Evaluation of the integral of $e^{-(x^2+y^2)}$ over a disk

Question

Show that $$\renewcommand{\intd}{\,\mathrm{d}} \iint_{D(R)} e^{-(x^2+y^2)} \intd x \intd y = \pi \left(1 - e^{-R^2}\right)$$ where $D(R)$ is the disc of radius $R$ with center $(0,0).$

I have never been asked to calculate a double integral without a defined region, so I don't even know where to start. I don't know the boundaries.

This is my guess:

$$0 < r < R\\ 0 < \theta < 2\pi $$

Is this correct?

Answer 1

Yes, using polar coordinates the boundaries are: $$0 \leq r \leq R \\ 0 \leq \theta \leq 2 \pi$$

Since $D(R)$ is the disk of radius $R$ with center at $(0,0)$: $D(R)=\{(x,y): x^2+y^2 \leq R^2\}$

So we have the following:

$$x = r\cos \theta, y = r \sin\theta \\ \renewcommand{\intd}{\,\mathrm{d}} \intd x \intd y = r \intd r \intd \theta$$ $$ \begin{align} \iint_{D(R)} e^{-(x^2+y^2)} \intd x \intd y &= \int_0^{2 \pi} \int_0^R e^{-r^2} r \intd r \intd \theta \\ &= \int_0^{2 \pi} \int_0^R \frac{\mathrm{d}}{\intd r} \left( -\frac{1}{2} e^{-r^2} \right) \intd r \intd \theta \\ &= \int_0^{2 \pi} \left( -\frac{1}{2} e^{-r^2} \right )_0^R \intd \theta \\ &= \int_0^{2 \pi} -\frac{1}{2} \left( e^{-R^2} - 1 \right) \intd \theta \\ &= -\frac{1}{2} \left( e^{-R^2} - 1 \right) 2 \pi \\ &= -\pi \left( e^{-R^2} - 1 \right) \\ &= \pi \left( 1 - e^{-R^2} \right) \end{align}$$

Answer 2

Hint: Let $x = r\cos \theta$ and $y = r\sin \theta$, then $$I = \int_{0}^{2\pi} \int_{0}^R r e^{-r^2} \,\mathrm{d}r \,\mathrm{d}\theta$$

Answer 3

You have a defined region to integrate about, which is $D(R)=\{(x,y) \in \mathbb{R}^{2}| x^{2}+y^{2} \leq R^{2}\}$ as the disc with radius $R$.

At first, you should choose an adequate parametrization for making the problem easier. Here you might use the transformation $(r, \theta) \mapsto (x, y)=(r \cos \theta, r \sin \theta)$ with $r \in [0, R]$, $\theta \in [0, 2\pi)$.