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Got me a multivariable calculus exam coming up soon. Most of the material is pretty simple – lagrange multipliers, for instance (I’ll be writing something about those soon) – but I’ve got one particular problem regarding integrating with radial coordinates that’s hard and interesting.

I’m asked to find the area of a single loop of a rose described in radial coordinates by

r = 5 \cos (3 θ)

. This is difficult to visualize, but luck hath smiled upon us, and Desmos supports radial coordinates:

Not too bad. The maximum extent would appear to be 5, which makes sense, because cosine maxes out at 1 and we multiply by 5. I was a bit curious to see if the multiplier of theta controls the number of loops – as it turns out, the relation is a bit complicated, and I won’t get into that now.

It’s looking to me like the best solution here is to integrate for

\frac{- π}{2} \leq θ \leq \frac{π}{2}

, to grab that petal in the first and fourth quadrants, but I’m worried that that isn’t a general enough solution: other roses that I’ve got to graph might not be so conveniently situated. The more general way I can think of is integrating for all theta and dividing by the number of petals – which isn’t necessarily easily predictable for all theta, but is a damn sight better than just hoping my hand-drawn sketches appropriately capture the useful quadrants. Regardless, let’s do the first idea first, and see how it works out. In this case, we know there are 3 petals, so the integral should be

\int_{\frac{- π}{2}}^{\frac{π}{2}} \int_{0}^{5 \cos (3 θ)} r dr d θ

. Why

r dr d θ

instead of just

1 dr d θ

? Because of the shape of the mesh we’re integrating; I won’t elaborate, but only because Paul's Online Math Notes already did.

Integrating this isn’t too hard. The inner integral is simply

\int_{0}^{5 \cos (3 θ)} r dr \to {\frac{r^{2}}{2} |}_{0}^{5 \cos (3 θ)} \to \frac{25}{2} \cos^{2} (3 θ)

, giving us

\frac{25}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} \cos^{2} (3 θ) d θ

. This is kind of a nasty trig integral, one which I very much don’t want to derive myself, but that’s what Wolfram is for! Wolfram kindly tells me that my disgusting integral is

{\frac{25}{24} (6 θ + \sin (6 θ)) |}_{\frac{- π}{2}}^{\frac{π}{2}}

. Because

\sin (3 π) = 0

, this works out nicely to…

\frac{25}{24} 6 π = \frac{25}{4} π

. This is wrong.

I’ve kept the wrong work up because it’s subtly wrong, and (I think) does a pretty good job of illustrating one of the most annoying problems with polar coordinates. Do you see it?

The problem lies in how cosine behaves with multipliers.

\cos (θ)

makes a full trip from 1 to 0 in

0 \leq θ \leq π

;

\cos (3 θ)

makes three such trips in the same span. And our rose has three peaks.

By some horrible mathemagic, integrating what should be a half-turn integrates the whole pattern. As it turns out, this is true for every cosine rose: regardless of the multiplier, a full span of pi (not 2pi!) covers the entire shape. Because we only want one petal, we have to cut our bounds in third:

\frac{25}{2} \int_{\frac{- π}{6}}^{\frac{π}{6}} \cos^{2} (3 θ) d θ \to {\frac{25}{24} (6 θ + \sin (6 θ)) |}_{\frac{- π}{6}}^{\frac{π}{6}} \to \frac{25}{12} π

. And that’s right!

What if we integrated the entire shape and divided by three, as mentioned above? This gives us actually the same integral, with a different multiplier and different bounds:

\frac{1}{3} \frac{25}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} \cos^{2} (3 θ) d θ \to {\frac{1}{3} \frac{25}{24} (6 θ + \sin (6 θ)) |}_{\frac{- π}{2}}^{\frac{π}{2}} \to \frac{25}{12} π

. Exactly the same! Math is weird like that.

To avoid boundary hell in the future, I’m just going to integrate whole shapes and divide by petals. It’s much easier to wrap my head around why that actually works.