Application of Riemann Integral

By Ng Tze Beng

Part 1. Arc Length

Take a function f : [a, b] ® R defined on a closed and bounded interval [a, b]. Suppose f is continuous on [a, b]. Then the graph of the function f is a curve in R² and is usually said to be given by the equation y = f (x) . What is the length of this curve? We can consider an estimate of the length by taking points on this curve and take the length of a polygonal curve passing through theses points. To do this we take a partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for the interval [a, b]. Let P₀ be the point (x₀ , f (x₀)) and P_i = (x_i , f (x_i)). Then the length of the polygonal curve P₀P₁¼P_n is an approximation of the arc length P₀P_n

.

The length of the polygonal curve P₀P₁¼P_n is given by

|P₀P₁| + |P₁P₂| + ­ + |P_n-1P_n| or .

Now the length of each line segment |P_i-1P_i| is the length of the line joining (x_i-1 , f (x_i-1)) to (x_i , f (x_i)). Thus by the Pythagorean Theorem

.

Therefore, the length of the polygonal curve P₀P₁¼P_n is given by

------------------ (1)

We define the arc length of the curve given by y = f (x), for a £ x £ b to be the limit of all possible polygonal approximation as given by (1), if it exists. If it exists, the curve is called a rectifiable curve, otherwise it is not. Not all continuous curves on a closed and bounded domain are rectifiable.

We may thus write the arc length as

----------------- (2)

Here in (2) as usual, ||D|| demotes the norm of the partition D, i.e., the maximum of all the lengths of the subintervals [x_i-1 , x_i] as defined by the partition D.

The limit (2) does not seem to be easily computable or at all convenient as a calculable process. We shall rewrite (2) under additional condition, in a form that we can apply an integral formula. That is, we shall write (1), if possible as a Riemann sum.

Now for each i,

Suppose now f is also differentiable on (a, b), then by the Mean Value Theorem, for each 1 £ i £ n there exists h _i in (x_i-1 , x_i ) such that

because f is continuous on [x_i-1 , x_i ] and differentiable on (x_i-1 , x_i ). Thus the length of the polygonal curve P₀P₁¼P_n is given by

------------------- (4)

The expression (4) is then a Riemann sum for the function with respect to the partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b. Therefore, if is Riemann integrable on [a, b], then the limit (2) is then the Riemann integral

. ------------------------------------- (5)

If f ' (x) is defined and Riemann integrable on [a, b], then is Riemann integrable on [a, b] and the arc length is given by the integral formula (5).

Remark.

1. Limit (2) exists if and only if there exists a constant K > 0 such that

for any partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for [a, b]. This can be deduced by observing that

This condition is the definition of f being of bounded variation on [a, b]. (Ref: my article on the Calculus web , Monotone function, Function of bounded variation and the Fundamental Theorem of Calculus.)

2. If f ' (x) is defined and Riemann integrable on [a, b], then is Riemann integrable on [a, b]. (Ref: Theorem 2 in my article on Calculus web, Composition and Riemann integrability.) In this case the arc length is given by the integral formula (5). This is true, in particular, when f ' is defined and continuous on [a, b].

3. If f is absolutely continuous on [a, b], even though f ' (x) may not be defined everywhere on [a, b], the arc length formula (5) still holds but with Riemann integral replaced by Lebesgue integral. (For the definition of absolute continuity, see my article "Change of Variable or substitution in Riemann and Lebesgue Integration" ). The definition of Lebesgue integral is more advanced and a good reference will be the book "Real analysis" by Royden. It is enough to note that some Lebesgue integral is given by an improper Riemann integral. (see Example 2 below). Also, if f is differentiable everywhere on [a, b] and f ' is Lebesgue integrable, then f is absolutely continuous on [a, b].

4. In general, if f is continuous and of bounded variation, then the arc length is given by the arc length formula (5), with Riemann integral replaced by Lebesgue integral plus another term which is the total variation of a singular function g associated with f . Indeed g is given by the function f (x) - F(x), where and the integral here is the Lebesgue integral.

Example 1. Let f (x) = x² . Then the graph of f is the parabola with equation y = x² . The length L of the part of the parabola from (0,0) to (1,1) is given by the Riemann integral

Now , where

.

Therefore, .

Note that in this example, the derived function f ' is continuous on [0, 1].

Example 2. Consider the unit circle defined by x² + y² =1. The arc length L of the minor arc from (0,1) to (1,0) is a quarter of the circumference of the circle. We shall calculate this using our integral formula. Let Then the length of L is given by the integral

, where .

Note that f '(x) is not defined at x = 1. The integral above is an improper integral. The arc length is thus given by

An explanation is in order here. We can also think of the minor arc from (0,1) to (1,0), as the limit of the arc from as t tends to 1 from the left. Thus the arc length L is then the limit of the arc length of the minor arc from which is given by since f ' is continuous on [0, t] for 0 < t < 1. (See Remark 2.)

Note also that f ' is Lebesgue integrable on [0, 1] and f can be expressed as an indefinite (Lebesgue) integral of f ' and so f is absolutely continuous and by remark 3 the arc length is given by the integral formula (5) involving the Lebesgue integral of on [0, 1] which is the improper integral of on [0,1] and equals

Part 2. Volume of solid of revolution.

Suppose f : [a, b] ® R is defined on a closed and bounded interval [a, b]. We take the region bounded by the curve y = f (x), the x-axis , the lines x = a and x = b and form the solid obtained by revolving this region about the x-axis. This solid is called the solid of revolution.

