How to calculate the volume of the following hyperboloid using integrals?

Question

I need to find the volume of the hyperboloid defined by the equation

${x^2\over30^2}+{y^2\over30^2}-{(z-160)^2)\over120^2}=1$ and bounded by the planes $z=0$ and $z=190$.

I decided to use cylindrical coordinates and wrote the next integral:

$\int^{190}_0 \int^{2pi}_0\int^T_0 ({r} )drd\theta dz$

Where $T$ is the radius $r=\sqrt{30^2+{(z-160)^2\over120^2}}$

I don't know if I defined the integral correctly.

Answer 1

hint

Your volume which is of revolution, is given by

$$V=\int_0^{190}S(z)dz$$ with $$S(z)=\pi r^2(z)$$ and $$r^2(z)=30^2(1+\frac{(z-160)^2}{120^2})$$

Answer 2

$$\int\limits_{z = 40}^{190} \pi 30^2 \left(1 + \frac{(z - 160)^2}{120^2} \right)\ dz = \frac{34125 \pi}{2}$$

where you have to solve for the lower limit on $z$ by setting $x = y = 0$.