I have a question about setting up the bounds on the theta integral of a polar double integral.
When you have a circle such as $(x-1)^2 + y^2 = 1$, the equation of the circle can be shown to be $2\cos(\theta)$. So $r$ will go from $0$ to $2\cos(\theta)$. Fair enough.
But the theta. I have watched one excellent online video (Prof. Leonard) who has the theta range of this integral as $0$ to $\pi$. On the other hand, the Stewart Calculus book has this same problem with a limit of $-\pi/2$ to $\pi/2$.
I know these may give the same answers, but which is the correct set up?
Thanks so much (and, yes, I've reviewed the other questions on the forum before posting this).
As it pointed in Jyrki Lahtonen's comment if we consider $(x-1)^2 + y^2 = 1$, then polar coordinates gives $0 \leqslant r \leqslant 2 \cos \theta$. So we have region for $\theta$ from inequality $$0 \leqslant \cos \theta$$ If initially we choose $ \theta \in (-\pi, \pi]$, then we obtain solution $ \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.