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[✓] Calculate a double Integrate over an ellipse region?

GROUPS:

Hello! I have a simple question? How do I double integrate a function over an Elipsis Region? Something similar to a Circle Region, like

enter image description here

Thanks in advance.

POSTED BY: Thiago Benine
Answer
2 months ago

Hello

 Integrate[1, {x, y} \[Element] Circle[{0, 0}, {a, b}]]

but Mathematica 11.3 gives unevaluated answer:

enter image description here

it should be: $4 a E\left(1-\frac{b^2}{a^2}\right)$

You can use:

ArcLength[Circle[{0, 0}, {a, b}]]
(* 4 b EllipticE[1 - a^2/b^2] *)

Regards,MI

POSTED BY: Mariusz Iwaniuk
Answer
2 months ago

Works in my copy of 11.3

In[6]:= Assuming[a > 0 && b > 0, Integrate[1, {x, y} \[Element] Circle[{0, 0}, {a, b}]]]

Out[6]= 2 (b EllipticE[1 - a^2/b^2] + a EllipticE[1 - b^2/a^2])
POSTED BY: Ilian Gachevski
Answer
2 months ago

Yes with Assumptions work in my copy to.

Thanks.

POSTED BY: Mariusz Iwaniuk
Answer
2 months ago

It also works without Assumptions, but returns (understandably) a much larger result and takes quite a bit of time. It might return unevaluated on a slower machine, I suppose.

POSTED BY: Ilian Gachevski
Answer
2 months ago

Indeed it does:

enter image description here

Sorry for the late posting, but apparently every contribution to any thread only shows up much later for me than for many others on the Community.

Cheers,

Marco

POSTED BY: Marco Thiel
Answer
2 months ago

Well, these might work, too:

Integrate[1, {x, y} \[Element] RegionBoundary[Ellipsoid[{0, 0}, {r1, r2}]], Assumptions -> {r1 > 0, r2 > 0}]

enter image description here

This gives the same answer (on 11.3)

Integrate[1, {x, y} \[Element] Circle[{0, 0}, {r1, r2}], Assumptions -> {r1 > 0, r2 > 0}]

Cheers,

Marco

POSTED BY: Marco Thiel
Answer
2 months ago

Thanks for the response! But, I have another doubt, I was expecting a number as a result (like a volume). What Do I Have to do with EllipticE?

POSTED BY: Thiago Benine
Answer
2 months ago

You will have to provide values for the two radii:

Integrate[1, {x, y} \[Element] Circle[{0, 0}, {1, 2}]]
(*8 EllipticE[3/4]*)

and

NIntegrate[1, {x, y} \[Element] Circle[{0, 0}, {1, 2}]]
(*9.68845*)

Cheers,

Marco

PS: Note that this is the circumference not the "volume" as you suggest in your question. The area is easy:

Integrate[1, {x, y} \[Element] Disk[{0, 0}, {r1, r2}], Assumptions -> {r1 > 0, r2 > 0}]
(*\[Pi] r1 r2*)

and

NIntegrate[1, {x, y} \[Element] Disk[{0, 0}, {1, 2}]]
(*6.28319*)

The volume of an Ellipsoid would be

Integrate[1, {x, y, z} \[Element] Ellipsoid[{0, 0, 0}, {r1, r2, r3}], Assumptions -> {r1 > 0, r2 > 0, r3 > 0}]
(*4/3 \[Pi] r1 r2 r3*)

or e.g.

NIntegrate[1, {x, y, z} \[Element] Ellipsoid[{0, 0, 0}, {1, 2, 3}]]
(*25.1327*)
POSTED BY: Marco Thiel
Answer
2 months ago

I don't know if I undestood correctly. How would you calculate this, for example?

enter image description here

POSTED BY: Thiago Benine
Answer
2 months ago

Hi,

this forum is not actually suppose to solve homework problems, but

Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 - 2 y <= 0, {x, y}]]
(*\[Pi]*)

So, the answer in this case is Pi.

Cheers,

Marco

PS: Also, could you change the title of this thread to "Double Integral"? Doube has a different meaning....

POSTED BY: Marco Thiel
Answer
2 months ago

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