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

Parametrize the graph by $\mathbf{r}(x,y)=(x,y,f(x,y))$ over its projection $D$ onto the $xy$ plane. The tangent vectors are $(1,0,f_x)$ and $(0,1,f_y)$, so their cross product is $(-f_x,-f_y,1)$. Hence the scalar surface element isandThe printed subscript $S$ is shorthand for integrating over the projected parameter region: $dxdy$ belongs to $D$, while the square-root factor converts projected area into actual surface area. The corresponding upward vector surface element of a graph is $(-f_x,-f_y,1)dxdy$.

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.

The plane cuts the positive coordinate axes at $(3,0,0)$, $(0,3,0)$ and $(0,0,3/2)$. Choose its unit normal vector $(1,1,2)/\sqrt{6}$, pointing away from the origin. In the parametrization $(x,3-x-2z,z)$, the vector surface element for this orientation is $(1,1,2)dxdz$. Thusand the triangular projection givesFor the opposite orientation the answer is $-9/4$. The normal direction must be stated because this open surface has no prescribed orientation in the question.