Integrate $f(x,y) = e^{-x^2-y^2}$ in a circle with radius $1$ and center at $(0,0)$

Question

I am asked to integrate $f(x,y) = e^{-x^2-y^2}$ in a circle with radius $1$ and center at $(0,0)$.

The setup of the integral (in my solution) is

$$ \int_{0}^{2\pi} \int_{0}^{1} e^{-r^2} r dr d\theta$$

which gives me

$$- \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{-1} e^u du d\theta = \frac{1}{2} \int_{0}^{2\pi} \left( 1-1/e \right) d\theta = \frac{2\pi}{2} \left( 1-1/e \right) = \frac{\pi(e-1)}{e}$$

as a solution.

The answer from the book is

$$\pi(e-1)$$

Where did I go wrong? Or is the textbook wrong?

Answer 1

Let $u = -r^2$, then $du = -2r \, dr$. And the limits change from $r=0$ and $r=1$ to $u = 0$ and $u=-1.$ So you get:

$$ \int_0^{2\pi} \int_0^{-1} -\frac{1}{2} e^u \, du \, d\theta, $$ which you correctly evaluated.

Answer 2

Assuming you are to integrate over the disk, not the circle (which is just the perimeter), you are doing fine. Alpha agrees with you that $$\int_0^1 e^{-r^2}r\ dr=\frac{e-1}{2e}$$

Answer 3

You have to compute $$ \int_{0}^{1}re^{-r^2}\,dr $$ which can't be negative. The substitution $-r^2=u$ gives $$ -2r\,dr=du $$ so the integral becomes $$ -\frac{1}{2}\int_0^{-1}e^u\,du=-\frac{1}{2}\Bigl[e^u\Bigr]_0^{-1}= -\frac{1}{2}(e^{-1}-1)=\frac{1-e^{-1}}{2} $$ Thus your original integral is $$ 2\pi\frac{1-e^{-1}}{2}=\pi(1-e^{-1}) $$