 We can also use integration to find the length of an arc given in polar coordinates. As in rectangular coordinates, arc length in polar coordinates relies on finding the length of one side of a right triangle. And so we'll find a representative length and sum it. So we might find our representative length between a point r theta and a nearby point r plus dr theta plus d theta. And so we'll have three sides of the right triangle, the hypotenuse which will run from r theta to r plus dr theta plus d theta, one leg which runs from r theta to r theta plus d theta, and the other leg which will run from r theta plus d theta to r plus dr theta plus d theta. So we can apply the Pythagorean theorem. The square of the representative length is equal to the sum of the squares of the other two sides. So the length of this other leg is just dr. For the length of the first leg we have to go back to geometry, and remember that if we have an arc on a circle of radius r with the central angle of t, the arc length s is just going to be r times t. So this arc which corresponds to the arc of a circle with radius r has central angle d theta, so this arc length is going to be r d theta. And so our representative length is going to be square root r d theta squared plus dr squared, and we can simplify that by factoring out a d theta. And this leads us to the following theorem. Let r equals f of theta define a curve that is continuous and differentiable over some interval. The length of the arc between theta equals a and theta equals b is going to be given by the definite integral from a to b as the square root of r squared plus the square of dr d theta. So anytime we have a shiny new toy in mathematics, one of the things we'd like to do is to make sure that it works where we already know the answer. So let's use our formula to find the circumference of the circle r equals 2. So the circle is the graph of r equals 2 over the interval between 0 and 2 pi. Our formula relies on the derivative of r with respect to theta, so we'll find that, which is 0 because r is a constant, and so we'll substitute that into our arc length formula. Our differential variable is theta, so we have to get rid of r, evaluate our integral, and we find a value of 4 pi for the arc length. Since this circle has radius 2, then its circumference is going to be 4 pi. And so we have some confidence that this theorem giving the arc length is actually correct. So let's say we want to find the length of the curve oracles cosine theta over the interval between 0 and 2 pi. And because we want to be good calculus students, we'll graph this curve, and we see that it's a circle with radius one-half. And so we might say to ourselves, self, this is a circle with radius one-half. So the length of the curve is just going to be the circumference of a circle with radius one-half, which will be pi. But let's go ahead and use our calculus. So we'll set up our formula for the arc length. It'll be the integral from 0 to 2 pi of square root r squared plus the square of dr d theta. So I need dr d theta, so I know r equals cosine theta, so we'll differentiate. And that gives me dr d theta's minus sine theta. I'll substitute that in. I'll do a little algebra. I'll do a little trigonometry. I'll do a little bit of calculus. And so we get an answer of 2 pi, which is exactly what we expected to get. Wait a minute. We only expected to get an arc length of pi. How did we get this answer of 2 pi? Did we make a mistake someplace here? So there's something going on here that's a little bit different. So let's add a little bit to this question. Let's try and explain why the answer for the arc length is not actually the circumference of the circle. To answer this question, we have to go back to the graph of r equals cosine theta over the interval between 0 and 2 pi. So let's take a closer look at the graph over this interval. And we see that if we graph r equals cosine theta over this interval, we actually trace the circle twice. So our arc length is twice the circumference of the circle. There's no contradiction here, even though the graph is that of a circle. Our curve traces the circle twice, and so we do get twice the circumference of the circle as our arc length. As with finding the arc lengths of graphs in rectangular coordinates, most of the time we'll set up an integral that we can't actually evaluate or an integral that will be very complicated to evaluate. Fortunately, we have a ways of evaluating an integral that don't require us finding the right substitution to make. What that does mean is that it's more important to set up the integral than to evaluate the integral. So let's consider the problem of setting up the integral to find the length of a half turn of the spiral r equals theta. The previous example should emphasize the importance of graphing to make sure that we're not retracing a portion of our curve. So let's graph r equals theta. Now if we want a half turn, that's going to be the arc over the interval from 0 to pi. So we'll set up our arc length integral. We need dr d theta, so since r equals theta, we'll differentiate. So we can replace dr d theta. Since our differential variable is theta, the only allowable variable in the integrand is theta, so we have to replace r. And if we wanted to go further, we'd have to evaluate this integral which will require some sort of trigonometric substitution followed by several other painful steps. We'll omit those steps because those steps are things that can be followed by a computer program or other object that is incapable of thinking.