We have seen how Riemann sums can be used to approximate areas and volumes. The definite integral, as the limit of a Riemann sum as the slice width goes to zero and the number of slices goes to infinity, provides a way to find the actual area or volume.

We can use the same technique to find the length of the graph of a function. Obviously, if the function's graph is a straight line, we can just used the distance formula to find the length of a piece of the line. However, we can approximate a curve by using straight line segments and can use the distance formula to find the length of each segment. Then, as the segment size shrinks to zero, we can use a definite integral to find the length of the arc of the curve.

See About the calculus applets for operating instructions. |

The applet initially shows an arc that is part of the graph of a parabola. Initially, we approximate the length of this arc by a straight segment connecting the end points. This is clearly not a very good approximation, but we can do better by increasing the number of segments. Move the intervals slider to increase the number, and see how the black set of segments more closely approximates the magenta curve.

Note that the approximate length gets closer and closer to the actual length. If we let Δ

Select the second example from the drop down menu. This is essentially the same example, except that now *x* is a function of *y*. The general formula is just
and the specific example in this case is
The length is the same, since this is really the same as the first example, just mirrored through the line *y* = *x*.

Select the third example from the drop down menu, showing a parametric curve. The curve is a semicircle (if it looks squashed a bit, click the Equalize Axes button). You can move the intervals slider to make the approximation closer to the actual answer, which we know from geometry is π. Since our independent variable in parametric equations is *t*, we can go back to the original distance formula
and factor a Δ*t *out of the radical to get
Summing this up and taking the limit as Δ*t *goes to zero gives us the integral
Note that here *a* and *b* are *t* limits, not *x* limits. To evaluate for our example we just plug in the derivatives of the two parametric equations to get
Note that this integral can be evaluated exactly, using the Pythagorean Identity to simplify the integrand.

Select the fourth example, showing a polar curve. In this example, *th* is used instead of *θ* to make it easier to type from the keyboard. Move the intervals slider see how the approximation gets better as the number of segments increases. To find the arc length, first we convert the polar equation *r = f* (*θ*) into a pair of parametric equations *x* = *f* (*θ*)cos*θ* and *y* = *f* (*θ*)sin*θ*. We then use the parametric arc length formula
where the two derivatives are of the parametric equations. Our example becomes
which is best evaluated numerically.

(You can greatly simplify the radicand by finding the derivatives, expanding the terms, and simplifying with the Pythagorean Identity, but the result is still not a simple enough integrand).

You can explore your own arc lengths by selecting the example that matches the type of curve (normal, inverse, parametric, polar) in which you are interested, then setting *a* and *b* and zooming/panning as usual. Note that if you set *a* > *b* the resulting length comes out negative. Since length is usually defined to be positive, you should keep *a* < *b*.

- Areas by Slicing
- Volumes of Revolution
- Volumes of Known Cross Section
- Arc Length
- Area of Polar Curve

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