Find the surface area of the portion S of the cone $z^2=x^2+y^2$, where z≥0, contained within the cylinder $y^2+z^2≤81$.
The work I have is that I parameterized the cone into polar coordinates.
$$x(r,\theta)=r\cos\theta$$
$$y(r,\theta)=r\sin\theta$$
$$z(r,\theta)=r$$
then taking the cross product of the partial derivatives in terms of $r$ and $\theta$ I get $|\Phi_r \times \Phi_\theta| = \sqrt{1+r^2}$
I got the upper bound of r by plugging in the parameterized $y$ and $z$ into
$$y^2+z^2≤81$$
$$(r\sin\theta)^2+r^2≤81$$
$$r^2(\sin^2\theta + 1)≤81$$
$$r≤\sqrt{\frac{81} {\sin^2\theta + 1}}$$
so then I am calculating $\int\int\sqrt{1+r^2}drd\theta$ with $0<\theta<2\pi$ and $0<r<\sqrt{\frac{81} {\sin^2\theta+1}}$ which I get as 190.004 but my online homework tells me that is incorrect. Where did I go wrong?
Best Answer
Polar coordinates? Don't you think you need spheric coordinates? The way you did it, you ignored the elevation angle. It is constant ($\pi/4$) but it leads to differences. You can see, if you do a reality check on your upper integration border for $r$. The way you set it up, you get $9$ as a maximum, but the maximum should be $\sqrt{162}$. Those are the points above the center axis of the cylinder ($y=0$, $z=9$). $9$ is actually the minimum you get for $r$.
Try again in spheric coordinates, where you can set the elevation angle to $\pi/4$, then you should be able to obtain the correct result.