# Find  int int \ e^(x^2+y^2) \ dA  over the region bounded by y = sqrt(9-x^2)?

Mar 29, 2017

$\frac{\pi \left({e}^{9} - 1\right)}{2}$

#### Explanation:

We want to find:

$I = \int {\int}_{R} \setminus {e}^{{x}^{2} + {y}^{2}} \setminus \mathrm{dA}$

Where $R$ is this region, also bounded by the $x$-axis:
graph{ sqrt(9-x^2) [-10, 10, -5, 5] }

If we examine the region $R$ we have;

$y = \sqrt{9 - {x}^{2}} \implies {y}^{2} = 9 - {x}^{2} \implies {x}^{2} + {y}^{2} = {3}^{2}$

which is a circle centre $O$, radius $3$ (we are only interested in the +ve semi-circle)

We can readily (and logically) perform this integration by converting to Polar coordinates, which leads to the following transformation. The region $R$ is:

an angle from $\theta = 0$ to $\theta = \pi$
a ray from $r = 0$ to $r = 3$

And as we convert to Polar coordinates we get:

$\setminus \setminus \setminus x = r \cos \theta$
$\setminus \setminus \setminus y = r \sin \theta$
$\mathrm{dA} = \mathrm{dx} \setminus \mathrm{dy} = r \setminus \mathrm{dr} \setminus d \theta$

And so we get (by performing the inner integration first, followed by the outer):

$I = {\int}_{0}^{\pi} {\int}_{0}^{3} \setminus {e}^{{\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2}} \setminus r \setminus \mathrm{dr} \setminus d \theta$
$\setminus \setminus = {\int}_{0}^{\pi} {\int}_{0}^{3} \setminus {e}^{{r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta} \setminus r \setminus \mathrm{dr} \setminus d \theta$
$\setminus \setminus = {\int}_{0}^{\pi} {\int}_{0}^{3} \setminus {e}^{{r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)} \setminus r \setminus \mathrm{dr} \setminus d \theta$
$\setminus \setminus = {\int}_{0}^{\pi} {\int}_{0}^{3} \setminus {e}^{{r}^{2}} \setminus r \setminus \mathrm{dr} \setminus d \theta$
$\setminus \setminus = {\int}_{0}^{\pi} {\left[\frac{1}{2} \setminus {e}^{{r}^{2}}\right]}_{0}^{3} \setminus d \theta$
$\setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\pi} \left({e}^{9} - {e}^{0}\right) \setminus d \theta$
$\setminus \setminus = \frac{1}{2} \left({e}^{9} - 1\right) \setminus {\int}_{0}^{\pi} \setminus d \theta$
$\setminus \setminus = \frac{1}{2} \left({e}^{9} - 1\right) \setminus {\left[\theta\right]}_{0}^{\pi}$
$\setminus \setminus = \frac{1}{2} \left({e}^{9} - 1\right) \setminus \left(\pi - 0\right)$
$\setminus \setminus = \frac{\pi \left({e}^{9} - 1\right)}{2}$