0
$\begingroup$

I have the following set

$$A=\{ (x,y,z) : x > 0, x < y, 1 < x^2 + y^2 < 3, 1 < z < 5 \}$$

I know how to transform $x^2 + y^2$ and $z$ to cylindrical coordinates:

$$1 < x^2 + y^2 < 3 \implies 1 < r < \sqrt{3}$$

$1 < z < 5$ just stays the same.

But what about $x > 0$ and $x < y$?

$\endgroup$
2
  • 1
    $\begingroup$ $x>0$ and $x<y$ gives you restriction on the angular coordinate $\theta$. You might find it helpful to try and sketch it. Basically, these two conditions tells you that the given region is some slice of the "tube"; you already find out the "tube" by the way, it has inner radius 1 and outer radius $\sqrt{3}$, restricted between $z=1$ and $z=5$. $\endgroup$
    – Chee Han
    Feb 10, 2017 at 6:01
  • $\begingroup$ I could write out the answer, but it's more fun if you try to figure it out by sketching ! $\endgroup$
    – Chee Han
    Feb 10, 2017 at 6:04

1 Answer 1

1
$\begingroup$

Okay so since $x > 0$, $x < y$ means:

enter image description here

and $\theta \in [0, 2 \pi[$ is the rotation counterclockwise, then since the blue area corresponds to $[\frac{\pi}{4}, \frac{\pi}{2}]$, then $[\frac{\pi}{4}, \frac{\pi}{2}[$ is the range of $\theta$ given the restrictions $x > 0$ and $x < y$.

$\endgroup$
3
  • $\begingroup$ Yes that's right. One thing: $\pi/4$ is not included since $x<y$ is a strictly inequality. $\endgroup$
    – Chee Han
    Feb 10, 2017 at 7:18
  • $\begingroup$ Hmm, but does the domain of integral over $d \theta$ still become $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}$? Or perhaps the upper bound is supposed to be indefinite and taken using some limit? $\endgroup$
    – mavavilj
    Feb 10, 2017 at 9:57
  • $\begingroup$ It doesn't affect the volume (triple) integral, since a line segment has zero area so similarly a surface has zero volume. $\endgroup$
    – Chee Han
    Feb 10, 2017 at 19:11

You must log in to answer this question.

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