Double Integral in Polar Coordinates - Confused by angle

Question

I need to compute the integral $$\int_{-1}^{0} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2} }x dy dx $$

by converting to polar coordinates.

Now, since $-\sqrt{1-x^{2}}\leq y \leq \sqrt{1-x^2}$, we are integrating over the unit circle. However, since $-1 \leq x \leq 0$ - i.e., $-1 \leq \cos \theta \leq 0$, I was under the impression that $\theta$ should range from $\displaystyle \frac{\pi}{2}$ to $\pi$.

But, in fact, I was informed that in this problem, we are integrating over the left half of the unit circle (so $\theta$ should range from $\pi/2$ to $3 \pi/2$ [and backwards??]), and not just the upper left-hand corner of the circle. Could somebody please explain to me why that is/how I should be able to tell that from the integral?

Thank you.

Answer 1

Draw the canonical unit circle, then: the integration domain is the left half unit disk, and from here the integral is:

$$\int_0^1\int_{\pi/2}^{3\pi/2}r^2\cos\theta\,d\theta\,dr$$

You can check this integrating both ways:

$$\int_{-1}^0\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} x\,dy\,dx=\int_{-1}^02x\sqrt{1-x^21}dx=\left.-\frac23(1-x^2)^{3/2}\right|_{-1}^0=-\frac23(1)=-\frac23$$

$$\int_0^1\int_{\pi/2}^{3\pi/2}r^2\cos\theta\,d\theta\,dr=\left.\int_0^1r^2\sin\theta\right|_{\pi/2}^{3\pi/2}\;dr=-2\int_0^1r^2\,dr=-\frac23$$

Explanation about the original integral: as $\;-1\le x\le0\;$ , it is clear we're on the left half circle, and the limits for the integral of $\;dy\;$ show we're ranging between the lower circle and the upper one there.

Answer 2

Here's a purely analytical explanation that does not rely on pictures. The domain of integration is described by the inequalities \begin{align} \tag{1} r|\sin \theta|&\le \sqrt{1-r^2\cos^2\theta}, \\ \tag{2} -1&\le r\cos \theta\le 0, \end{align} where $(r, \theta)\in [0, \infty)\times [0, 2\pi]$ are the polar coordinates.

Squaring the first inequality and using that $\sin^2\theta+\cos^2\theta=1$, we see that (1) is equivalent to $r^2\le 1$. On the other hand, since $r\ge 0$, (2) implies that $\cos \theta\le 0$, so necessarily $\theta\in [\tfrac\pi 2, \tfrac{3 \pi}2]$. At this point, this is only a necessary condition. However, for each value of $\theta$ in this range, there exist some $r\in [0, 1]$ such that (2) holds. Therefore, we conclude that the two inequalities (1) and (2) are equivalent to $$ 0\le r\le 1, \qquad \frac\pi2\le \theta\le \frac{3\pi}{2}, $$ which is the desired result.

---

DISCLAIMER. I advice using pictures every time it is possible. However, reasoning like we did in this answer might be a useful exercise sometimes.