Suppose construction workers needed to remove dirt in a hill in order to run a road though it. The swath needed to be excavated is [latex]10[/latex] meters wide, and the hill is parabola shaped. Here is a picture:
How much dirt needs to be removed? What do we need to know to solve the problem?
That’s right — we need to know the volume of the region we’re excavating. Here is a picture of the region we are removing:
Such an object is called a “prism” — it’s basically two dimensional shape that has been given a little thickness. Finding the volume involves finding the area of the shape times its thickness:
In this case, the area is labeled [latex]A[/latex] in the diagram, and the width is [latex]10[/latex]. So the amount of dirt being hauled away is [latex]10A[/latex]. But what is [latex]A[/latex]? How do we find it? That is where integration comes in.
[latex]A[/latex] is a parabola like shape, but let’s say it is bounded on top by the function [latex]y = x(50-x)/125[/latex]. Thus you get a picture like this:
How do we find the area of a curved shape like this? Well, we don’t know how to find the area of an arbitrary curved shape. But we DO know how to find the area of a rectangle: it is height times width. So let’s use rectangles:
So it’s not a great approximation yet, but bear with me. The area of the boxes does approximate the area under the curve to some extent. But how do we figure out the area of the rectangles? Well, that’s height times width, and we can see that the width of each box is [latex]10[/latex]. So we just need the height of each rectangle.
When I drew each rectangle, I made it so that the right side of the rectangle was exactly the height of the function. The height of the function is something we can find! That’s because we know the hill follows the function [latex]y = x(50-x)/125[/latex]. The height of the hill, and the height of the rectangles can then be found. For example, the height of the first rectangle is just plugging [latex]x = 10[/latex] into our function. So the height is [latex]y = 10(50-10)/125 = 400/125 = 16/5 = 3.2[/latex]. The height of the second rectangle is [latex]y = 20(50-20)/125 = 600/125 = 24/5 = 4.8[/latex]. The third rectangle has height [latex]y = 30(50-30)/125 = 600/125 = 4.8[/latex], the fourth rectangle has height [latex]y = 40(50-40) = 400/125 = 3.2[/latex], and the fifth has height [latex]y = 50(50-50)/125 = 0/125 = 0[/latex].
If we take these heights and multiply by the width, we get the area of each rectangle.
Adding up all the area from all the rectangles, we have [latex]32 + 48 + 48 + 32 + 0 = 160[/latex] m[latex]^2[/latex]. This is an approximation of the area.
Now you might say that this approximation might not be very good, and you’re correct. It certainly gives the rough idea, which might be all that we need. But if we need a very precise answer, [latex]160[/latex] isn’t good enough. So what can we do? Let’s use more rectangles!
The idea of using rectangles like this to get better and better approximations of area is called a Riemann Sum. Now, we could find the height and width of each of those rectangles, but it would be a bit tedious. Since mathematicians are lazy, we often have a computer do the work for us. For example, website “MathWorld” has a Riemann Sum calculator: http://mathworld.wolfram.com/RiemannSum.html. Using this calculator, we can see that when using [latex]20[/latex] rectangles, the sum becomes [latex]166.25[/latex]. This is called numeric integration (If you use more and more rectangles, you can actually find a very precise answer of [latex]166.667[/latex] — this is how the program computes the “actual area”. However, this is still an approximation, not an exact value.)
What does this have to do with the dirt problem? Remember, we were trying to find the volume of this slice:
What we have done with the Riemann sums is find [latex]A = 166.667[/latex]. To find how much dirt is hauled away, we just need to multiply by the width of the prism, which is [latex]10[/latex] m. Hence [latex]1666.67[/latex] cubic meters of dirt must be hauled away.