The question asks $\iint_R (3x+4y^2)\; dA$ where $R$ is the region in the upper half plane bounded by the circles $x^2+y^2=1$ and $x^2+y^2=4$ $$\int_0^\pi \int_1^2 (3r\cos \theta + 3r^2 \sin^2 \theta) r \;dr d\theta $$ Now the question is since $r=-1$ is the same as $r=1$ by the convention if we replace the bounds of $r$ by $-1,-2$ or $-1,2$ it will give a different answer! What's wrong here?
In the past the first time I learned about $r$ in polar coordinate we did not allowed negative values why in the books like Calculus Early Transcendentals by James Stewart have this convention?
In complex analysis $r$ is equal to $|z|$ which always positive.