Evaluate the double integral of e^(-(x^2 + y^2)) over the region x^2 + y^2 <= 1.

Step-by-step solution

1. 1

   See why Cartesian coordinates are hopeless. In $x$ and $y$ the inner integral would be $\int e^{-x^2}dx$, and that antiderivative simply does not exist in elementary form. The integrand and the region both depend only on the distance from the origin, so polar coordinates are not merely convenient here — they are what makes the problem solvable.

2. 2

   Convert integrand, region, and area element together. With $x=r\cos\theta$, $y=r\sin\theta$:

   $x^2+y^2=r^2,\qquad dA=r\,dr\,d\theta,\qquad x^2+y^2\le 1\iff 0\le r\le 1$

   so

   $\iint_{x^2+y^2\le 1}e^{-(x^2+y^2)}dA=\int_0^{2\pi}\!\!\int_0^1 e^{-r^2}\,r\,dr\,d\theta$

   The extra $r$ from the Jacobian is the whole trick: it is exactly the factor needed for a $u$-substitution.

3. 3

   Do the radial integral with $u=r^2$. Let $u=r^2$, so $du=2r\,dr$ and $r\,dr=\tfrac12\,du$, with $u$ running from $0$ to $1$:

   $\int_0^1 e^{-r^2}r\,dr=\frac12\int_0^1 e^{-u}du=\frac12\left[-e^{-u}\right]_0^1=\frac12\left(1-e^{-1}\right)$

4. 4

   Integrate over the angle. The radial result is a constant, and the full circle contributes $2\pi$:

   $\int_0^{2\pi}\frac12\left(1-e^{-1}\right)d\theta=2\pi\cdot\frac12\left(1-e^{-1}\right)=\pi\left(1-e^{-1}\right)$

5. 5

   Evaluate numerically and check the bounds.

   $\pi\left(1-\frac{1}{e}\right)\approx \pi(1-0.36788)\approx 1.9858$

   On the unit disk the integrand ranges from $e^{-1}\approx 0.368$ (at the rim) to $1$ (at the centre), and the disk has area $\pi\approx 3.1416$. So the integral must lie between $0.368\pi\approx 1.156$ and $\pi\approx 3.142$, and $1.9858$ does $\checkmark$. Extending the same computation to $r\to\infty$ gives $\pi$, the classic result behind $\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt\pi$.