Triple integral - cylindrical coordinates problem

Question

I have to figure out this here integral:

$$\iiint \sqrt{x^2+y^2} dxdydz$$

in the boundaries $x^2+y^2=z^2$, $z=1$, $z=2$

Now, I know that the intersection of the two planes and the conic are circles. The area itself is the area between those planes.

Now, if I introduce cylindrical coordinates:

$x=r\cos\phi$

$y=r\sin\phi$

$z=z$

If I plug this in, I get that

$r^2 = z^2$, which after substituting the two values of $z$ I get that $r \in [1,2]$

$\phi \in [0, 2\pi]$

But what about the boundaries for $z$? Surely they can't be $z \in [1,2]$!

I'm at a loss here, because if I express the boundaries for $z$ to be $ \in [1,2]$, what about $r$?

Could anyone help?

Answer 1

Yes, you can take $z\in[1,2]$. For each such $z$, since $x^2+y^2\leqslant z^2$, you have $r^2\leqslant z^2$, which means (since $r\geqslant0$) that $r\leqslant z$.

So, compute$$\int_0^{2\pi}\int_1^2\int_0^zr^2\,\mathrm dr\,\mathrm dz\,\mathrm d\theta.$$You should get $\frac{5\pi}2$.

Answer 2

You should notice that if you integrate wrt $dz$ first, you are bound between planes $z = 1$ and $z = 2$ for $r \leq 1$ and for $1 \leq r \leq 2$, you are bound between the cone and the plane $z = 2$. See the projection in xz-plane.

But if you integrate wrt $dr$ first (horizontal strips), you are bound by the cone for all $z$.

So if you go in the order $dz$ first,

$\displaystyle \int_0^{2\pi}\int_0^1 \int_1^2 r^2 \ dz \ \ dr \ d\theta + \int_0^{2\pi} \int_1^2\int_{r}^2 r^2 \ dz \ dr \ d\theta$

But if you go in the order $dr$ first, it is one integral and that is of course simpler.