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I have an integral $$\int_C xy^2dx-4x\sin y\,dy$$ where $C$ bounded with some constrains, for instance inside $x^2+y^2=1$ and below $y=x^2$.

I can integrate of one variable and also with some calculations do above integral, but I want to know how can Mathematica do it. Is it something of the form

Integrate[f[x]-g[y], {x, 0, 2 π}, {y, 0, 2 π}]

but it doesn't work, or can software do that!

How can I do this type of integrals? Thanks.

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  • $\begingroup$ I would rewrite it in polar coordinates... $\endgroup$
    – MsTais
    Nov 3, 2017 at 0:33
  • $\begingroup$ Thanks. but I want just this form! $\endgroup$
    – Nosrati
    Nov 3, 2017 at 0:34
  • $\begingroup$ From the documentation of Integrate, you can specify a region. Example from the doc: Integrate[1, {x, y, z} \[Element] Sphere[]]. $\endgroup$
    – anderstood
    Nov 3, 2017 at 0:36
  • $\begingroup$ Thanks. I think this is triple integral but the question is about some weird with $x$ and $y$ variable! $\endgroup$
    – Nosrati
    Nov 3, 2017 at 0:38
  • $\begingroup$ Integrate[x y^2 - 4 x Sin[y], {x, y} [Element] Circle[]]... "..weird x and y.." gives you a circle... $\endgroup$
    – MsTais
    Nov 3, 2017 at 0:40

2 Answers 2

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Using Green's theorem is simplest. Here is L, M and the region:

L[x_, y_] := x y^2
M[x_, y_] := -4 x Sin[y]
region = ImplicitRegion[x^2 + y^2 < 1 && y < x^2, {x, y}];

Visualize region:

Region @ region

enter image description here

Perform integral:

    NIntegrate[D[M[x, y], x] - D[L[x, y], y], {x, y} ∈ region]

2.09163

Answers using the line integral approach can be compared to this answer.

Update for M9 users

In M9, one can use Boole:

NIntegrate[
    (D[M[x, y], x]-D[L[x, y], y]) Boole[x^2+y^2<1 && y<x^2],
    {x,-1,1},
    {y,-1,1}
]

2.09163

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  • $\begingroup$ Thanks. I restart my software but it doesn't work (version 9.0). Is it need to execute a package (like ancient versions) for your codes? $\endgroup$
    – Nosrati
    Nov 3, 2017 at 1:47
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Sorry the code line gets screwed up in comments...

Integrate[x y^2 - 4 x Sin[y], {x, y} \[Element] Circle[]]

with zero output.

Reference is here ,and here for how to define the unit circle...

Another way to express it is

Integrate[x y^2 - 4 x Sin[y], {x, y} \[Element] Circle[{0, 0}, 1]]
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  • $\begingroup$ How do I add the condition $y\geq x^2$.? $\endgroup$
    – Nosrati
    Nov 3, 2017 at 0:59
  • $\begingroup$ condition is there. See edits, pls. $\endgroup$
    – MsTais
    Nov 3, 2017 at 1:14

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