Transforming integral limits from cartestian to polar coordinates I'm asked to find the integral of $\iint_Df(x,y)\,dA$ where $D$ is limited by the circle $x^2+y^2=1$, $y=x$ and $x=1$. $f(x,y)$ is given in the original task and simplifies the integral a lot when doing this in polar coordinates. My question however is about the limits. 
Setting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get that from $y=x$, the limits for $\theta$ must be $0 \leq \theta \leq \frac{\pi}{4}$.
From $x=1$, I obtain $r\cos(\theta) = 1 \iff r = \frac{1}{\cos(\theta)}$. I initially thought that the limits for $r$ in this case must then be $0 \leq r \leq \frac{1}{\cos(\theta)}$, but it turns out the correct answer is 
$$1 \leq r \leq \frac{1}{\cos(\theta)}.$$ It's not hard to draw this sketch by hand, so I know what the area looks like. However I seem to be having a difficult time visualizing why $r$ must begin at $1$.

The domain of integration is outer with respect to the circle $x^2+y^2=1$ then of course we have that 
$$1 \leq r \leq \frac{1}{\cos(\theta)}$$
otherwise we would integrate over the triangle bounded by $y=x$ and $x=1$ in the first quadrant.
In other words, the domain of integration $D$ is given by


*

*$x^2+y^2\ge 1$

*$0\le x \le 1$

*$0 \le y\le x$

