In this section, we'll study a few more geometric objects using the definite integral. First, we'll ask how to determine the distance we've travelled if we drive along a curvy graph. Then we'll tackle roughly the same question for surfaces of revolution by computing the surface area.

Arc LengthConsider the graph of a function and imagine that it is a road along which we drive our car. We would like to compute the distance it would take to travel along this road from to .We know how to find the distance if the road were straight: in that case, we could just use the Pythagorean Theorem. However, when the road is winding, we do not have a clear way to find the distance. The idea will be to approximate the graph by a sequence of straight lines and add up the distance along those line segments. This is demonstrated below:

Here, you can see that as the approximating line segments are made shorter, they provide a more accurate approximation of the graph. This means that the distance along the line segments more accurately reflects the distance along the graph.

We will then break the interval into smaller pieces of width . When is very small, the graph looks very much like a straight line and so we can form the triangle in the picture. The length of this hypotenuse is approximately the length along the curve for this little segment. From the picture, we have

Notice that the ratio that appears in this expression, , is very suggestive: it is measuring the slope of this line which is approximating the graph. When this interval is very small, this line is the tangent line and is its slope. This means that as the notation suggests.

Now to find the total length along the curve

s,we simply add up all of the lengths of the very short line segments:

Some examples

- First, we'll find the circumference of a circle of radius
R.Of course, we expect to find the result that , but here we can verify the result.The circle can be described as which means that, as a graph, the upper semi-circle is

We will find the length along this upper semi-circle which is half the circumference of the circle. Notice that

Then we can evaluate the arc length along the upper semi-circle as

This means that the circumference of the circle, which is twice the length along the upper semi-circle, is as expected.

- Now we'll compute the length along the parabola between and . This is, in fact, the curve shown in the demonstration above. Through our approximation there, we expect that the arc length is approximately 1.479...
Let's begin:

This means that the arc length is given by . This is a difficult integral to evaluate using the Fundamental Theorem of Calculus: an antiderivative for the integrand is not immediately clear through one of the techniques we've discussed. However, if we look on a table of integrals, we can really save ourselves some work. In particular, there you will notice that

This is not exactly the integral we want to evaluate, but it is pretty close. In fact, if we use the substitution , we find that

This is the same result that we found in the demonstration above.

Surface AreaNow we would like to find the surface area of a surface obtained by revolving the portion of the graph betweenx = aandx = babout thexaxis. Again, this is difficult to understand directly, but we can approximate the surface area by small pieces whose surface area can be computed.To that end, let's again approximate the curve by dividing up the interval into little pieces of width . The picture is the same one we saw above.

When we revolve this straight line about the

xaxis, we have a portion of a cone whose base radius is .

Now imagine that we cut this cone and lay it out flat. What we would have would almost be rectangular and so the area would be approximately .

Again, if we add up all the areas, we find that the surface area is

Examples

- First, we'll find the surface area of the sphere of radius which can be found by revolving the graph about the
xaxis. We have already computed above that .This means that

This may be a familiar formula but now you verified it for yourself.

- Now we'll compute the surface area obtained by revolving the graph of between
x = 0andx = 1about theyaxis.