1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.