Multiple IntegralsHard

Double integral in polar coordinates

Calculate

\displaystyle\iint_D (x^2+y^2)\,dA

on the disc x²+y²≤4.

Given data

D: x²+y²≤4

Steps

0 / 3

Switch to polar coordinates: x²+y² = r²; D* = [0,2]×[0,2π].
Write your working — the numerical result must appear in the text (logica)
Integrate in θ: ∫₀^{2π} dθ.
Integrate in r: ∫₀²r³ dr and multiply by 2π.

Full worked solution

Switch to polar coordinates: x²+y² = r²; D* = [0,2]×[0,2π].
$\iint = \int_0^{2\pi}\int_0^2 r^2\cdot r\,dr\,d\theta$
In polar coordinates: $x^2+y^2 = r^2$ , $dA = r\,dr\,d\theta$ . The integrand becomes $r^2 \cdot r = r^3$ , over $r \in [0,2]$ , $\theta \in [0,2\pi]$ .
Integrate in θ: ∫₀^{2π} dθ.
$\int_0^{2\pi}d\theta = 2\pi$
$\int_0^{2\pi} d\theta = \mathbf{2\pi}$ . The $\theta$ integral separates from the $r$ integral, giving a factor of $2\pi$ .
Integrate in r: ∫₀²r³ dr and multiply by 2π.
$2\pi\int_0^2 r^3\,dr = 2\pi\cdot4 = 8\pi \approx 25.13$
$\int_0^2 r^3\,dr = \left[\frac{r^4}{4}\right]_0^2 = \frac{16}{4} = 4$ . Result: $2\pi \times 4 = \mathbf{8\pi} \approx \mathbf{25.13}$ .

Result:∬(x²+y²) dA = 8π ≈ 25.13.

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