Find the volume of the solid bounded by $x^{2}+y^{2}-\frac{z^{2}}{4}=1$ and $z=2$ ,$z=-2$

Question

Find the volume of the solid bounded by $x^{2}+y^{2}-\frac{z^{2}}{4}=1$ and $z=2$ ,$z=-2$.

I think the volume is $$8\int_{0}^{\sqrt{2}}\int_{0}^{\sqrt{2-x^{2}}}\int_{0}^{1}\,dz\,dy\,dx$$

Bu on the other hand the set $$E=\left\{\left(x,y,z\right):0\le x\le\sqrt{2},0\le y\le\sqrt{2-x^{2}},0\le z\le1\right\}$$

Is a cylinder which is on the first first octant , and not a hyperbola of one sheet, So I think I need to describe $z$ as a function of $x,y$,but then $$8\int_{0}^{\sqrt{2}}\int_{0}^{\sqrt{2-x^{2}}}\int_{0}^{2\sqrt{x^{2}+y^{2}-1}}\,dz\,dy\,dx$$ Which is undefined.

Answer 1

$x^2 + y^2 = 1 + \frac{z^2}{4}$. So as you can see, $x^2 + y^2 = 1$ is the smallest cross-section at $z = 0$. If we are going $dz$ first,

For $0 \leq x^2 + y^2 \leq 1$, you simply have the volume of a cylinder between $-2 \leq z \leq 2$. But for $1 \leq x^2 + y^2 \leq 2$, the lower bound of $z$ cannot be $0$. For volume above xy-plane, it will be $2 \sqrt{x^2 + y^2 - 1} \leq z \leq 2$. Similarly below xy-plane, it will be $-2 \leq z \leq - 2 \sqrt{x^2 + y^2 - 1}$.

So we will have to split it into two integrals considering symmetry above and below xy-plane.

But it is easier to integrate wrt to $dz$ last, esp. using polar coordinates,

$V = \displaystyle \int_{-2}^2 \int_{- \sqrt{1 + z^2/4}}^{\sqrt{1 + z^2/4}} \int_{- \sqrt{1 - y^2 + z^2/4}}^{\sqrt{1 - y^2 + z^2/4}} \: dx \ dy \ dz$

Using polar coordinates,

$V = \displaystyle \int_{-2}^2 \int_0^{2\pi} \int_0^{\sqrt{1 + z^2/4}} \: r \ dr \ d\theta \ dz$

Answer 2

For $z \in [-2,2]$ let

$$Q(z):=\{(x,y) \in \mathbb R^2: x^2+y^2=1+\frac{z^2}{4}\}.$$

$Q(z)$ is a disc with radius $ \sqrt{1+\frac{z^2}{4}}.$ Hence the area of $Q(z)$ is given by

$$A(z)= \pi (1+\frac{z^2}{4}).$$

Cavalieri now gives that the volume in question is given by

$$ \int_{-2}^2 A(z) dz.$$