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Integrate f(x,y) over a disk

asked 2017-12-21 13:30:14 -0500

Kolja gravatar image

I have a homogeneous polynomial $F(x,y)$, and I'd like to integrate it over a disk $B_1(0)$, or approximate by integrating over a regular polygon. I just need the numerical value or an approximation of $$\int_{B_1(0)}F(x,y)dxdy$$ I looked up everywhere but I can't seem to find a way to do it. I used

integral ( integral ( F , x , -1 , 1 ), y , -1 , 1 )

for integrating over a rectangle, but I couldn't even find a way to integrate over regular polygons, and such an approximation would suffice.

Is there a way to integrate directly over disks ?

Thanks

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Doesn't integral(integral(F,x,-sqrt(1-y^2),sqrt(1-y^2)),y,-1,1) do the trick? You shouldn't expect too nice an answer from this, of course: for F=1 the value of the integral is already a transcendental number. The standard calculus bag of tricks applies, so depending on F there may be smart substitutions that might get you a closed-form solution that is otherwise not easily attained. Rewriting in polar coordinates would be an obvious thing to try too.

nbruin gravatar imagenbruin ( 2017-12-21 16:31:02 -0500 )edit

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answered 2017-12-22 10:28:38 -0500

dan_fulea gravatar image

A possibility to use the fact that $F$ is homogeneous is to pass to polar coordinates, $x=r\; \cos t$, $y=r\; \sin t$, so $$dx\wedge dy = (\cos t\; dr - r\; \sin t\; dt)\wedge (\sin t\; dr + r\; \cos t\; dt) =r\; dr\wedge dt\ .$$ If $F$ is homogeneous of degree $k$, then we have: $$ \int_{B_1(0)}F(x,y)\; dx\; dy = \int_0^1\int_0^{2\pi} r^k\; F(\cos t, \sin t)\; r\; dr\; dt = \frac 1{k+2}\int_0^{2\pi} F(\cos t, \sin t)\; dt\ .$$

For example, for the function $F(x,y)=x^8+y^8$ we have $k=8$ and we can compute exactly, and numerically:

var( 'x,y,t' )
F(x,y) = x^8 + y^8
k = 8
J = 1/(k+2) * integrate( F( cos(t), sin(t) ), (t,0,2*pi) )
print "J = %s ~ %s" % ( J, J.n() )

This gives:

J = 7/64*pi ~ 0.343611696486384

Using the idea of nbruin:

sage: integral( integral( F(x,y), (y,-sqrt(1-x^2),+sqrt(1-x^2)) ), (x,-1,1) )
7/64*pi

The transformation to polar coordinates also works for a "positively homogeneous" function like in the following example:

F(x,y) = abs(x)^(3/5) * abs(y)^(7/5)
k = 2
J = 1/(k+2) * integrate( F( cos(t), sin(t) ), (t,0,2*pi) )
print "J is %s" % J
print "J is numerically %s" % J.n()

Result:

J is 1/4*integrate(abs(cos(t))^(3/5)*abs(sin(t))^(7/5), t, 0, 2*pi)
J is numerically 0.534479668623671

The integral could not be computed, but the numerical value is accessible. Alternatively, we can use the numerical integral, the following example shows which is the error, the trap i am falling in every time:

sage: x=0.12345; numerical_integral( lambda y: F(x,y), (-sqrt(1-x^2),+sqrt(1-x^2)) )
(0.23319001033141668, 1.3392247519283806e-09)

(My error is that i always forget about the error... The result of the numerical_integral is a tuple, not a number.) In order to get a number for this particular value of $x$, we have to take the pythonically zeroth part of the tuple.

sage: x=0.12345; numerical_integral( lambda y: F(x,y), (-sqrt(1-x^2),+sqrt(1-x^2)) )[0]
0.23319001033141668

So the numerical integral using Fubini is:

sage: numerical_integral( lambda x: numerical_integral( lambda y: F(x,y), (-sqrt(1-x^2),+sqrt(1-x^2)) )[0], (-1,1) )[0]
0.5344796700960925
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Asked: 2017-12-21 13:30:14 -0500

Seen: 44 times

Last updated: Dec 22 '17