Solución
Esta es una integral impropia porque estamos integrando sobre una región R 2 R^2 R 2 R 2 R^2 R 2 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ ∞ 0 \le \theta \le 2\pi, 0 \le r \le \infin 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ ∞
∬ R 2 e − 10 ( x 2 + y 2 ) d x d y = ∫ θ = 0 θ = 2 π ∫ r = 0 r = ∞ e − 10 r 2 = r d r d θ = ∫ θ = 0 θ = 2 π ( lim a → ∞ ∫ r = 0 r = a e − 10 r 2 r d r ) d θ = ( ∫ θ = 0 θ = 2 π d θ ) ( lim a → ∞ ∫ r = 0 r = a e − 10 r 2 r d r ) = 2 π ( lim a → ∞ ∫ r = 0 r = a e − 10 r 2 r d r ) = 2 π lim a → ∞ ( − 1 20 ) ( e − 10 r 2 ∣ 0 a ) = 2 π ( − 1 20 ) lim a → ∞ ( e − 10 a 2 − 1 ) = π 10 \begin{aligned}
\iint_{R^2}e^{-10\big(x^2+y^2\big)}dxdy &= \int_{\theta=0}^{\theta= 2\pi}\int_{r=0}^{r=\infin}e^{-10r^2}=rdrd\theta = \int_{\theta=0}^{\theta= 2\pi}\bigg(\lim\limits_{a \to \infin}\int_{r=0}^{r=a}e^{-10r^2}rdr\bigg)d\theta\\
&= \bigg(\int_{\theta=0}^{\theta= 2\pi}d\theta\bigg)\bigg(\lim\limits_{a \to \infin}\int_{r=0}^{r=a}e^{-10r^2}rdr\bigg)\\
&= 2\pi\bigg(\lim\limits_{a \to \infin}\int_{r=0}^{r=a}e^{-10r^2}rdr\bigg)\\
&= 2\pi\lim\limits_{a \to \infin}\bigg(-\frac{1}{20}\bigg)\bigg(e^{-10r^2}\bigg|_0^a\bigg)\\
&= 2\pi\bigg(-\frac{1}{20}\bigg)\lim\limits_{a \to \infin}\bigg(e^{-10a^2}-1\bigg)\\
&= \frac{\pi}{10}
\end{aligned} ∬ R 2 e − 10 ( x 2 + y 2 ) d x d y = ∫ θ = 0 θ = 2 π ∫ r = 0 r = ∞ e − 10 r 2 = r d r d θ = ∫ θ = 0 θ = 2 π ( a → ∞ lim ∫ r = 0 r = a e − 10 r 2 r d r ) d θ = ( ∫ θ = 0 θ = 2 π d θ ) ( a → ∞ lim ∫ r = 0 r = a e − 10 r 2 r d r ) = 2 π ( a → ∞ lim ∫ r = 0 r = a e − 10 r 2 r d r ) = 2 π a → ∞ lim ( − 20 1 ) ( e − 10 r 2 ∣ ∣ 0 a ) = 2 π ( − 20 1 ) a → ∞ lim ( e − 10 a 2 − 1 ) = 10 π