How to calculate this double integral by using polar?

Question

$$\iint_S \sqrt{4-x^2 - y^2} dA $$

How to calculate the above double integral where S is an area inside the first quadrant of a circle $x^2 + y^2 = 4$, and bounded by line $y = 0$ and $y = x$.

What I know is $x^2 + y^2 = r^2$, $y = r \sin \theta$, $x = r \cos \theta$, $dA = r dr d\theta$.

I have tried put converting the equation by converting $x, y$, and $dA$ to match the polar equation but still confused in determining the boundaries of $r$ and $\theta$, it would be very great if anyone would give the explanation for determining the boundaries and the meaning of $S$ below the double integration in the above question.

Answer 1

In polar coordinates, $S=\{(r,\theta): 0\leq r\leq 2, 0\leq \theta\leq \pi/4\}$. Your integral is $$\int_0^{\pi/4}\int_0^2 r\sqrt{4-r^2}\,dr\,d\theta.$$

Answer 2

The function $\sqrt{4-x^2-y^2}$ can be rewritten as $\sqrt{4-r^2}$.

After that, it's just a matter of expressing your region using bounds on $r$ and $\theta$. This is most easily done if you can picture it, or draw a picture of it. You will find that both polar variables are bounded by constants: $a\le r\le b$ and $c\le \theta\le d$. Can you tell what values we need to use for $a,b,c,d$?