Finding $\iint_D \frac{10}{ \sqrt {x^2 + y^2}}dx\,dy$ for $D = \{(x,y) \mid 0 \leq x \leq 1, \sqrt{1-x^2} \leq y \leq x\}$

Question

As I said title, $$\iint_D \frac{10}{ \sqrt {x^2 + y^2}}\,dx\,dy$$ for $$D = \{(x,y) \mid 0 \leq x \leq 1, \sqrt{1-x^2} \leq y \leq x\}$$

I tried it using integration by substitution by $(x,y) = (r\cos\theta, r\sin\theta)$

Then Only just we left to find the $\iint_{D'}10 \,dr \,d\theta$ for the domain $D'$ whose variable $r$ and $\theta$

We can surely say $0\leq \theta \leq {\pi \over 4}$ But I couldn't find the range of the $r$.

The answer sheet said $1 \leq r \leq {1 \over \cos\theta}$. So Why does the range of the $r$ should that be?

Any help would be appreciated. Thanks.

Answer 1

Putting polar coordinates to inequaility $\sqrt{1-x^2} \leqslant y$ gives exactly $1 \leqslant r $ and putting to $x \leqslant 1$ gives $ r \leqslant {1 \over cos\theta}$. By the way, restrictions on $\theta$ is obtained on same way and plot on $(r,\theta)$ is very nice.