Find the area of the region inside: $r= 6\sin(\theta)$ but outside of $r = 1$

How do we find the area of the region inside $r = 6 \sin(\theta)$, but outside $r = 1$?

So, here's my work thus far:

First off, we know: $r^2 = x^2 + y^2$ and $\mathrm{sin}(\theta) = y/r$

Therefore,

$r = \sqrt{x^2+y^2}$ and $6\mathrm{\sin}(\theta) = \frac{6y}{r}$

The original equation turns into: $\sqrt{x^2 + y^2} = 6y/r$ ; or equivalently $x^2 + y^2 = 6y$

Completing the square turns this into $x^2 + (y-3)^2 = 9$, a circle centered at $(0,3)$ with a radius of $3$

The area of that circle is $9\pi$, so the answer is going to be $9\pi$ - something. I'm having a hard time figuring out the something.

The other circle is $x^2 + y^2 = 1$, a circle centered at the origin with a radius of $1$.

Well, I was thinking, let's find the points of intersection by setting the two equations equal to eachother. So set $x^2 + y^2 - 1 = x^2 + y^2 - 6y$ (because they both = 0).

So cancel out the $x^2$ and $y^2$, so we get $-1 = -6y$ or $y = 1/6$. Plug that back in and find the two x coordinates. Plugging it back into $x^2 + y^2 = 1$, so you get +/- the $\sqrt{35/36}$. So the two intersection points are $(-\sqrt{35/36} , 1/6)$ and $(\sqrt{35/36} , 1/6)$. So I'm thinking, alright, just integrate the following from $-\sqrt{35/36}$ to $\sqrt{35/36}$: $x^2 + y^2 - 1 - (x^2 + y^2 - 6y)dx.$ But I get the integral of $(6y - 1)dx$. I don't understand please help!

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There's actually this formula (equation 8) that allows you to work directly with polar coordinates instead of having to go Cartesian... – J. M. Jul 28 '11 at 20:33
One strategy: consider only the first quadrant, and take the area bounded by the horizontal axis and the two circles. Subtract twice that area from the area of a semicircle... – J. M. Jul 28 '11 at 20:36
So, using that formula, wouldn't it be (1/2) * ( integral of 0 to pi of (6sin(theta) - 1 ) d(theta) ) right? Or, from 0 to pi/2 of the same thing and times it by 2... but I get 6 - pi/2 which doesn't sound right – user13327 Jul 28 '11 at 20:41
...but before you use that, make sure it's covered in whatever book you're using; shocking the teacher with weird formulae isn't always a good thing. ;) – J. M. Jul 28 '11 at 20:43
+1 for showing work and where you got stuck – Ross Millikan Jul 28 '11 at 22:02

We show how to do the area calculation using polar coordinates. There is a potential minor complication. There are two different conventions about the meaning of a polar equation when $r<0$. Some say the curve is then not defined. Some say it is defined, but one must reflect across the origin. For our particular curve, the two conventions give the same region, so we don't need to worry. But for other curves, you may need to know what convention your course uses.

The curve $r=6\sin \theta$ and the circle $r=1$ meet where $\sin\theta=1/6$. Let $\theta_0=\arcsin(1/6)$. From $\theta=\theta_0$ to $\theta=\pi-\theta_0$, the curve $r=6\sin\theta$ is "outside" the circle $r=1$. By symmetry, we can just look at the part from $\theta=\theta_0$ to $\theta=\pi/2$, and double the resulting area.

The area from $\theta=\theta_0$ to $\theta=\pi/2$, bounded by our curve $r=6\sin\theta$, is $$\int_{\theta_0}^{\pi/2} (1/2)r^2 d\theta.$$ From this we must subtract the area inside the circle $r=1$, which is $$\int_{\theta_0}^{\pi/2} (1/2)\,1^2 \,d\theta.$$ Subtract, and multiply by $2$ to take care of the other half of the region. The desired area is therefore $$\int_{\theta_0}^{\pi/2} (36\sin^2\theta-1)\, d\theta.$$

Now integrate. Using the identity $\cos 2\theta=1-2\sin^2\theta$, we find that $36\sin^2\theta=18-18\cos 2\theta$. Thus our area is $$\int_{\theta_0}^{\pi/2} (17-18\cos 2\theta)\, d\theta.$$

The rest is routine. We will need $\sin 2\theta_0$. When you calculate this, you will bump into the $35$ that you already met in your rectangular coordinates analysis. The ultimate answer is not "nice," because $\arcsin(1/6)$ is not a pleasant number. Too bad that the polar curve was not $r=2\sin\theta\:$!

Comment: We can also obtain the result by switching to rectangular coordinates, as you did. The switching in this case was straightforward, and showed we were dealing with a circle. However, in many cases the rectangular coordinates version of a curve given in polar coordinates can be very unpleasant. It is therefore useful to learn to work directly in polar coordinates.

In situations where there is strong circular symmetry, polar coordinates come up very naturally. For example, the force of gravity exerted by the Earth on a small object is directed towards the center of the Earth. In studying motions of satellites, polar coordinates are the way to go.

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This is what I did to find the solution! Thank you! – user13327 Jul 28 '11 at 22:30
@Silver: You are welcome. I am a little surprised at the choice of $6$ in front of $\sin\theta$. Most exercises of this kind set up things so that we get pleasant "exact" answers. But of course the real world is not like that. – André Nicolas Jul 28 '11 at 22:37

You want the region where $r>1$. That's the same as $\sin\theta > 1/6$. The boundary of that region is where $\sin\theta=1/6$. That happens when $\theta = \arcsin(1/6)$ and when $\theta = \frac\pi 2 - \arcsin(1/6)$. So you want $$\int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} \frac{r^2}{2} \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 18\sin^2\theta \, d\theta = \int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} 9 - 9\cos(2\theta)\,d\theta$$ $$= \left[ 9\theta - \frac92\sin(2\theta) \right]_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)}$$ To find $\sin(2\arcsin\frac 1 6)$ use the double-angle formula for the sine. For that, you want to remember that $\cos(\arcsin x) = \sqrt{1-x^2}$. To find $\sin(2(\frac\pi 2 - \arcsin(1/6))) = \sin(\pi - 2\arcsin(1/6))$, remember that $\sin(\pi - \theta)$ is the same as $\sin\theta$.

And remember that $\cos(\pi-\theta) = -\cos(\theta)$.

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Tricky, a little confusing, but I think I understand. Thanks :) – user13327 Jul 28 '11 at 22:30
....corrected now, I hope..... – Michael Hardy Jul 29 '11 at 5:19
@Michael Hardy: We have $\sin\theta=1/6$ at $\arcsin(1/6)$ and also $\pi-\arcsin(1/6)$, not $\pi/2-\arcsin(1/6)$. So the upper limits of integration are not right. Also, after the integration, we need to subtract the area inside $r=1$, from $\arcsin(1/6)$ to $\pi-\arcsin(1/6)$. This is because we want area outside $r=1$ but inside $r=6\sin\theta$. – André Nicolas Jul 29 '11 at 12:00