dx.\\$$
I can't find a general formula for
$$\int_\ \frac{1}{\sqrt[3]{x^3+c}} dx.\\$$
If c is polynomial and its degree will be less then the x term before it.
How would I do a general expression for:
$$\int_\ \frac{1}{\sqrt[n]{ax^n+bx^{n-1}+\cdots+z}} dx.\\$$
General indefinite integrals of the form $$\int \frac{dx}{\sqrt[n]{a_n x^n + \cdots + a_1 x + a_0}}, \qquad a_n \neq 0,$$ do not have closed forms when $n > 2$.
For the particular integral $$\int \frac{dx}{\sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = \frac{c}{1 - u^3}$$ gives $$\int \frac{dx}{\sqrt[3]{x^3 + c}} = \int \frac{u \,du}{1 - u^3} ,$$ which can be handled by standard techniques.
On the other hand, as $x \to \infty$, $\frac{1}{\sqrt[n]{a_n x^n + \cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$\frac{1}{\sqrt[n]{a_n}} \frac{1}{x},$$ and so the definite integrals of the form $$\int_M^{\infty} \frac{dx}{\sqrt[n]{a_n x^n + \cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.