So the first thing I did was to try and parametrize the parabloid as:
$$r(\theta,z)=\sqrt{z}\cos(\theta)i+\sqrt{z}\sin(\theta)j+zk$$
Then I found $||r_\theta\times r_z||=\frac{1}{2}\sqrt{4z+1}$.
Hence the surface area is $\int\int _SdS=\int \int_D \frac{1}{2}\sqrt{4z+1}dA$. Here, $D$ is the region with $0\leq \theta \leq 2\pi$ and $0 \leq z\leq 4-x=4-\sqrt{z}\cos(\theta)$.
But Here is my problem, I can't find the upper limit in of z in this integral, the lower limit is 0. But I can't find an upper limit $g(\theta)$. How do I get rid of the $z$ from the limit? Any help would be appreciated
I also tried as suggested below to change to a new parametrization but I always go back to the same integral, any hints here?
Best Answer
You need to solve the equation $$ z= 4-\sqrt{z}\cos\theta $$ Let $Z=\sqrt{z}$, and solve $$ Z^2+Z\cos\theta-4 =0 $$
This will give you a complicated integral though. Consider parametrizing differently with $$ r(x,y)=xi+yj+(x^2+y^2)k $$ You will find that $||r_x\times r_y ||=\sqrt{1+4x^2+4y^2}$, easy to integrate in polar coordinates on the area defined by $x^2+y^2\le 4-x$.