The integral to be calculated is $$ \int_0^1\int_0^{2\pi} r \;\mathrm{d}\theta\,\mathrm{d}r = \pi. $$

Note that the inner integral is over $\theta$ and the outer is over $r$. Therefore, the call to dblquad should call the function $f(r,\theta)=r$ as lambda theta, r: r (note the order of the arguments).

In [x]: dblquad(lambda theta, r: r, 0, 1, lambda r: 0, lambda r: 2*np.pi)
Out[x]: (3.141592653589793, 3.487868498008632e-14)

Alternatively, swap the order of the integration:

dblquad(lambda r, theta: r, 0, 2*np.pi, lambda theta: 0, lambda theta: 1)
(3.141592653589793, 3.487868498008632e-14)