Past exam of the mathematical methods II course of the University of Oxford / 2016 / 10 / b / i / Solution

Use the cone parametrization $\mathbf{r}(x,\theta )=(x,x\cos \theta ,x\sin \theta )$ for $2\le x\le 3$, $0\le \theta \le 2\pi$. Its surface element is $\sqrt{2}xdxd\theta$. Since $x^2{y}^2=x^4{\cos }^2\theta$, the scalar surface integral isThis uses only the curved conical surface. There are no closing discs in the specified integration surface.