Shell surface of hyperboloid

Question

I need to calculate the shell surface of a hyperbola rotating around the $x$-axis and therefore generating a hyperboloid. To calculate the shell surface I should use the integral formula for the shell surface. The equation of the hyperbola is: $$\frac{x^2}{9}-\frac{y^2}{16}=1.$$ The border values of the integral reach from $-5$ to $5.$

First of all I need to get y and the derivation of y. Accordiung to the hyperbola equation I mentioned above $y$ should be: $y=4\sqrt {(x^2/9)-1}$.

At this point I got a little bit stuck. I tried some things but none of them ended up in giving me the expected solution ($\approx 198$).

Answer 1

It remains to apply the formula for the surface of revolution: $$2\cdot 2\pi\int_3^5f(x)\sqrt{1+f'(x)^2}\, dx= 16\pi\int_3^5\sqrt{\left(\frac{5x}{9}\right)^2-1}\,dx $$ where $f(x)=4\sqrt{\frac{x^2}{9}-1}$ is the function that you have found. In order to evaluate the integral, take a look at Indefinite integral of $\int\sqrt{x^2-1}dx$. You can check the final result at WA.

Notice the interval of integration: by symmetry we consider $0\leq x\leq 5$ such that $\frac{x^2}{9}-1\geq 0$, i.e. $x\geq 3$ and we double the final result.