 We can also do calculus in polar coordinates. First, let's find some areas. When we found the area under a curve in rectangular coordinates, we partitioned our region into a bunch of rectangles. And so in rectangular coordinates, a representative area was a rectangle, which we summed to find the area of the region. So how do we find the area in polar coordinates? We'll do the same thing. We'll partition our region into a number of subregions. But this time, a representative area is a wedge. Now these areas are actually circular sectors. And so we know from geometry that the area of a circular sector with radius r and central angle t is going to be one-half r squared t. Since these wedges have central angle very small bit of theta, the area of a representative wedge will be one-half r squared d theta. So let's try it out. And because this is a new method, let's make sure that it actually works in a case where we know what the answer is supposed to be. So let's use integration to find the area of the region bounded by the curve r equals 2 over the interval theta between 0 and 2 pi. So first, we'll graph the region. And since r is always equal to 2, then our distance from the pole is always 2, which means that this region will be a circle with radius 2, and we know the area. To find the area using integration, we'll graph a representative wedge. And the area of the wedge is going to be one-half r squared d theta. That's the area of one of these representative wedges. To find the area of the entire region, we need to sum these areas from 0 to 2 pi. Since our differential variable is theta, this is the only allowable variable. So we need to replace r with something else. And since r equals 2, we can replace it and do the integral, which gives us our area of 4 pi, which is what we expected it to be because we know that the region is in fact a circle. So now we know that our formula works in a situation where we know the answer. Let's apply it to a situation where we don't know the answer and to find the area of the region enclosed by the graph of r squared equals sine of 2 theta. To get the wrong answer quickly, we can skip the graph and simply evaluate the integral. I know that I'm going to be evaluating one-half r squared d theta over the region between 0 and 2 pi, so I'll drop that into my integral and evaluate it. And we get the answer of 0, which we happily circle and say this is the area of the region. And if your goal is to very quickly get the wrong answer, this is exactly how you should proceed. On the other hand, suppose you're doing something where it actually matters whether you get the correct answer. Then it's better to take a little bit more time to make sure that the final answer that you write is the correct one. So to begin with, we should graph this curve. We see that our graph is going to consist of two identical loops. So we can find the area of one loop and double it. One of those loops occupies the region between theta equals 0 and theta equals pi over 2. We'll take a look at that region and we'll draw our representative wedge. The area of the representative wedge is one-half r squared d theta, and we'll sum these wedges over the region between 0 and pi over 2. Since our differential variable is theta, that is the only allowable variable, so we'll need to replace this r with some function of theta. Fortunately, we know that r squared equals sine of 2 theta, so we can make that substitution, and now we can evaluate our integral. So remember, this is just the area of one loop. Since there are two loops, the area is 2 times a half or 1.