Volume of a Hyperboloid using triple integration/shadow method

Question

Could someone help me find the volume of the hyperboloid $$\frac{x^2}{1817}+\frac{y^2}{1817}-\frac{z^2}{10914}=1$$ with the limits in the $z-axis$ of $-130$ to $43$? I tried using triple integration, but I'm not sure how to convert the variable the in limit of the innermost integral into a number. I would also be open to any other methods of finding the volume.

Answer 1

Usually triple integration is used to find weight of a volume with the density function d(x,y,z). However you can still adjust your density d(x,y,z) = 1 in order to obtain the volume.

I suggest you to draw the volume by a mathematical tool to see the natural bounds before integration.

I think the double integral you are using is on XY-plane. Thus in order to define limits of integration you should put z=0. Then you get a circle equation

$\frac{x^2}{1817} + \frac{y^2}{1817} = 1$

Then if your differential is dxdy then the x values should be inside

$[-\sqrt(1817 - y^2) , \sqrt(1817 - y^2)]$

The outer integration bound (which bounds y values) is

$[-\sqrt(1817) , \sqrt(1817)]$

Finally you will have;

$\int_{-\sqrt(1817)}^{\sqrt(1817)}\int_{-\sqrt(1817 - y^2)}^{\sqrt(1817 - y^2)}F(x,y).dxdy$

Note that as you have limits on z-axis, you will arrange the integrand function $z = F(x,y)$ by these limits when the z-value crosses them.