Find the area of the top half of the polar curve :
$$r=3-2\cos \theta$$
I'm not sure how to tackle this. I need to find the area of a unique shape (almost a semi-circle), but I don't know where to start.
Find the area of the top half of the polar curve $r=3-2 \cos \theta$
calculuscurvesdefinite integralsintegrationpolar coordinates
Related Solutions
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
The big issue you have to watch out for is when $r$ is negative. Whenever that happens your curve is actually on the other side of the origin than the angle theta says it is. Check this out: Go to Wolfram Alpha and tell it to "plot r = 3cos(theta)". You will get a nice circle. Note that it plotted the curve from $0$ to $2\pi$. But the circle is entirely to the right of the origin. There is no part of the curve in the region with $\frac \pi 2 < \theta < \frac {3\pi}2$. Wolfram Alpha says it plotted the curve for those angles, but there is nothing over there. In fact, if you tell Wolfram Alpha "plot r = 3cos(theta) from 0 to pi)", it will produce the exact same graph, and will again if you plot it "from pi to 2pi".
What happens is this: the curve starts off at $(x,y) = (1,0)$ with $\theta = 0$ and $r = 1$. As $\theta$ increases, the cosine, and $r$ decrease until at $\theta = \frac \pi 2, r = 0$, and we are at the origin. This traces out the top half of the circle. As $\theta$ continues to increase, $r$ becomes negative, so instead of being in the 2nd quadrant, the curve moves symmetrically opposite in the 4th quadrant instead. By the time $\theta = \pi, r = -1$. So the direction is to the left, but the negative $r$ value places us back on the right at $(1,0)$ where we started. Therefore we've traced out the entire circle in just the angles from $0$ to $\pi$. As $\theta$ ranges from $\pi$ to $2\pi$, it repeats the process, mapping the same circle out a 2nd time.
Now if we want the area of the circle, we only want it once, not twice, so we only need to trace around the circle once in the integration. So we only integrate from $0$ to $\pi$.
How to handle this in general? Note that the problem comes in because $r$ changes sign. If $r$ stayed positive, then for each $\theta$, the point on the curve would be in the expected direction from the origin. If $r$ were always negative, then the curve would always exactly opposite the expected direction. Assuming that $r$ is a simple closed curve, you need a full $2\pi$ radians to get back to the start. But when $r$ swaps signs midway through, then it can pull this doubling stunt. However, every time a continuous $r$ switches signs, it has to pass through $0$, and there is the trick: Find the angles where $r = 0$ and then integrate from one zero to the next. You will have to examine the different ranges between zeros to determine whether the curve is just repeating itself or is doing something new. In the $r = 3\cos \theta$ example, it is only because of the regularity of the cosine that the curve repeats itself from $\pi$ to $2\pi$. But even if the curve does not repeat itself, you will want to think about each section as being a separate curve, and how you handle it depends heavily on exactly what region you are supposed to integrate.
Applying this to $r = 3\cos \theta$, we see that the intervals between zeros are $\left(-\frac \pi 2, \frac \pi 2\right)$ and $\left(\frac \pi 2, \frac {3\pi} 2\right)$. Either one would provide a full circle for the integration (as would any other interval of length \pi by periodicity of cosine, but we only need one interval of integration, not every possible one).
These is another issue. It can take more $2\pi$ to get back around where you started. For example $r = \sin\frac\theta 2$. But when this happens, the curve generally crosses itself and forms several different regions (try it in Wolfram Alpha if you are not familiar with this curve), so clarification would be needed about what region you want find the area for.
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Best Answer
Draw tha graph of the cartesian equation $y=3-2\cos x$. With this, it is easy to draw (without the need of a computer) the graph in polar coordinates of $r=3-2\cos \theta$, it would look something like that :
Now notice that the top of the figure is obtained when $0\leq \theta \leq \pi$. So, the area is given by $$\frac{1}{2} \int_0^\pi r^2 d\theta = \frac{1}{2} \int_0^\pi (3-2\cos \theta)^2 d\theta$$