Change the Cartesian integral into an equivalent polar integral, then solve it.
$$\int_1^{\sqrt{3}}\int_{1}^x\,dy\,dx$$
i could find $\theta$ goes through $\pi/6$ to $\pi/4$ using $\tan^{-1}\theta$=$\frac{y}{x}$ and points: ($\sqrt{3}$,1) and ($\sqrt{3}$,$\sqrt{3}$) or from graph and area of triangle= 2-$\sqrt{3}$
x=rcos$\theta$ , y=rsin$\theta$ ,When x=1 and x=$\sqrt{3}$ r=sec$\theta$ and r=$\sqrt{3}$sec$\theta$
When y=1 and y=$\sqrt{3}$, r=csc$\theta$ and r=$\sqrt{3}$csc$\theta$
Which equations of r should i use and why?
what confuses me is that y=1 not 0, if y was from 0 to x i would choose r to be r=sec$\theta$ and r=$\sqrt{3}$sec$\theta$ and $\theta$ from 0 to $\pi/4$ and area=1
For a given angle $\theta$ (between the bounds that you have found), the corresponding ray from the origin enters the integration domain when it intersects the line $y=1$, i.e. when $r \sin\theta=1$, and it exits when it intersects the line $x=\sqrt3$, i.e. when $r \cos\theta=\sqrt3$. So you get $$ \int_{\theta=\pi/6}^{\pi/4} \int_{r=1/\sin\theta}^{\sqrt3/\cos\theta} \cdots $$