 Given any integral, we can interpret it as the area of a region, but given a region, we can find its area by summing representative rectangles that run vertically, giving us an integral in terms of dx, or we can sum representative rectangles that run horizontally, giving us an integral in terms of dy. Since we have a choice, we have to decide which direction is better, and as a general rule, there is no general rule. But if you can't evaluate the integral as written, see a switching direction help, and avoid rectangles with opposite ends on the same curve. For example, let's set up an integral to find the area of the region between y equals arc sine x and the x-axis over the interval between 0 and 1. So we'll graph our region. Now if we run our rectangles vertically, we'd use if we run our rectangles horizontally, we see that the length of the rectangle, end, minus beginning, the end is always over here at x equals 1, and the beginning is some value x, and the height of this rectangle is a little portion of the y-axis, dy, and so our integral would be. Now, we don't know the antiderivative of arc sine x, so we can't evaluate the first integral. Since we don't know the antiderivative of arc sine x, we'll evaluate the second integral. So remember, the differential variable is controlling. So if our differential variable is y, we have to rewrite everything in terms of y. Since y equals arc sine x, then x is sine of y. For our limits of integration, x equals 0 and x equals 1, equals means replaceable, so sine y is 0, and so y is 0, and sine y is 1, so y is i halves, and so our integral becomes, which we evaluate. Now, since finding an area is a geometry problem, it's important to always sketch the graph. This is particularly important if the region crosses the y-axis. So consider the area between y squared equals 8x and x equals 2. We'll graph the region, and note that if we run a representative rectangle vertically, it runs between two parts of the same curve, and while we could find the area this way, it's generally harder. If we run our representative rectangles horizontally, the length of each rectangle, end minus beginning, where our end is always at x equals 2 and our beginning is some x coordinate, and the height is some small portion of the y-axis, dy, and so our integral will be, and remember the differential variable is controlling, so everything has to be written in terms of y. Our original interval ran from x equals 0 to x equals 2, so we have y squared equals 8x, and so we find solving for y gives us, and notice that we have three solutions, and because we graph the region, we see our limit should be y equals negative 4 to y equals 4. Since 8x equals y squared, x is 1 eighth y squared, so our integrand becomes, and we can evaluate.