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I started with:

inside $r_1=5 \sin(θ)$ and outside $r_2=2+\sin(θ)$ and was told to sketch curve in the same polar coordinate system

I first set both equal to $0$ and solved to get $\pi$, $2\pi$, and $0$ for the inside equation and "No Solution" for the second equation. I don't really see how doing any of that is relevant to the rest of the question though.

I then set up a table of values and chose $0$, $\displaystyle \frac{\pi}{2}$, $\pi$, $\displaystyle \frac{3\pi}{2}$, and $2\pi$ as my $θ$ values and solved for my $r_1$ and $r_2$ values respectively. I obtained those numbers.

Now I am stuck. These values tend to form some sort of shape, correct? From what I understand the $θ$ values represent the angle and the $r$ values are the distance from the origin. How do I know if those $r$ values go on the $x$ or $y$ axis? If I have $\displaystyle \frac{3\pi}{2}$ and it gave me an $r_1$ value of "$-5$" how do I know where to plot that on the graph? This is the part I really struggle with for some reason, I'm hoping someone here can help make it "click" for me.

Part 2 of this question asks for me to setup an integral to evaluate the area inside those plotted above. I think I know what to do there except for how I know what my limits of integration will be there. How do I know what they will be? Do I just set my two equations equal to each other and solve for $θ$? I actually did that already and I obtained $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$, are those my limits of integration?

Thank you (sorry for the lengthy write up!).

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2 Answers 2

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I hesitate to pen even a part answer, since without pictures it will be hopelessly inadequate. Perhaps someone with drawing skills will give an answer, and make this one obsolete.

You really need a live person to show you what is going on. Or as a weak second best, google "Polar Coordinates." A few entries down, there are a couple of videos. Not overly impressive, but better than nothing.

But here is a partial reply to your question.

The rectangular coordinate system that you are familiar with is an "addressing scheme" for points in the plane. The polar coordinate system is an entirely different addressing scheme.

Let $O$ be the origin. Draw the usual axes, but concentrate on the positive $x$-axis. The address of a point $P$ in the polar coordinate system is an ordered pair $(r,\theta)$, where $r$ is the distance from $O$ to $P$, and $\theta$ is the angle through which you must rotate the positive $x$-axis counterclockwise so that it falls along the line $OP$.

Example: Let $P$ have polar coordinates $(4, \pi/6)$. So we need to rotate the positive $x$-axis through $30$ degrees, and go outward a distance $4$ from the origin. I hope you can calculate, using basic right triangle stuff, and see that $P$ has rectangular coordinates $(2\sqrt{3},2)$.

You should practice a bit, changing from polar coordinates to rectangular coordinates, and vice-versa. Your text will have plenty of exercises.

Drawing $r=2+\sin\theta$: The points on this curve are all the points whose polar address is $(r,\theta)$ and such that $r=2+\sin\theta$.

As you read what follows, please draw.

Let's start with $\theta=0$. So we have not rotated away from the positive $x$-axis at all. And we want the distance from the origin to be $2+\sin(0)$. So we get the point which has rectangular coordinates $(2,0)$.

Now take $\theta=\pi/6$. So $2+\sin\theta=2.5$. Now our point $P$ is at distance $2.5$ from the origin, and $PO$ makes an angle of $30$ degrees with the positive $x$-axis.

Continue in this way, plotting points. But that's not how I would think of it. I would imagine $\theta$ growing, that is, the angle with the positive $x$-axis growing, so my pencil is moving counterclockwise. Then $2+\sin\theta$, the distance from the origin, grows, at least for a while. It grows all the way to when $\theta$ reaches $\pi/2$, when $r=3$. So the point with rectangular coordinates $(0,3)$ is on the curve. To get from $\theta=0$ to $\theta=\pi/2$, draw a curve that starts at distance $2$ from the origin, and as you travel counterclockwise, gets further and further from the origin, until the distance reaches $3$ when you hit the positive $y$-axis.

Let's continue past $\pi/2$. Now $\sin\theta$ is decreasing, and hits $0$ at $\theta=pi$. So as you continue counterclockwise, $r$ is shrinking, until it hits $2$ at $\theta=\pi$. We end up at the point with rectangular coordinates (-2,0)$.

Continue past $\pi$. Now $\sin\theta$ is negative, and getting more negative. So $r$ is dipping below $2$, and reaches $1$ at $\theta=3\pi/2$. We have reached the point which has rectangular coordinates $(0,-1)$.

Now continue from $\theta=3\pi/2$ to $\theta=2\pi$. You should be able to link up with the first point we drew on the curve.

After all this, we end up with what kind of looks like an upside-down Valentine's Day heart, or, for the less romantically inclined, like a fat pointy-headed man sitting.

Drawing $r=5\sin\theta$: There is a complication here, unfortunately. There are two conventions about how to deal with negative $r$. Convention $1$ says that negative $r$ makes no sense. Convention $2$ says that for example, to plot the point with polar coordinates $(-8,\theta)$, you imagine plotting $(8,\theta)$, and reflect the result across the origin, or equivalently rotate through $180$ degrees. I do not know what convention your course/book uses.

Imagine plotting $r=5\sin\theta$ using Convention $1$. Start at $\theta=0$. Then $r=0$, so we are at the origin. Now let $\theta$ grow towards $\theta=\pi/2$. As $\theta$ grows, $\sin\theta$ grows, so as our angle with the positive $x$ axis increases, our distance from the origin increases, until at $\theta=\pi/2$ the distance reaches $5$, and we end up at $(0,5)$. Now let $\theta$ increase from $\pi/2$ towards $\pi$. Our distance from the origin is decreasing, and reaches $0$ at $\theta=\pi$, so we join up with our beginning point. As we go past $\pi$, $\sin\theta$ becomes negative, and by Convention $1$ we get no curve.

If instead we use Convention $2$, as we go from $\pi$ to $2\pi$, we trace out the curve again (remember the stuff about reflecting through the origin). So visually the two different conventions give curves that look the same. That will not always be true.

Area: I will not do this. But with careful drawing, you should be able to identify the region talked about. And you were right in thinking the points of intersection important. You found the relevant $\theta$ correctly.

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Thank you for the great write up. You were right about needing visuals to do it though, I found a great website which I mentioned below and it finally clicked for me. –  Ryan May 24 '11 at 4:32
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Have you looked at Wikipedia? It has a discussion of plotting points in polar coordinates and some examples. The x and y axes are not important-r is the distance from the center and theta is the angle counterclockwise from the right-pointing horizontal. You should probably use many more theta values, then it will be very clear what the curve is.

Then to find the area between them, you need to use that the area of a narrow "pie-slice" in polar coordinates is $\frac{r^2}{2} \; d\theta$, so you will integrate $\frac{r_2(\theta)^2-r_1(\theta)^2}{2} d\theta$ where $r_1$ and $r_2$are your two curves from one crossing to the other.

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This website actually did a great job of explaining it, I think I finally get it now: intmath.com/plane-analytic-geometry/ans-8.php?a=1 As for the integral, so I don't need to state my limits of integration then? From my notes, the examples in there appeared to have them. –  Ryan May 24 '11 at 0:58
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