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%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
from scipy import integrate
The 2d polar integral of a scalar function $f(r, \theta)$ is defined as:
$$ I(r_{max}) = \int_0^{r_{max}} \int_0^{2\pi} f(r, \theta) r drd\theta $$Write a function integrate_polar(f, rmax)
that performs this integral numerically using scipy.integrate.dblquad
.
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def integrate_polar(f, rmax):
"""Integrate the function f(r, theta) over r=[0,rmax], theta=[0,2*np.pi]"""
integrand = lambda r,t: r*f(r,t)
rmin = 0
rmax = rmax
tmin = lambda r : 0
tmax = lambda r : 2*np.pi
I = integrate.dblquad(integrand, rmin, rmax, tmin, tmax)
return I
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integrate_polar(lambda r,t: 1, 1.0) #This is returning 2pi instead of pi and I don't know why
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integrate_polar(lambda r, t: np.exp(-r)*(np.cos(t)**2), np.inf) #this doesn't equal pi either
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assert np.allclose(integrate_polar(lambda r,t: 1, 1.0), np.pi)
assert np.allclose(integrate_polar(lambda r, t: np.exp(-r)*(np.cos(t)**2), np.inf), np.pi)
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