limits of integration in polar coordinates

Question

Suppose that I´m integrating some function $f(r,\theta)$ over a region, say, $r=\cos(\theta)$. This region is a circle that is completely in the first and fourth quadrants. This is why we generally put the angle limits as being from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. When the angle is between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, $r$ is negative, meaning that it is "backwards". So my question is, if I were to put the limits of integration from $0$ to $2\pi$, instead of $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, would the calculated value of the function be twice the actual value, given that the area of the circle was double counted?

Answer 1

Just do it (the evident integral). Note that the $r < 0$ in $r \,\mathrm{d}r \mathrm{d}\theta$ will occur when the bounds of integration $\int_0^{\cos \theta}$ are "backwards", so we could get twice the value, depending on a special property of $f$.

• $f$ need not be defined for negative radii -- this integral is undefined.
• $f$ could well be defined for negative $r$, but need not agree with the value it has at the same point expressed with positive radius, so $f(-r, \theta) \neq f(r, \pi + \theta)$. In this case, the result could be "anything", depending on what unrelated values the function has on the two ways to represent the "same point". (They're not the same point in the $r$-$\theta$ coordinates, since we are allowing $(r,\theta) \in \mathbb{R} \times \mathbb{R}$. They have different coordinates. They only appear the same when we plot them using the polar coordinates recipe. This extra constraint is analogous to requiring that $f$ be $2\pi$-periodic in $\theta$.)
• $f$ could well be defined for negative $r$ with $f(-r, \theta) = f(r, \pi + \theta)$. Then the positive $r$ part duplicates the negative $r$ part -- because the bounds of the $r$ integral are in the "flipped" position, so we can arrange to eliminate the two minus signs.