# Sketching Domain in Spherical Coordinates

$$D$$ 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?

• $0 \le r = \sqrt{x^2 + y^2} \le 1$ describes a disk, not a square. Nov 1, 2023 at 13:12
• @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.
– Dam
Nov 1, 2023 at 13:15
• 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. Nov 1, 2023 at 13:21
• @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.
– Dam
Nov 1, 2023 at 13:23

$$x^2 + y^2 = 1$$ 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.

• 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.
– Dam
Nov 1, 2023 at 13:15
• @Dam I misunderstood the question. If you want to sketch a graph in the Z - r plane, it will indeed be a square. Nov 1, 2023 at 15:52
• It's okay. I figured it out now. Thanks for your help.
– Dam
Nov 2, 2023 at 9:52
• @Dam Good to know that. Nov 3, 2023 at 13:26