Evaluate the double integral of e raised to the power minus (x squared plus y squared), over the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

Step-by-step solution

1. 1

   Do not reach for polar coordinates. Polar is the standard trick for Gaussians, but it only helps when the region is a disk or the whole plane. Here the region is a square, so the boundary $r$ would depend on $\theta$ in a piecewise way. The right move is separation instead.

2. 2

   Factor the integrand. The exponent is a sum, so the exponential is a product:

   $e^{-(x^2+y^2)} = e^{-x^2} e^{-y^2}$

   Because the limits are constants and each factor involves only one variable, the double integral splits into a product of single integrals:

   $\int_0^1 \!\! \int_0^1 e^{-x^2} e^{-y^2}\,dy\,dx = \left(\int_0^1 e^{-x^2}dx\right)\left(\int_0^1 e^{-y^2}dy\right) = \left(\int_0^1 e^{-t^2}dt\right)^2$

3. 3

   Name the one-dimensional integral. The function $e^{-t^2}$ has no elementary antiderivative, so the answer is expressed with the error function, defined by

   $\operatorname{erf}(a) = \frac{2}{\sqrt{\pi}}\int_0^a e^{-t^2}\,dt$

   Rearranging gives

   $\int_0^1 e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}\operatorname{erf}(1)$

4. 4

   Square it.

   $\int_0^1 \!\! \int_0^1 e^{-(x^2+y^2)}\,dA = \left(\frac{\sqrt{\pi}}{2}\operatorname{erf}(1)\right)^2 = \frac{\pi}{4}\operatorname{erf}(1)^2$

5. 5

   Evaluate numerically and sanity-check. With $\operatorname{erf}(1) = 0.8427008$:

   $\frac{\pi}{4}(0.8427008)^2 = 0.5577463$

   Direct quadrature over the square gives $0.55774629$, matching to eight digits. The bound check also works: the integrand runs from $e^{-2} \approx 0.135$ to $1$ on a region of area $1$, and $0.558$ lies between.