Double Integration (polar)

Wize Concept
Double Integration (polar). If D⊂R2D \subset \mathbb{R}^2 is defined by
g1(θ)≤r≤g2(θ),a≤θ≤b,g_1(\theta) \le r \le g_2(\theta),\quad a\le \theta \le b,
where g1(θ)g_1(\theta) and g2(θ)g_2(\theta) are continuous for a≤θ≤ba \le \theta \le b and f(r,θ)f(r, \theta) is continuous on DD, then
∬Df(r,θ)dA=∫ab(∫g1(θ)g2(θ)f(r,θ)rdr)dθ\iint\limits_{D}f(r,\theta)dA = \int_a^b\Bigg(\int_{g_1(\theta)}^{g_2(\theta)}f(r,\theta)rdr\Bigg)d\theta
If D⊂R2D \subset \mathbb{R}^2 is defined by
h1(r)≤θ≤h2(r),c≤r≤d,h_1(r) \le \theta \le h_2(r),\quad c\le r \le d,

where h1(r)h_1(r) and h2(r)h_2(r) are continuous for c≤r≤dc \le r \le d and f(θ,r)f(\theta, r) is continuous on DD, then
∬Df(θ,r)dA=∫cd(∫h1(r)h2(r)f(r,θ)dθ)rdr\iint\limits_{D}f(\theta,r)dA = \int_c^d\Bigg(\int_{h_1(r)}^{h_2(r)}f(r,\theta)d\theta\Bigg)rdr

→\to Notes:
  1. If the function and/or region displays some symmetry around the origin, it may be worthwhile converting from rectangular coordinates to polar.
  2. Don't forget the extra factor of rr in the integrand.

Practice: Double integration (polar)

Compute ∬DxdA\displaystyle\iint\limits_{D}xdA, where DD is the region bounded by x2+y2≤4x^2 + y^2 \le 4 and x≥0x \ge 0.