Working with triple integrals and cylindrical coordinates. How would you go about sketching the function $(r-2)^2 + z^2 \le 1$? I'm fairly confident plotting functions like this in Cartesian coordinates e.g. $x^2+y^2 \le 1$ would be a cylinder along the $z$-axis but not sure about the cylindrical ones.


1 Answer 1


The easy way to think of this is to look in the $xz$ plane, then rotate around $z$ axis. So when $y=0$ the given condition becomes $$(x-2)^2+z^2\le 1$$This is a disk of radius $1$, with center at $x=2$ and $z=0$. If you rotate around the $z$ axis, the location of the center will be a circle of radius $2$ in the $xy$ plane, centered on the origin. So the figure you get is a torus, centered on origin, inner radius $2-1=1$, outer radius $2+1=3$.

  • $\begingroup$ This is super helpful, thank you. One question, does it matter wether I view it as the x and z axis or the y and z axis? The figure would surely look the same and it's not really specified wether the (r-2)^2 is the X or Y coordinate. $\endgroup$
    – C. K.
    Mar 16, 2020 at 18:44
  • 1
    $\begingroup$ No, it does not. You get the same result since your equation does not depend on the angle. $\endgroup$
    – Andrei
    Mar 16, 2020 at 18:46

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