Volume between a cone and and an Hyperboloid I'm trying to use integration in several variables to find put what is the volume between  the cone $x^2+y^2=z^2$ and the hyperboloid $x^2+y^2=3+z^2$ I'm having a hard time with this problem, as the two surfaces actually never meet. I'm open to your suggestions, maybe even a convenient variable change.
 A: You don't need a variable change. But I do recommend that you first draw a picture. You should recognize this as a shape where the hyperboloid surrounds the cone, never meeting. Let's do this problem in two different ways: using multivariable calculus, and then using single-variable calculus.
Multivariable Calculus
We will use cylindrical coordinates, $r, \theta, z$. At each height $z$, the hyperbola has a circular cross-section with equation $r^2 = z^2 + 3$, and the cone has cross-section $r^2 = z^2$. And at each height, we want the area between the two. The hyperbola cross-sections contribute area
$$ \int_0^{2\pi} \int_0^{\sqrt{z^2 + 3}}rdrd\theta = \int_0^{2\pi} \frac{1}{2}(z^2 + 3) d\theta = \pi(z^2+3),$$
and the cone cross-sections contribute
$$ \int_0^{2\pi}\int_0^{\sqrt{z^2}} rdrd\theta = \int_0^{2\pi} \frac{1}{2}z^2 d\theta = \pi z^2.$$
Let's say we integrate from the center, the singular point of the cone, up to height $H$. Then the hyperbola has volume from $z=0$ to $z=H$ given by
$$ \int_0^H \int_0^{2\pi} \int_0^{\sqrt{z^2 + 3}}rdrd\theta dz= \int_0^H\pi(z^2+3)dz = \frac{\pi}{3}H^3 + 3\pi H,$$
and the cone has volume
$$\int_0^H\int_0^{2\pi}\int_0^{\sqrt{z^2}} rdrd\theta dz =\int_0^H \pi z^2 dz = \frac{\pi}{3}H^3.$$
Let's pause for a moment to see if we believe this. We've just said that the volume of a cone of height $H$ and radius $H$ is $\frac{\pi}{3}H^3$. We know the volume of a cone. Is this right? (yes, it is)
So the difference is $3\pi H$. So the "total volume" between the two is unbounded.
Single-Variable Calculus
Everything is rotationally symmetric, so we can use volumes of revolution. In particular, on the $z,x$ axes, where we think of $z$ as being the independent variable, the cone has equation $x = z$ and the hyperbola $x = \sqrt{z^2 + 3}$. Using the "disk method" of integration, the volumes will be given by
$$\pi \int_0^H (\text{radius at } z)^2 dz,$$
which amounts to
$$\pi \int_0^H z^2 dz = \frac{\pi}{3} H^3$$
and
$$\pi \int_0^H z^2 + 3 dz = \frac{\pi}{3}H^3 + 3\pi H.$$
In fact, if you think about it, you'll see that these are fundamentally the same.
