Page 1 of 1

A Calculus Primer Part VII- Applications of Integrals

#1 macosxnerd101   User is offline

  • Games, Graphs, and Auctions
  • member icon

Reputation: 12648
  • View blog
  • Posts: 45,822
  • Joined: 27-December 08

Posted 21 May 2011 - 09:38 PM

This tutorial will introduce some applications of integrals, including the area between curves, volumes of functions, and volumes of revolving solids.

Area Between Curves
An integral by definition is used to find the area under the curve of a function. It can also be used to find the area between two functions. First, let's discuss the integral a little more. The form Integral(a, b, f(x) dx) can be defined as the area of f(x) between a and b, which comes out to F(b) - F(a). Or another definition is Integral(0, b, f(x) dx) - Integral(0, a, f(x) dx). Basically, it is the area under f(b) that is not also under f(a). Understanding this tells us what information to look for when finding the area between multiple curves.

The general formula for the area between two curves is: Area = Integral(a, b, (f(x) - g(x)) dx).

The limits a and b are determined by the points where f(x) and g(x) intersect. This can be determined by setting f(x) = g(x) and solving. Since area is always positive, it is either necessary to take the absolute value of the integral or determine which of the two functions is on top, setting it so f(x) is on top and g(x) is on the bottom.

Let's look at an example, to find the area between f(x) = 3x - 2, and g(x) = x^2. Setting the two equal to each other produces:
x^2 = 3x - 2
x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x = 1, 2

So evaluating |Integral(1, 2, 3x -2 - x^2, dx)| yields |(3x^2)/2 - 2x - (x^3)/3| evaluated from 1 to 2. The area between these curves is 1/6.

There are two types of volumes to be concerned with in Calculus: the volume between two curves, and volumes of revolving solids. The first is pretty straight-forward. It works the same as finding the area between two curves, simply squaring f(x) and g(x) in the integral, leaving: Volume = Integral(a, b, (f^2(x) - g^2(x))dx).

Volumes of revolving solids are a little more tricky. The key to remember though is that areas are added up to produce a volume. There are two methods for finding volumes of revolving solids: discs and shells. The discs method uses the area of a circle (A = pi * r^2), and the shells method uses the volume of a cylinder (A = 2pi * r * h).

Discs is a pretty easy method to use because volumes, like areas, are always positive. So simply integrating pi * |Integral(a, b, (f^2(x) - g^2(x)) dx)| will produce the volume of the solid revolved around the x axis (denoted by the dx). If the volume is revolved around the y axis, then f(x) and g(x) will have to be solved as functions of y instead, and the partition will be dy instead of dx. Again, set f(y) = g(y) and solve for the values of y where they are equal.

The shells method requires a little more analysis than the discs method. As was discussed above, the shells method uses the area of a cylinder A = 2pi * r * h. The other thing to keep in mind with shells is that if the area is being revolved over the x-axis, the function must be in terms of y. And if the area is being revolved over the y-axis, the function must be in terms of x.

Let's look at an example, finding the volume of the function y = sqrt(x) bounded by x = 0 and x = 4, revolved around the x-axis. To start, let's re-write the function in terms of y: x = y^2. Next, let's fill in the parameters for the area formula. The 2pi is a constant, leaving the radius and the height. In this function, height varies with y, so that is easy. The radius changes with respect to x. Lastly, the limits of integration are 0 and 2, the y-bounds, since the function is in terms of y. So the integral comes out to Volume = Integral(0, 2, y(4-y^2) dy). This results in (2y^2 - y^3/3) evaluated from 0 to 2, yielding Volume = 16/3.

Is This A Good Question/Topic? 1
  • +

Replies To: A Calculus Primer Part VII- Applications of Integrals

#2 elgose   User is offline

  • D.I.C Head

Reputation: 102
  • View blog
  • Posts: 228
  • Joined: 03-December 09

Posted 26 June 2011 - 10:25 PM

Good for a primer. A couple things that stick out on a quick read-through are the fact that you can find the area between two curves from two points other than the intersections, and quite often when dealing with higher order functions you'll have to find an area between the curves when there are >2 points of intersection. Might be something fun to cover.

Also, terminology wise, "the volume between two curves" and volume of solid of revolution are not two different things. What you are explaining is a solid of revolution bounded by two curves and finding the volume using the washer method, basically an extension of the disk method.

I believe there are alternatives to using just the disk and shell methods. Not sure if you do this in other primers, but rotation around lines other than the x- or y-axis often come up and are important (and, I think, fascinating) topics.

Keep it up!
Was This Post Helpful? 0
  • +
  • -

Page 1 of 1