 A useful idea in geometric problems is tape and scissors. You can put together a geometric figure by using tape to join two pieces, or using scissors to remove parts you don't want. So for example, let's say we have the polar graphs of r equals f of theta and r equals g of theta. And suppose the two graphs intersect at 2 pi thirds, write an expression to find the area between the two curves. And notice the area we want can be found by taking all the area enclosed by r equals f of theta, then removing the area enclosed by r equals g of theta. So to begin with, the area enclosed by r equals f of theta, well that's one half r squared d theta, and our limit of integration is going to go from zero to pi thirds. And in this case, r is f of theta. So the area between the two curves will be the difference where our first area is one half integral from zero to pi thirds f of theta squared d theta. Meanwhile, the area enclosed by r equals g of theta, one half r squared d theta, and our limits of integration are again going to go from zero to pi thirds. This time, r is g of f theta, and we'll subtract this amount. And unless we know what f of theta and g of theta are, we can't do much more than set up the expression. So maybe we know that this is the graph of r equals four cosine theta, and r equals six over pi theta. And again we know that the graphs intersect at two pi thirds. Let's find the area between the two curves. Since we've already found an expression for the area, let's ignore it and redo the problem from the start. Wait, wait. Since we've already found an expression for the area, let's use it. So again, the area that we want could be found by taking all of the area and then removing the area enclosed by r equals g theta. And so our expression was this. So the area enclosed by r equals four cosine theta, we can calculate that. And similarly, the area enclosed by r equals g of theta, which gives us the area of the region between the two curves.