6
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Notice that the intersection curve $4-x = x^2+y^2$ is a circle. Try using polar coordinates centered at the center of this circle (instead of polar coordinates centered at the origin). This should eliminate your troubles describing the domain. $\endgroup$
    – user197427
    Dec 9, 2016 at 20:58
  • $\begingroup$ You can show the intersection of the surfaces to be the cylinder $\left(x+\frac12\right)^2+y^2=\frac{17}4$, then parameterize $S$ by$$\vec r(u,v)=\left(u\cos(v)-\frac12\right)\,\vec\imath+u\sin(v)\,\vec\jmath+\left(u^2-u\cos(v)+\frac14\right)\,\vec k$$with $(u,v)\in\left[0,\frac{\sqrt{17}}2\right]\times[0,2\pi]$. The integral is still a problem, though. $\endgroup$
    – user170231
    Dec 13, 2021 at 16:17

1 Answer 1

0
$\begingroup$

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 $x^2+y^2\le 4-x$.

$\endgroup$
7
  • $\begingroup$ But aren't we then doing the same thing here again? For example, to solve for the area of $x^2+y^2 \leq 4-x$ We sub in polar coordinates to get $r^2 \leq 4-r\cos(\theta)$ Which is again the same inequality I was trying to solve above and so the limit will again give me a complicated integral no? $\endgroup$ Dec 7, 2016 at 3:58
  • $\begingroup$ What I mean is $Z^2+Z\cos(\theta)-4=0$, solving for $Z$ is exactly solving for $r$ here to get the upper bound right? $\endgroup$ Dec 7, 2016 at 3:59
  • $\begingroup$ yep, good point. $\endgroup$
    – Kuifje
    Dec 7, 2016 at 13:32
  • $\begingroup$ You are right, both parametrizations lead to the same complicated integral. I am not sure there is any around it. In all cases your methodology is correct, and the trick is to solve the equation $Z^2+Z\cos\theta-4=0$ to find the bounds. $\endgroup$
    – Kuifje
    Dec 7, 2016 at 13:58
  • $\begingroup$ I get $\iint_S \; dS =41.02945733$. $\endgroup$
    – Kuifje
    Dec 7, 2016 at 13:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .