<s>[INST] Calculate $\iint \sqrt{2-x^2-y^2} \, dx \, dy$ over the circle $x^2+y^2=x\sqrt{2}$ so as my question states I have to calculate
$$\iint_\limits A \sqrt{2-x^2-y^2}\,dx\,dy$$ where $A$ is the circle defined by
$$x^2+y^2=x\sqrt{2}$$
After introducing polar coordinates $(x = r\cos\theta, y= r\sin\theta)$, I get that:
The left side of the equation of the circle is always nonnegative, so the right side also has to be nonnegative. This means that $x \ge 0$, or
$$\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].$$
Also we have that
$$r=\sqrt{2} \cos\theta$$
Here's where I have a slight problem. If I think about it, $2-x^2-y^2$ has to be $\ge 0$ for the square root to be real. Now, that means that $r \le \sqrt{2}$.
How do I determine the boundaries of $r$? I think that the boundaries for $r$ are $r \in [\sqrt{2}\cos\theta, \sqrt{2}]$. If yes, could anyone explain why exactly? I'm confused because I don't have an inequality in $r=\sqrt{2}\cos\theta$!
 [/INST] Your are almost there. You need to integrate over the circle $x^2 + y^2 \leq \sqrt2 x$.
So as you said, $0 \leq r \leq \sqrt2 \cos\theta, - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
The integrand is $\sqrt{2-x^2-y^2} = \sqrt{2-r^2}$
and as $r \leq \sqrt2 \cos\theta, \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \ , \  $  we have $\ r \leq \sqrt2$
So the integrand is defined in the entire region and the integral in polar coordinates is,
$$ \int_{-\pi/2}^{\pi/2} \int_0^{\sqrt2 \cos\theta} r \ \sqrt{2-r^2} \ dr \ d\theta$$
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