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Surface Area problem with double integral confusion

Posted 1 month ago

I am a student currently in Calc 3 and I got my homework back and one of the problems confused me and my professor. It is a Surface Area calculation over the function (e^-x) * (siny) and bounded by the region R = [(x,y) : x^2 + y^2 <= 4]. My question comes with the symmetry of the problem. The first problem on the attached photo is the equation setup before integration without symmetry. The second is the first degree of symmetry just effecting the y term. However when I do the same on the last part of the equation. I am given a wrong answer. I have no idea why this is happening so any help is much appreciated. Cheers!


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The function is symmetric with respect to y only, not to x. However, when I tried by mistake this:

NIntegrate[Sqrt[1 + Exp[-2 x]], Element[{x, y}, Circle[{0, 0}, 2]]]

I got 34.27, which is the LINE INTEGRAL, not the two-variable integral, which should be zero, because the circle has zero two-dimensional measure. This automatic switching from NIntegrate to NLineIntegrate without warning is unexpected and annoying to me.

POSTED BY: Gianluca Gorni

But you are integrating along a circle - which is a line! You probably meant to write:

NIntegrate[Sqrt[1 + Exp[-2 x]], Element[{x, y}, Disk[{0, 0}, 2]]]
(*  Out:    25.4452   *)
POSTED BY: Henrik Schachner

Yes, I wrote Circle by mistake instead of Disk. My objection stands: NIntegrate in two variables should do the integral with respect to the two-dimensional Lebesgue measure. The line integral should be reserved to NLineIntegrate. At the very least, I would expect a warning.

POSTED BY: Gianluca Gorni

Yes, yes, yes - as it occurred to me in a sleepless night, you are absolutely right, sorry Gianluca! The correct result of

NIntegrate[Sqrt[1 + Exp[-2 x]], Element[{x, y}, Circle[{0, 0}, 2]]]

should definitely be 0. And one can generate more questionable results (at least with my version 13.3), like:

Integrate[n, x \[Element] Point[{2}]]
(* Out:   n  *)
Integrate[1, x \[Element] Circle[]]
(* Out:   2 \[Pi]  *)
Integrate[1, {x, y} \[Element] Circle[]]
(* Out:  2 \[Pi]  *)
Integrate[1, x \[Element] Disk[]]
(* Out:   \[Pi]  *)
POSTED BY: Henrik Schachner

At a second thought, the result

Integrate[1, x \[Element] Disk[]]
(* Out:   \[Pi]  *)

is consistent with x being interpreted as a two-dimensional vector, as in

Integrate[Norm[x], x \[Element] Disk[]]

The following gives bizarre asnwers, though:

EuclideanDistance[x, {0, 0}]
Integrate[EuclideanDistance[x, {0, 0}], x \[Element] Disk[]]
POSTED BY: Gianluca Gorni

Thank you Gianluca!

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