I am answering sample exams for my Calculus class and my attention was caught by the following item.

Set-up the definite integral or sum of definite integrals equal to the area of the region above the polar axis, inside the limaçon $r = 3 + 2 \sin \theta$ and outside the lemniscate $r^2 = 32 \cos 2\theta$ given that the two curves intersect at $(4,\frac{\pi}{6})$.

plots of the polar curves $r = 3 + 2 \sin \theta$ (blue) and $r^2 = 32 \cos 2\theta$ (violet) from $0$ to $2\pi$ (generated by _Mathematica_)

At first, I thought that the area is given by $$\dfrac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{[(3 + 2\sin \theta)^2 - (32 \cos 2\theta)] \mathrm{d}\theta}$$ but I know that the area of the lemniscate is tricky so I may have given a smaller area.

My question is this: How do you know the limits of integration for lemniscates? (I know that the limits of integration for the area of the lemniscate alone is from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$, but how about for small portions of the curve?)

I'll appreciate any help. Thank you so much.


There was lemniscate trouble, as you feared. Up to $\theta=\frac{\pi}{4}$, everything was fine. But from $\frac{\pi}{4}$ and for quite a while (up to $\frac{3\pi}{4}$), $\cos 2\theta$ is negative, so $r^2$ is negative and there is no curve. The integral doesn't know and doesn't care: it cheerfully "adds up" these negatives, giving the wrong answer.

So for the limaçon part, if you do things in your style, you will have to break up the limaçon integral into two integrals, $\frac{\pi}{6}$ to $\frac{\pi}{4}$ and then $\frac{3\pi}{4}$ to $\frac{5\pi}{6}$.

Since the integrals for the lemniscate and the limaçon are over different intervals, we cannot express their difference as a single integral.

What I would do is to use the symmetry, take the right-half of the region and multiply the resulting area by $2$. The limaçon part to be subtracted uses the integral from $\frac{\pi}{6}$ to $\frac{\pi}{4}$.

So there is $\frac{1}{2}$ as in your formula, but ultimately we multiply by $2$, so our area is $$\int_{\pi/6}^{\pi/2}(3+2\sin \theta)^2\,d\theta-\int_{\pi/6}^{\pi/4}32\cos 2\theta\,d\theta.$$

More pleasant, fewer minus signs to worry about!

  • $\begingroup$ Would you say I got it when I said the limits of integration has something to do with the polar tangents? $\endgroup$ – Gelo Jacamile Aug 6 '12 at 12:53
  • 1
    $\begingroup$ @GeloJacamile: First please note I had a typo (corrected) in the displayed integral, for the limaçon we start at $\pi/6$. We do have that the line $\theta=\frac{\pi}{4}$ is tangent to the curve, but the real issue is that there is no curve, but $r^2$ is still meaningful to the integral. $\endgroup$ – André Nicolas Aug 6 '12 at 13:06
  • $\begingroup$ Thank you. This really helped. $\endgroup$ – Gelo Jacamile Aug 7 '12 at 7:27

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