First we have to make sure we know what *R* is. Draw the graph *y* = *x*^{2} and the line *y* = 4: Then shade in the region they surround: We could slice this area either way. Sometimes it is easier to evaluate the integral if we slice it one way instead of the other. We will see what we get from slicing each way here. (a) If we slice vertically, the variable of integration is *x*. The variable *x* tells the distance of the slice from the *y*-axis: The height of a slice is 4 – *x*^{2}: This means the area of a slice is (4 – *x*^{2}) Δ *x*. Since we're integrating with respect to *x*, the limits of integration are whatever the least and greatest reasonable values are for *x*. The line *y* = 4 intersects the graph *y* = *x*^{2} when *x* = ± 2. So the limits of integration are from -2 to 2. The area of *R* is . There's another option that's even easier. Since the region *R* consists of two identical pieces, we can integrate from 0 to 2 and multiply the result by 2. The area of *R* is . This area is simple to integrate, but we won't do it here. (b) If we slice horizontally, then *y* is the variable of integration. The variable *y* tells us the "height" of the slice above the *x*-axis. The slice at height *y* has right endpoint (*x*,*y*) on the graph of *y* = *x*^{2}, so we could also write the coordinates of this endpoint as . The width of the slice is twice the *x*-coordinate, or : This means the area of the slice is . The slice at the bottom of *R* is at "height" *y* = 0, and the slice at the top is at "height" *y* = 4. So the limits of integration are from 0 to 4, and the area of *R* is . This area is more difficult to integrate than the vertical slices. We could do it here, but we should be glad we don't have to. We dodged a bullet. Sometimes a region is shaped so oddly that we have to turn our heads and rub our eyes just to make sure we aren't hallucinating. Yes, we will still have to integrate it. We just have to be smart about the way you slice it, because it affects whether or not we can express its area with just one integral. We might have to split the region into pieces and find the area of each piece separately. In these cases, an "integral expression" may consist of more than one integral. |