Right now I am working on a problem that involves finding the area enclosed by a single loop given the equation $r=4\cos(3\theta)$. I know that the cosine is bounded from zero to $\pi$, but when using a lower limit of $0$, and a upper limit of $\pi/3$, I get the wrong answer (the answer is $4\pi/3$).

Should I instead use zero to $2\pi/3$, since that would be a full period? I tried drawing the graph as well, but had no luck with that. What is the best way of finding the upper and lowers limits for these kind of problems?

  • $\begingroup$ "I tried drawing the graph as well, but had no luck with that" For this part, see this online tool, which might also make the mathematical answer to the rest of your question quite transparent. $\endgroup$
    – Did
    Dec 15, 2016 at 9:28
  • $\begingroup$ Use limits of theta (T) from 0 to (pi/6), since 3T will have total limit from 0 to (pi/2). $\endgroup$
    – Bhaskar
    Dec 15, 2016 at 9:46
  • $\begingroup$ Using those limits worked, but why are those the limits of integration. Why not -pi/6 to pi/6, since the given function would be zero there. $\endgroup$
    – Fyree
    Dec 15, 2016 at 17:46

1 Answer 1


It helps if you have an idea of the graph, but even if you don't: it should be clear that at $\theta = 0$, you have $r(0) = 4$ so you're not at the beginning of a loop. A loop starts when $r=0$ and a single loop closes at the next root of $r(\theta)$.

The function $\cos x$ has a root at $-\tfrac{\pi}{2}$ and the next one is at $\tfrac{\pi}{2}$. For this polar curve $r = 4 \cos(3\theta)$, you get (with $x=3\theta$): $$3\theta = \pm \frac{\pi}{2} \Rightarrow \theta = \pm \frac{\pi}{6}$$ so you go through exactly one loop if you let $\theta$ run from $-\tfrac{\pi}{6}$ to $\tfrac{\pi}{6}$. Using the formula for area: $$\int_{\theta_1}^{\theta_2} \tfrac{1}{2}r^2 \,\mbox{d}\theta \to \int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$ Note that because the function is even, it is slightly easier to (manually) calculate : $$\color{blue}{2}\int_{\color{red}{0}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \int_{0}^{\tfrac{\pi}{6}} \left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$

  • $\begingroup$ That helps, thanks! When I integrate it though I am getting 8pi/3, not 4pi/3. $\endgroup$
    – Fyree
    Dec 15, 2016 at 17:27
  • $\begingroup$ It's probably a mistake in your calculation, the answer is indeed $4\pi/3$. $\endgroup$
    – StackTD
    Dec 15, 2016 at 21:59
  • 1
    $\begingroup$ Yes, I was not dividing by two when I needed to. I looked it over again, and got 4π/3. $\endgroup$
    – Fyree
    Dec 16, 2016 at 20:27

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