 We can find the area under y equals f of x by finding the area of a representative rectangle y dx, then summing these areas. If the boundary curve is defined parametrically, the representative rectangle has height y of t with x prime of t dt, and we sum over the t values. Or can we? Well, I don't know. Let's try and find out. So, suppose we want to find the area below x equals cosine t, y equals sine of t, above the x-axis, and over the interval between 0 and pi. Now, remember to get street cred among mathematicians, you should do these problems in the hardest way possible. And so, to do geometry problems in the hardest way possible, don't draw a picture. And our formula says that the area under a curve defined parametrically can be found by summing the areas y of t, x prime of t, dt. And we find the derivative of x will be, and now we can evaluate our integral. And so, our area is negative pi halves. Well, let's actually look at the graph this time. So, if we consider our curve over this interval, we do see that it marks out the unit circle, but it does so counter-clockwise. And this means that we're summing the region from right to left. Now, remember, if you reverse the order of integration, you also change the sign of the integral. The integral from a to b of f of x dx is the negative of the integral from b to a of f of x dx. And so, we're actually integrating in some sense in the reverse direction, and so it's not surprising that we got a negative value for the area. But how do we fix it? Since we're finding the area, we want the height and the width to be non-negative. So, we have two choices. We can either do a re-parameterization so that these are non-negative, or we could use absolute values. And while we should re-parameterize, it's often easier to use the absolute value of our differential. So, again, we find our derivative x prime is negative sine t, but over the interval between 0 and pi, our derivative will be negative. So, we want to use the absolute value of negative sine t. And again, we're only interested in what happens at the interval between 0 and pi, and in that interval, the absolute value of negative sine of t is sine of t. And so, the correct area will be given by the integral sine of t, sine of t dt, or sine squared of t, which we can evaluate to get pi halves. So, for example, we might try to find the area in the first quadrant bound by the curve x of t equals t plus sine t and y of t equals sine t over the interval between 0 and pi. Since we're given the interval, we should trust that it's the interval we want. Maybe let's sketch the graph over the region and confirm that it generates a region in the first quadrant. Okay, that looks good. So, now that we are definitely in the first quadrant over this interval, we find x prime of t, and we note that this is non-negative in this interval, so we don't have to worry about absolute values here. And so, we find our integral from 0 to pi y of t, x prime of t will be, and that gives us our area.