Evaluation of double integration by changing it into polar co-ordinates

Evaluation of double integration by changing it into polar co-ordinates

1. Why Polar

When $D$ is a circle, sector, annulus, cardioid or integrand involves $x²+y²$, polar simplifies limits and $f$. Transformation:

with $r≥0, θ$ range. Then $x²+y² = r²$, and $∬_{D} f(x,y) dA = ∬_{D_{rθ}} f(r cosθ,r sinθ) r dr dθ$.

2. Limits in Polar

• Disc $x²+y² ≤ a²$: $r 0→a, θ 0→2π$.
• Semicircle $y≥0$: $θ 0→π$.
• Cardioid $r = a(1±cosθ)$: $r 0→a(1±cosθ), θ 0→2π$ or $0→π$ by symmetry.
• Sector between $θ=α$ to $β$, $r = r1(θ) to r2(θ)$.

3. Worked Examples

(a) ∬_{x²+y²≤a²} e^{-(x²+y²)} dA = ∫_{0}^{2π}∫_{0}^{a} e^{-r²} r dr dθ = 2π·[ -e^{-r²}/2]_{0}^{a}= π(1-e^{-a²}) ✓
    As a→∞, integral →π (Gaussian integral trick leads to ∫ e^{-x²}dx =√π).

(b) Area of circle r=2a cosθ (diameter on x-axis): D: r 0→2a cosθ, θ -π/2→π/2
    Area = ∫_{-π/2}^{π/2}∫_{0}^{2a cosθ} r dr dθ = ∫ 2a² cos²θ dθ = π a²? Actually diameter length 2a ⇒ radius a ⇒ area πa² ✓

(c) ∬_{D} √(x²+y²) dA where D is quadrant disc x≥0,y≥0,r≤a:
    = ∫_{0}^{π/2}∫_{0}^{a} r·r dr dθ = (π/2)·a³/3 = π a³/6 ✓

(d) I = ∫_{0}^{a}∫_{0}^{√(a²-x²)} (x²+y²) dy dx ; convert: x=r cosθ,y=r sinθ, 0≤r≤a,0≤θ≤π/2
    = ∫_{0}^{π/2}∫_{0}^{a} r²· r dr dθ = (π/2)·a⁴/4 = π a⁴/8 ✓

(e) ∫_{0}^{1}∫_{0}^{√(1-x²)} e^{x²+y²} dy dx? Actually easier polar: =∫_{0}^{π/2}∫_{0}^{1} e^{r²} r dr dθ = (π/2)(e-1)/2 ✓

4. Jacobian Reminder

The factor $r$ is essential: $dA = |J| dr dθ$ where $J = r$. For $∫∫ f(x,y)dx dy$, forgetting $r$ is common error.

5. Exam Points

• Convert $∬_{D} f(x,y) dA$ over disc/sector/cardioid to $∫∫ f(r,θ) r dr dθ$ — 5-7 marks, very frequent.
• Evaluate $∬ e^{-(x²+y²)}$, $∬ √(x²+y²)$, $∬ (x²+y²)$ over circle/quadrant — 7 marks, very frequent.
• Find area via $∬ 1·r dr dθ$ — 5 marks, frequent.