# Length of $r=3\sin(\theta)$

I have a general understanding of calculating arc length, but this one's a real curve ball.

So, I need to find the exact length of $r=3\sin(θ)$ on $0 ≤ θ ≤ π/3$

So the way I've thought of approaching it is by using some handy formulas:

$$r^2 = x^2 + y^2$$

and $$y=r\sin(θ)$$

So we start off with

$$r=3\sin(θ)$$

$$r^2 = 3r\sin(θ)$$ $$(x^2 + y^2)^{1/2} = 3y$$

$$x^2 + y^2 = 9(y^2)$$

$$x^2 = 8y^2$$

$$(1/8)^{1/2}x = y$$

Then, from that I use the integration formula to find length. Am I even close to being right on this one?

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After $r^2=3r\sin\theta$, the next line should be $x^2+y^2=3y$, without the square root, since the left-hand side was $r^2$. – joriki May 27 '11 at 19:34

I think this is easier to do directly in polar coordinates. We have

$$\mathrm ds^2=\mathrm dr^2+r^2\mathrm d\theta^2\;,$$

and thus

$$\begin{eqnarray} \int\mathrm ds &=& \int_0^{\pi/3}\sqrt{\left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2+r^2}\;\mathrm d\theta \\ &=& \int_0^{\pi/3}\sqrt{(3\cos\theta)^2+(3\sin\theta)^2}\;\mathrm d\theta \\ &=& \int_0^{\pi/3}3\mathrm d\theta \\ &=& \pi\;. \end{eqnarray}$$

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Of course!! Wow, I feel silly now. Thanks for both the clarity AND the noted correction. – Fred W. May 27 '11 at 19:48
@Fred: Nothing to feel silly about -- live and learn :-) – joriki May 27 '11 at 19:54

This is not useful in general, but for this specific case:

$r = 3 \sin \theta$ is a circle of diameter 3, above the $x$ axis. $\theta = 0$ to $\pi/3$ is just one third of this circle. The length must then immediately be $\pi$.

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As a couple of others have pointed out, the graph of $r = 3 \sin \theta$ is actually a circle of diameter $3$. But why is this? You can see it fairly easily by converting to Cartesian coordinates, as you tried to do initially. Your mistake was pointed out by joriki in his comment, so we'll start there. Rewrite $x^2+y^2 = 3y$ as $x^2 + y^2 - 3y = 0.$ Completing the square gives $$x^2 + y^2 - 3y + \frac{9}{4} = \frac{9}{4} \Longrightarrow x^2 + \left(y - \frac{3}{2}\right)^2 = \left(\frac{3}{2}\right)^2.$$ Thus the graph of $r = 3 \sin \theta$ is a circle of radius $\frac{3}{2}$ centered at $(0, \frac{3}{2})$. Running $\theta$ from $0$ to $\pi$ traces the circle out once at a uniform rate, so running $\theta$ from $0$ to $\pi/3$ traces out a third of the circle. Thus the answer you're looking for is one-third of the $\pi*$diameter result for the circumference of a circle, or $3\pi/3 = \pi$.

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Nice demonstration that it is a circle. I do quibble with the assertion that "The circle is completely traced out once by running θ from 0 to π, so running θ from 0 to π/3 gives a curve length of one-third of the usual ...". Both sides of the "so" are true, but in general the first need not imply the second. – wnoise May 28 '11 at 16:45
@wnoise: You're right; you also need the fact that the circle is traced out uniformly as $\theta$ runs from $0$ to $\pi$; I'll add that in. – Mike Spivey May 28 '11 at 20:07

[Edit - whoops, missed the $3$]

In general, $x=r \cos \theta$ and $y=r \sin \theta$. Given $r= 3v\sin \theta$, that means that $x = 3v\sin \theta \cos \theta = \frac {3\sin {2\theta}} 2$ and $y= 3\sin \theta \sin \theta = \frac {3(1-cos 2 \theta)} 2$

Now, the length is $$\int_{0}^{\frac{\pi}3} \sqrt { (\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 } d\theta$$

I'll let you figure out what the integrand is here.

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The approach through rectangular coordinates will work, but it is more tedious, and there are many opportunities for error. It is undoubtedly intended that you work directly with polar coordinates. A standard version of the formula for arc length in polar coordinates is: $$\int_{\theta=a}^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$

It so happens that if you use this, everything collapses, in a nice way. That is no accident, it so happens that you are finding the arc length of part of the circle $x^2+y^2=3y$.

As a general heuristic, I would suggest that if you are given an equation in polar coordinates, and want arc length, or area, you not try to change to rectangular coordinates. In principle, with care, it will work. But it will likely be messy, and probably you will get an integral for which a trigonometric substitution is required. You will end up evaluating an integral much like the one you could have written down almost immediately if using polar coordinates!

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