Calculus 2 integration , Lecture notes of Educational Mathematics

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APPLICATIONS OF INTEGRALS

Arc Length

We are going to look at computing the arc length of a function. We want to determine the

length of the continuous function y  f ( x ) on the interval [ a , b ]. We’ll also need to

assume that the derivative is continuous on [ a , b ].

Initially we’ll need to estimate the length of the curve. We’ll do this by dividing the

interval up into n equal subintervals each of width (^)  x and we’ll denote the point on the

curve at each point by Pi. We can then approximate the curve by a series of straight lines

connecting the points. Here is a sketch of this situation for n  9.

Now denote the length of each of these line segments by pi (^)  1 Pi and the length of the

curve will then be approximately,

and we can get the exact length by taking n larger and larger. In other words, the exact

length will be,

Now, let’s get a better grasp on the length of each of these line segments. First, on each

segment let’s define ^ yi  yi  yi  1  f (^ xi ) f ( xi  1 ). We can then compute directly the

length of the line segments as follows.

By the Mean Value Theorem we know that on the interval [ xi (^)  1 , xi ]there is a point

i

x so

that,

Therefore, the length can now be written as,

The exact length of the curve is then,

However, using the definition of the definite integral, this is given by,

A slightly more convenient notation is the following.

In a similar fashion we can also derive a formula for x  h ( y )on [ c , d ]. This formula is,

Therefore, the formula for the length of a curve can be written as;

where

Note that no limits were put on the integral as the limits will depend upon the ds that

we’re using. Using the first ds will require x limits of integration and using the second ds

will require y limits of integration.

Example

Determine the length of y ln(sec x )between 4

0



 x .

Solution

Let’s also get the root out of the way since there is often simplification that can be done

and there’s no reason to do that inside the integral.

Note that we could drop the absolute value bars here since secant is positive in the range

given. The arc length is then,

Example

Determine the length of^2

3

( 1 ) 3

x  y  between 1  y  4.

Solution

We first compute the derivative and the root

Thus:

This could also be done as follows:

The root in the arc length formula

The limits are:

The integral that will give the length.

We can use the u-substitution

Using this substitution, the integral becomes,

Example

Determine the length of

2

2

x  y between 2

0  x . Assume that y is positive.

Solution

We first compute the derivative and the root

and

The integral for the arc length is then

This integral will require the following trig substitution.

The length is then,

Surface Area

We want to find the surface area of the solid of revolution.

We look at rotating the continuous function y  f ( x ) in the interval [ a , b ] about the x-

axis. We’ll also need to assume that the derivative is continuous on [ a , b ]. Below is a

sketch of a function and the solid of revolution we get by rotating the function about the

x-axis.

We can derive a formula for the surface area as follows. We’ll start by dividing the

interval into n equal subintervals of width  x. On each subinterval we will approximate

the function with a straight line that agrees with the function at the endpoints of each

interval. Here is a sketch of that for our representative function using n  4.

Now, rotate the approximations about the x-axis and we get the following solid.

The approximation on each interval gives a distinct portion of the solid and to make this

clear each portion is colored differently. Each of these portions are called frustums and

we know how to find the surface area of frustums.

The surface area of a frustum is given by,

where

and l is the length of the slant of the frustum.

For the frustum on the interval [ xi (^)  1 , xi ]we have,

and we know from the previous section that,

Before writing down the formula for the surface area we are going to assume that x is

“small” and since f ( x ) is continuous we can then assume that,

So, the surface area of the frustum on the interval [ xi (^)  1 , xi ]is approximately,

The surface area of the whole solid is then approximately,

and we can get the exact surface area by taking the limit as n goes to infinity

We could also derive a similar formula for rotating x  h ( y ) on [ c , d ]about the y-axis.

This would give the following formula.

Therefore, the formulas for calculating the surface area are:

where,

There are a couple of things to note about these formulas. First, notice that the variable in

the integral itself is always the opposite variable from the one we’re rotating about.

Second, we are allowed to use either ds in either formula. This means that there are, in

some way, four formulas here. We will choose the ds based upon which is the most

convenient for a given function and problem.

Example

Determine the surface area of the solid obtained by rotating

2 y  9  x ,  2  x  2 about

the x-axis.

Solution

The formula that we’ll be using here is,

Since we are rotating about the x-axis and we’ll use the first ds in this case because our

function is in the correct form for that ds and we won’t gain anything by solving it for x.

Let’s first get the derivative and the root taken care of.

The integral for the surface area is,

The dx means that we shouldn’t have any y’s in the integral. So, before evaluating the

integral we’ll need to substitute in for y as well. The surface area is then,

Example

Determine the surface area of the solid obtained by rotating

3 y  x , 1  y  2 about the

y-axis. Use both ds’s to compute the surface area.

Solution

a) The derivative and root.

We’ll also need to get new limits. To do this we plug in the given y’s into our equation

and solve to get that the range of x’s is 1  x  8. The integral for the surface area is

then,

Using the substitution,

the integral becomes,

b) The derivative and root.

The surface area is then,

We used the original y limits this time because we picked up a dy from the ds. Also

note that the presence of the dy means that this time, unlike the first solution, we’ll

need to substitute in for the x. Doing that gives,

Exercise

Read on the use of integration in probability