I need to evaluate
$$\iint_D \sqrt{x^2 + y^2}\, dxdy$$
where $D: x\le x^2 + y^2\le 1$.
I can change this using polar coordinates. That is,
$$\iint_D r\cdot r\,drd\theta$$
where $0\le r\le 1$. But I cannot guess the range of $\theta$.
2026-03-29 06:29:05.1774765745
Evaluating $\iint_D \sqrt{x^2 + y^2}\, dxdy$ using polar coordinates
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Hint. The inequality $x^2 + y^2\leq 1$ represents the disc centred at $(0,0)$ and radius $1$.
On the other hand, the inequality $ x\leq x^2 + y^2$, which is equivalent to $(x-1/2)^2+y^2\geq 1/4$, is the complement of the disc centred at $(1/2,0)$ and radius $1/2$.
Now make a drawing of $D$.