Hello! I have a simple question? How do I double integrate a function over an Elipsis Region? Something similar to a Circle Region, like
![]()
Thanks in advance.
Hello! I have a simple question? How do I double integrate a function over an Elipsis Region? Something similar to a Circle Region, like
![]()
Thanks in advance.
Hello
Integrate[1, {x, y} \[Element] Circle[{0, 0}, {a, b}]]
but Mathematica 11.3 gives unevaluated answer:

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
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])
Yes with Assumptions work in my copy to.
Thanks.
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.
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?
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*)
I don’t know if I undestood correctly. How would you calculate this, for example?

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…