1
$\begingroup$

$D$ is the portion of the unit cylinder $x ^2 +y ^2 ≤ 1$ which lies between $z = 0$ and $z = 1$.

I normally solve questions like this by sketching the $z-r$ plane. Obviously I draw $z=0$ and $z=1$, but I am unsure how to sketch $x ^2 +y ^2 ≤ 1$. I know this can be written as $r^2 \leq 1$. But then this would mean I would be drawing $r \leq 1$ which would give a square with corners at $(0,0) (0,1) (1,1) (1,0)$. Is this correct?

For the bounds I know that since the domain is axysymmetric, then $0 \leq \theta \leq 2 \pi$. I can figure things out like $p \leq sec( \phi)$ and $p \leq cosec(\phi)$, but I really need to sketch the domain in the $z-r$ plane to figure out how to split the integral up.

I know there are other methods of solving but I only really understand by sketching in the $z-r$ plane if someone could please help?

$\endgroup$
4
  • $\begingroup$ $0 \le r = \sqrt{x^2 + y^2} \le 1$ describes a disk, not a square. $\endgroup$
    – Abezhiko
    Nov 1, 2023 at 13:12
  • $\begingroup$ @Abezhiko Can you sketch that in the $z-r$ plane for me. It just seems like two vertical lines to me with shading in between them. $\endgroup$
    – Dam
    Nov 1, 2023 at 13:15
  • 1
    $\begingroup$ Excuse me, I misunderstood your question, I thought you were considering the sketch in $\Bbb{R}^3$. Indeed, in the z-r plane, you will end up with the square you mention. $\endgroup$
    – Abezhiko
    Nov 1, 2023 at 13:21
  • $\begingroup$ @Abezhiko I think I figured it out. In the $z-r$ plane we have that square, which I can split up with the diagonal with $\phi$ being $ \frac{\pi}{4}$, and then everything seems simple to me. Sorry I understand how to do it now I think, I was just being stupid. $\endgroup$
    – Dam
    Nov 1, 2023 at 13:23

1 Answer 1

0
$\begingroup$

$x^2 + y^2 = 1$ is the equation of a circle whose centre lies at the origin and which has a radius of one unit.

Essentially, $x^2 + y^2 \leq 1$ is the cross section of your cylinder, which lies between $z=0$ and $z=1$.

If you're having trouble sketch graphs of functions, you may use a graphing calculator, like Desmos, or GeoGebra.

$\endgroup$
4
  • $\begingroup$ I know that, but I want to sketch it in the $z-r$ plane which is my issue. The equation of that circle is $r=1$ also. $\endgroup$
    – Dam
    Nov 1, 2023 at 13:15
  • 1
    $\begingroup$ @Dam I misunderstood the question. If you want to sketch a graph in the Z - r plane, it will indeed be a square. $\endgroup$ Nov 1, 2023 at 15:52
  • $\begingroup$ It's okay. I figured it out now. Thanks for your help. $\endgroup$
    – Dam
    Nov 2, 2023 at 9:52
  • $\begingroup$ @Dam Good to know that. $\endgroup$ Nov 3, 2023 at 13:26

You must log in to answer this question.

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