2
$\begingroup$
integ1 = Integrate[x^2, Element[{x, y}, Region[Disk[{0, 0}, 1]]]]
(*  ==> Pi/4 *)
integ2 = Integrate[x^2, Element[x, Region[Disk[{0, 0}, 1]]]]
(*  ==> {Pi/4, Pi/4} *)

If the region integrated over is irregular, such as a Polygon, the integ2 will give two different values.

I was wondered by the behavior of integ2. Can someone give me a hint?

$\endgroup$
4
$\begingroup$

In the second example, x ∈ Disk[{0, 0}, 1] (I dropped the unnecessary Region wrapper) means that x is a 2-element list. In effect, the second integral is equivalent to:

Integrate[{x, y}^2, {x, y} ∈ Disk[{0, 0}, 1]]

{π/4, π/4}

$\endgroup$
  • $\begingroup$ A brief lesson! Thank you. $\endgroup$ – PureLine Oct 11 '17 at 6:57
3
$\begingroup$
Integrate[x^2, Element[{x, y}, Region[Disk[{0, 0}, {1, 2}]]]]
(*\[Pi]/2*)
Integrate[y^2, Element[{x, y}, Region[Disk[{0, 0}, {1, 2}]]]]
(*2 \[Pi]*)
Integrate[a^2, Element[a, Region[Disk[{0, 0}, {1, 2}]]]]
(*{\[Pi]/2,2 \[Pi]}*)

Change the integral area, and the difference maybe become obvious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.