Answer
$\bar f = \frac{1}{{\sqrt 2 \pi }}\mathop \smallint \limits_{x = - \sqrt 2 }^{\sqrt 2 } \mathop \smallint \limits_{y = - \sqrt {1 - {x^2}/2} }^{\sqrt {1 - {x^2}/2} } {{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x$
Using a computer algebra system, we evaluate this integral numerically and the result is
$\bar f \simeq 1.042$
Work Step by Step
The ellipse $\frac{{{x^2}}}{2} + {y^2} = 1$ is bounded left and right by $x = - \sqrt 2 $ and $x = \sqrt 2 $, respectively. These are obtained by substituting $y=0$ in the equation of the ellipse.
Using $\frac{{{x^2}}}{2} + {y^2} = 1$, we obtain $y = \pm \sqrt {1 - \frac{{{x^2}}}{2}} $. Thus, we can describe the region of the ellipse as vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)| - \sqrt 2 \le x \le \sqrt 2 , - \sqrt {1 - \frac{{{x^2}}}{2}} \le y \le \sqrt {1 - \frac{{{x^2}}}{2}} } \right\}$
We learn from Exercise 34 in Section 16.6, the area of the ellipse: ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$ is $\pi ab$. Therefore, the area of the ellipse $\frac{{{x^2}}}{2} + {y^2} = 1$ is
${\rm{Area}}\left( {\cal D} \right) = \sqrt 2 \pi $
By definition, the average value of $f\left( {x,y} \right)$ on a domain ${\cal D}$ is given by Eq. (8) in Section 16.2:
$\bar f = \frac{1}{{{\rm{Area}}\left( {\cal D} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A$
So,
$\bar f = \frac{1}{{\sqrt 2 \pi }}\mathop \smallint \limits_{x = - \sqrt 2 }^{\sqrt 2 } \mathop \smallint \limits_{y = - \sqrt {1 - {x^2}/2} }^{\sqrt {1 - {x^2}/2} } {{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x$
Using a computer algebra system, we evaluate this integral numerically and the result is
$\bar f \simeq 1.042$
