Calculate : $\int_{} \int_ {} e^{-(x^2+y^2)}dA$ While $R$ is the area in $x^2+y^2=1$

Question

Hello I need to calculate this

$$\iint_R e^{-(x^2+y^2)}d\mu$$

Where $R$ is the unit disk, given by $\left\{(x,y):x^2+y^2\leq 1\right\}.$

What I did :

$$\int_{-1}^1 \int_{-\sqrt{1+x^2}}^{\sqrt{1+x^2}} e^{-(x^2+y^2)}dydx=\int_0^{2\pi} \int_0^1 (e^{-1}*r)drd\theta=\int_0^{2\pi} 0.5(e^{-1}*1)d\theta=(0.5(e^{-1}*1))2\pi$$

But for some reason I don't get the same answer as the book any ideas?

Thanks in advance!

Answer 1

I guess you mean the integral area is $x^2+y^2\le 1$.

Take a polar transformation, we have: $$ \iint_Re^{-(x^2+y^2)}dA=\int_0^{2\pi}\int_0^1e^{-r^2}rdrd\theta=\frac{\pi(e-1)}{e} $$