I want to show $\int_0^1[\int_{1-x}^{\sqrt{1-x^2}} f(x,y)dy]dx=\int_0^{\pi /2}[\int_{1/(\cos t+\sin t)}^1f(r\cos t,r\sin t)rdr]dt$
Using polar transformation ; I see that $r\cos t=0 $ and $r\sin t =1$ when $y=1-x$ and $x=0$ ; and $r\cos t=1 $ and $r=1$ when $y=\sqrt{1-x^2} $ and $x=1$ ; so the range of $r$ is clear to me and it seems $0 \le \cos t \le 1$ , and at lower limit $\cos t=0$ and at upper $\cos t =1$ , but then how does $t$ run from $0$ to $\pi/2$ , since clearly $\cos 0=1 , \cos \pi / 2=0$ ? Please help . Thanks in advance
I think that a plot for the doamin of integration will clarify the idea. So consider the following plot for $ y=1-x$ and $y=\sqrt{1-x^2}$ with $0 \leq x \leq 1$
Trying to change into polar coordinates, it is clear from above that $t$ which It is represented in the plot by $\theta$ has a range $[0 , 2\pi]$.
On the other hand, we have the lower bound of the domain is the curve $y=1-x$, substituting $y=r\sin(t)$ and $x=r\cos(t) $ we get $ r\sin (t)=1-r\cos(t)$ and hence $ r\sin(t) +r\cos(t)=1 $ so $$r=\frac{1}{\sin(t)+\cos(t)}$$ Now, the upper bound of the domain is the curve $y=\sqrt{1-x^2}$ with both $x$ and $y$ positive, so we may square both sides and get $ r^2\sin(t)^2= 1-r^2 \cos(t)^2 $, implying $r^2=1$ and so $r=1$ ($r \geq 0$).