As in part 1, we take a partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for the interval [a, b]. For 1 £ i £ n, consider the solid disk D_i or cylinder formed by revolving the rectangular area of width (x_i - x_i-1 ) and length | f (x_i)|. This is a circular disk or solid cylinder of radius | f (x_i)| and thickness (x_i - x_i-1 ). Therefore, the volume of this disk D_i is Hence, an approximation to the volume of the solid of revolution is the sum of the volumes of these solid disks D_i ,1 £ i £ n. This is

.

Therefore, the volume of the solid of revolution is the limit

--------------------------- (6)

Here as usual ||D|| denotes the norm of the partition D, i.e., the maximum of all the lengths of the subintervals [x_i-1 , x_i] as defined by the partition D.

Note that is a Riemann sum for the function with respect to the partition D. It follows that if ( f (x))² is Riemann integrable on [a, b], then the volume of the solid of revolution is given by the Riemann integral

----------------------------------- (7).

Of course if f (x) is Riemann integrable on [a, b], then ( f (x))² is Riemann integrable on [a, b]. In particular if f is continuous on [a, b], it is then Riemann integrable on [a, b] and so the volume of the solid of revolution is given by (7).

Example 3. Let y = x² be the parabola. Then the volume of the solid of revolution obtained by rotating the region bounded by the parabola, the x-axis and the line x =1 is given by

, where f (x ) = x² , a= 0, b=1

.

Part 3. Area of surface of revolution.

Suppose f : [a, b] ® R is defined on a closed and bounded interval [a, b]. Suppose f is a non-negative function. Then the surface of revolution is obtained by rotating the graph of f or y = f (x) about the x-axis. Take a partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for the interval [a, b]. We can approximate the area of this surface of revolution by the sum total of the area of bands B_i obtained by rotating the line segment joining (x_i-1 , f (x_i-1)) to (x_i , f (x_i )) for i =1 to i = n. Each of these bands is a frustum of a cone. We shall first determine the area of a frustum of a cone before we examine the approximation.

Figure 4 shows a cone with a band B. If we slice the cone along a slant edge VQ we get the band in a sector of a disk with radius VQ which is ( l + l₂ ) and subtended by an angle q as shown in Figure 5.

Here the angle and so . Therefore, the area of the band B is given by

------ (8)

Now we apply formula (8) to our band B_i determined by rotating P_i-1P_i about the x-axis, with r₁ = f (x_i), r₂ = f (x_i-1) and . We get the area of B_i. is given by

Thus, assuming that f (x) ³ 0 on [a, b], an approximation to the area of the surface of revolution is

-------- (9)

Suppose f is differentiable on (a, b). Then by the Mean Value Theorem, for each 1 £ i £ n there exists h _i in (x_i-1 , x_i ) such that

And so (9) becomes

---------------- (10)

If f is continuous and is Riemann integrable on [a, b], (10) may be "approximated" by

----------------------------- (11)

Thus (11) is a Riemann sum for the function . Since by assumption both f (x) and are Riemann integrable, the function is Riemann integrable. Therefore,

.

Hence, if we have that f (x) is continuous and non-negative on [a, b] and if either f ' is defined and continuous on [a, b] or f ' is Riemann integrable on [a, b], then the area of the surface of revolution obtained by rotating the curve y = f (x) about the x-axis is given by

-------------------------- (12)

We shall explain after the following examples what we meant by (10) can be approximated by (11).

Example 4. Find the area of the surface of revolution obtained by rotating the curve y = x² , 0 £ x £ 1, about the x-axis.

The function is f (x) = x² . Obviously, f is continuous on [0, 1], f is differentiable on [0,1 ] and f ' (x) = 2x is continuous on [0, 1]. Therefore the surface area is given by (12)

----------------- (13)

Thus, we need to determine the Riemann integral (13).

Note that

, where 2x = tan(q)

Now by integration by parts, it can be shown that

.

Therefore,

Thus,

'

' .

Hence, the surface area

Example 5. Find the area of the surface of revolution obtained by rotating the curve , 0 £ x £ 1, about the x-axis.

Let Then for 0 £ x <1 and f is not differentiable at x = 1. Then the surface area is the left limit of the surface area S_t of that part of curve with 0 £ x < t rotated 2p radians about the x-axis, as t tends to 1 on the left.

By formula (12), S_t is given as below:

Therefore, the surface area is .

As a consequence the surface are of the unit ball is 4p .

Remark. Why (10) can be "approximated" by (11).

We shall assume that f is continuous and non-negative on [a, b] , f is differentiable on [a, b] and f ' is Riemann integrable on [a, b]. Thus, it follows that is Riemann integrable on [a, b]. The curve y = f (x) , a £ x £ b, is rectifiable and the arc length is given by . Note that since f is continuous on [a, b], it is uniformly continuous on [a, b]. Hence we have for any e > 0, there exists d₁ > 0 such that

------------- (14)

Thus, for each i from 1 to n , if ||D|| < d₁ , we have then that

and also that . This is because and that . We have then that

----------- (15)

Therefore,

Thus we have,

---------------- (16)

Since , there exists d₂ > 0 such that

Therefore, --------------------- (17)

Thus if , we have then from (16) and (17) that

---------------- (18)

This means that given any e > 0, there exists a d > 0 such that for any partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for [a, b],

Hence, can be "approximated" by . Consequently, can be "approximated" by .

Now we proceed to show that the integral formula (12) gives the surface area.

By our assumption that f is continuous and differentiable on [a, b] and f ' is Riemann integrable on [a, b], the function is Riemann integrable on [a, b]. Hence the integral exists and we can then suppose . Then by the definition of the Riemann integral, for any e > 0, there exists a d₃ > 0 such that for any partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for [a, b],

------------------ (19)

Now we take . Then by (18) and (19), for any partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for [a, b],

or equivalently,

This means

Thus .