# For solid volumes, why does the Integral behave as a summation?

When you take a definite integral, you can think about calculating the area under the curve (via Riemann rectangle slices approximation)

Now, when you take the volume of a 3D object, you sum the slices.
Volume = summation of all the little slices' volumes = A(x) * width

But why are you just slapping an integral as a way to do the summation? Not seeing the connection between integral being "area under the curve" and suddenly integral being "summation" of the various areas of bases.

• Because integration is a summation--it is not the "area underneath the curve". – Jared Apr 17 '14 at 22:35
• Right, b/c integral is the limit of all those rectangles as their number approaches infinity (width approaches 0) – JackOfAll Apr 18 '14 at 0:16

## 3 Answers

Area under the curve can be seen as a summation of small "volumes" of the type you alluded to: Namely the area under the curve is the sum of areas of rectangles of height $f(x)$ and small width when you split up your region of integration into small intervals, and evaluate $f(x)$ within each interval.

• Right, b/c integral is the limit of all those rectangles as their number approaches infinity (width approaches 0) – JackOfAll Apr 18 '14 at 0:19

Look at the definition of an integral:

$$\int\limits_a^b f(x)dx = \lim_{n\rightarrow\infty} \sum_1^n f\left(a + i\frac{\Delta x}{n}\right)\frac{\Delta x}{n} \text{, where } \Delta x = b - a$$

Integration is inherently a sum. We only say that the integral is the "area underneath the curve" because if you do that infinite sum that is what it gives you: Then you get:

$$A = \int dA = \int f(x)dx$$

Also note that the following integral doesn't exactly fit the definition of the "area underneath the curve":

$$\int\limits_0^{2\pi}\sin(x)dx = 0$$

So is there area between the sinusoid and the $x$-axis? How can an area be negative? Clearly the integral isn't directly computing the "area underneath the curve" since to make this correct you actually have to do:

$$\int\limits_0^{2\pi}\left|\sin(x)\right|dx = 2\int\limits_0^\pi\sin(x)dx = 4$$

An integral is a sum of infinitely many infinitely small quantities.

I am inclined to agree with the view that it makes more sense to teach students that they can use an area to approximate an integral than that they can use an integral to find an area.

For example:

• A spring is being stretched. Energy equals force times distance, BUT the amount of force it takes to continue stretching the spring keeps changing as the spring stretches. So you multiply the force at some point by the infinitely small distance you stretch the spring at that point, and then you add up all of those infinitely many infinitely small contributions to get the total energy. That is an integral.

• A small asteroid passes between the earth and the moon and then continues in its path around the sun (which path has been altered by its sojourn in the vicinity of the earth and the moon). It is acted on by the gravity of the earth, the moon and the sun, and the magnitudes and directions of those forces are continually changing as it moves. The speed and direction of the asteroid in constantly changing. Distance equals rate times time, but the rates keep changing. So you multiply the rate at some point in time by the infinitely small increment of time during which it moves at that rate, to get the infinitely small distance it travels. Then you add up all those infinitely many infinitely small distances to get the total distance it travels. That is an integral.

• Earth's gravity attracts an orbiting satellite toward the earth. That's why the satellite does not coast in a straight line, but rather its direction changes, always turning toward the earth, so it may follow a circle around the earth rather than a straight line. The force of gravity is inversely proportional to the square of the distance from the earth, but different parts of the earth are at different distances from the satellite. So you take the infinitely many infinitely small contributions of different parts of the earth and add them up. (Isaac Newton did that.) That is an integral.

• The probability that a random variable is between two values is the probability density times the distance between those two values. But the density is different at different places between those values. So you multiply the density in an infinitely small region by the infinitely small size of that region, and add up all those infinitely small probabilities to get the total probability. That is an integral.

• The length of an irregularly shaped curve can be viewed as the sum of infinitely many infinitely short line segments, each separately found via the Pythagorean theorem. That is an integral.

All of these can be represented as area under a curve.

A parallel question about derivatives might be this: Why are derivatives being used to represent things other than slopes of lines? But the value of a derivative is an instantaneous rate of change. All instantaneous rates of change can be viewed as slopes of curves, but finding instantaneous rates of change rather than finding slopes of curves is what the concept of derivative is about.

• When looking at graphs, it's useful to be able to visualize the area underneath the curve to give some meaning, but it should not be taught that this is the definition of an integral. It obviously leads to confusion illustrated by this question. And frankly, I disagree with your examples being "easy" to translate to areas underneath the curve. Most, in my opinion, are easier to right down first, the differential, e.g. $dE = Fdx = kxdx$, then integrate. – Jared Apr 17 '14 at 23:52