Solution
= Solution
Use the <cone> parametrization $\mathbf r(x,\theta)=(x,x\cos\theta,x\sin\theta)$ for $2\leq x\leq3$, $0\leq\theta\leq2\pi$. Its <surface element> is $\sqrt2x\,dx\,d\theta$. Since $x^2y^2=x^4\cos^2\theta$, the scalar <surface integral> is
$$
\boxed{\iint_Sx^2y^2\,dS=\sqrt2\int_2^3x^5\,dx\int_0^{2\pi}\cos^2\theta\,d\theta=\frac{665\pi\sqrt2}{6}.}
$$
This uses only the curved conical surface. There are no closing discs in the specified integration surface.