# Finding the surface area of $S={(r\cos\theta,r\sin\theta,3−r):0\leq r \leq 3, 0\leq \theta\leq2 \pi }$

So we've been given this set:

$$S={(r\cos\theta,r\sin\theta,3−r):0\leq r \leq 3, 0\leq \theta\leq2 \pi }$$

and I can see that this is part of a cone but I'm not too sure how to find the surface area. The question is:

Use a surface integral to find the surface area of $S$.

If anyone could show me how to do this I'd really appreciate it

According to Archimedes the surface $S$ is the union of infinitesimal isosceles triangles of base length $ds$ (on the rim of the cone) and height $3\sqrt{2}$. The total rim length is $s=6\pi$, so that we arrive at $${\rm area}(S)={1\over2}\cdot6\pi\cdot3\sqrt{2}=9\pi\sqrt{2}\ .$$ In order to compute this area as a surface integral we have to produce an essentially $1:1$ parametrization of the cone. You already have given such a parametrization: $$S:\quad(r,\theta)\mapsto{\bf r}(r,\theta):=(r\cos\theta,r\sin\theta, 3-r)\qquad(0\leq r\leq3, \ 0\leq\theta\leq2\pi)\ .$$ You now have to apply the standard formula $${\rm area}(S)=\int_0^3\int_0^{2\pi}|{\bf r}_r\times {\bf r}_\theta|\>d\theta\>dr\ .$$ Due to the rotational symmetry of $S$ the inner integration will be for free.