Question

Could the idea of calculating the area under a curve be expanded to the calculation of the volume of a sphere? Or a square to a cube? Consider the structural differences of potatoes, oranges, onions, and any other produce which comes to mind as a model. Please give a example.

Answer 1

Area Between Two Curves We will start with the formula for determining the area between y=f(x) and y=g(x) on the interval [a,b]. We will also assume that f(x)≥g(x) on [a,b]. We will now proceed much as we did when we looked that the Area Problem in the Integrals Chapter. We will first divide up the interval into n equal subintervals each with length, Δx=(b−a)/n Next, pick a point in each subinterval, x∗i , and we can then use rectangles on each interval as follows.  The height of each of these rectangles is given by, f(x∗i)−g(x∗i) and the area of each rectangle is then, (f(x∗i)−g(x∗i))Δx So, the area between the two curves is then approximated by, A≈n∑i=1(f(x∗i)−g(x∗i))Δx The exact area is, A=limn→∞n∑i=1(f(x∗i)−g(x∗i))Δx Now, recalling the definition of the definite integral this is nothing more than, A=∫baf(x)−g(x)dx The formula above will work provided the two functions are in the form y=f(x) and y=g(x) . However, not all functions are in that form. Sometimes we will be forced to work with functions in the form between x=f(y) and x=g(y) on the interval [c,d] (an interval of y values…). When this happens, the derivation is identical. First we will start by assuming that f(y)≥g(y) on [c,d] . We can then divide up the interval into equal subintervals and build rectangles on each of these intervals. Here is a sketch of this situation.

Following the work from above, we will arrive at the following for the area, A=∫dcf(y)−g(y)dy