## Illustration Here's a natural problem - what is the volume under the graph of 1-(x^2+y^2) and over the xy-plane. I describe this as "natural" because it is the function itself that describes the domain of integration. ### Illustration 2 Here's the view from above. You can see how the unit circle arises naturally. ### Illustration 3 Here's the domain of integration in the xy-plane with arrows (like we did last time) indicating how we might integrate with respect to y first. Note that we solve the equation x^2+y^2=1 for y to get +/- sqrt(1-x^2) for the top and bottom bounds of integration. ### Illustration 4 We can use symmetry to simplify this a bit. We take the quarter of the domain of integration in the first quadrant and then multiply by 4. Thus our lower bounds of integration are x=0 and y=0. The resulting integral still looks ridiculously complicated! ## Polar Coordinates As it turns out, the integral we just met can be computed very easily if we first translate it to polar coordinates. There's actually a very important and general principle here - some problems are most easily expressed in terms of one coordinate system as opposed to another. In this particular problem, both the function and the domain display circular symmetry so it makes sense that we might want to do this in polar coordinates. Of course, we might need a little refresher on what polar coordinates are - namely, just an alternative way to describe the position of a point in the plane. Rather than stating how far we must travel horizontally and then vertically from the origin to get to a point (as we do in Cartesian coordinates), we'll state the angle theta through which we must rotate to face the point and the distance r we must walk to get to the point. That yields the polar point (r,theta). ### Polar coordinates 2 Here are a few points with polar descriptions. You can see one big difference between Cartesian coordinates and polar coordinates right away - Cartesian coordinates are unique; i.e., given a point there is exactly one cartesian description of that point. Polar coordinates are not unique; every point has infinitely many polar descriptions. We can always add another 2pi onto theta to get back to the same point. ### Polar coordinates 3 Here are the relationships between Cartesian and polar coordinates. This allows us to translate back and forth pretty easily. ## Translation When we translate a Cartesian integral over a domain with circular symmetry to a polar integral, there are three parts we need to pay attention to: The integration symbols, that are tied to the bounds of integration, the function itself, and the area element dA ### Translation 2 In Cartesian coordinates, the simplest domains are rectangles because they are bound to the left and right by lines of constant x and they are bound below and above by lines of constant y. The simplest domains in polar coordinates are bound by rays of constant theta and arcs of constant r. Such a domain is shown here. ### Translation 4 To express a function of the Cartesian variables x and y in terms of the polar variables r and theta, we can simply replace all the xs with rcos(theta) and all the ys with rsin(theta). ### Transation 5 Often, the function itself will display some rotational symmetry. We can spot this algebraically by replacing any (x^2+y^2)s that we see with r^2. If we are left with a function that has no more xs or ys then result is independent of theta so it will have rotational symmetry. ### Translation 6 The final step is by far the easiest. The dA is simply replaced by r\,dr\,dtheta - always. While this is easy to do, it's a bit confusing to see where that extra r comes from. First off, I claim that it can't be just dr\,dtheta because dr has units of length while dtheta is unitless; thus, drdtheta has units of length, while we want units of area. rdrdtheta, by contrast has units of area, which is at least the correct units. ### Translation 7 This picture explains it a bit more completely. Note that any simple polar region can be divided into many smaller polar regions using rays of constant theta and arcs of constant r. They don't all have the same area, though. The regions closer to the origin have smaller area than the ones farther away. The r is there to scale appropriately. In fact, if we look at a little wedge bound inside by r and outside by r+delta_r and by rays that are delta_theta apart, then the area is just about r*delta_theta*delta_r. Again, the moral of this story is: always include an r,dr,dtheta in your polar integrals! ## Basic example Here's another look at the basic, natural example that we started with: the volume under the graph of $f(x,y)=1-(x^2+y^2)$ and over the $xy$-plane. Again, it's pretty easy to set up in Cartesian coordinates but not so easy to integrate. ### Basic example 2 When we translate it to polar coordinates, we actually get a pretty easy integral! ### Basic example 3 Here's the evaluation. Note that the integrand is completely independent of theta. Thus, I did the outer integral by simply multiplying by 2pi. ## Final example I can't overstate how awesome this final example is. First off, evaluation of this integral is of tremendous importance in probability and statistics. Second, the integrand e^{-x^2} has no elementary anti-derivative. As it turns out though, the extra r introduced in polar coordinates is exactly what we need to evaluate this. In addition, it provides a nice application of improper integrals. ### Final example 2 The trick here is to swap an x for a y in the second integral so that we can combine them into one double integral. ### Final example 3 Once we do that, the actual evaluation is not so bad. The integrand is again independent of theta so we can simply multiply by 2pi and we can do a u-substitution thanks to that r sitting outside.