Answer
$\iint_{\sigma} z^2\, dS = \frac{15\pi\sqrt{2}}{2}$
Work Step by Step
$\iint_{\sigma} f(x,y,z)\, dS $ where $ f(x,y,z) = z^2,$
and $\sigma$ is the portion of the cone $z=\sqrt{x^2 + y^2}$ between the planes $z=1$ and $z=2$.
Since $z = \sqrt{x^2 + y^2}$, we set
$ x = r \cos \theta$ $y = r \sin \theta$ $z = r, $
with the bounds
$1\leq r\leq 2$, $0\leq \theta\leq 2\pi $
The position vector is
$r(r,\theta)=(r\cos{\theta}, r\sin{\theta},r) $
Now
$\mathbf{r}_r = (\cos\theta,\; \sin\theta,\; 1), \qquad$
$\mathbf{r}_\theta = (-r\sin\theta,\; r\cos\theta,\; 0).$
Their cross product is
$\mathbf{r}_r \times \mathbf{r}_\theta = (-r\cos\theta,\; -r\sin\theta,\; r),$
and its magnitude is
$\|\mathbf{r}_r \times \mathbf{r}_\theta\| = r\sqrt{2}.$
Thus, the surface element becomes
$dS = r\sqrt{2}\; dr\; d\theta.$
Since $z = r$ on the cone, the integrand becomes
$f(x,y,z) = z^2 = r^2.$
Therefore,
$\iint_{\sigma} z^2\, dS = \int_{0}^{2\pi}\int_{1}^{2} r^2 \left(r\sqrt{2}\right)\, dr\, d\theta
= \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2} r^3\, dr\, d\theta.$
Evaluate the inner integral:
$\int_{1}^{2} r^3\, dr = \left.\frac{r^4}{4}\right|_{1}^{2} = \frac{16 - 1}{4} = \frac{15}{4}.$
Finally,
$\iint_{\sigma} z^2\, dS
= \sqrt{2} \cdot 2\pi \cdot \frac{15}{4}
= \frac{15\pi\sqrt{2}}{2}.$
$\boxed{\iint_{\sigma} z^2\, dS = \frac{15\pi\sqrt{2}}{2}}$
