Calculus II Class 02/21

The 'arc; of the graph of y = x^2 between x = 0 and x = 3 is approximated by the total length of the 'slope lines' at the tops of the trapezoids.

• For each 'slope line' we easily find the rise and the run.
• Knowing the rise and the run we easily find the length of the line, using the Pythagorean Theorem.

We obtain the following specific results:

• For the first trapezoid the run is 1 and the rise is 1 - 0 = 0 so the length of the slope line is sqrt( 1^2 + 1^2) = sqrt(2) = 1.4 approx.
• For the second  trapezoid the run is 1 and the rise is 4 - 1 = 3 so the length of the slope line is sqrt( 1^2 + 3^2) = sqrt(10) = 3.2 approx.
• For the third trapezoid the run is 1 and the rise is 9 - 4 = 5 so the length of the slope line is sqrt( 1^2 + 5^2) = sqrt(26) = 5.1 approx.

The total arc distance is therefore about 1.4 + 3.2 + 5.1 = 9.7.

Find the length of a 'slope segment' corresponding to slope m on a trapezoid of width 2. 

We again find the rise and the run of the 'slope segment', which is shown below.

• The run is the width 2 of the trapezoid.
• The rise is `dy = m * `dx = m * 2 = 2 m.
• It follows that the length of the segment is the hypotenuse sqrt( 2^2 + ( 2m)^2).
• This expression is easily simplified to the form 2 sqrt(1 + m^2).

Find the length of a 'slope segment' corresponding to slope m on a trapezoid of width `dx. 

We again find the rise and the run of the 'slope segment', which is shown below.

• The run is the width dx of the trapezoid.
• The rise is `dy = m * `dx,
• It follows that the length of the segment is the hypotenuse sqrt( 2^2 + ( m `dx)^2).
• This expression is easily simplified to the form sqrt(1 + m^2) * `dx

Find the length of a 'slope segment' corresponding a trapezoidal approximation of width `dx, located in the vicinity of coordinate x, for the function y = x^2. 

We again wish find the rise and the run of the 'slope segment', which is shown below.

• The run is the width `dx of the trapezoid.
• The slope is close to the slope at position x.  If y = x^2 then this slope is y ' = 2 x.
• The rise is `dy = slope * `dx = 2x * `dx
• It follows that the length of the segment is the hypotenuse sqrt( `dx^2 + ( 2x * `dx)^2).
• This expression is easily simplified to the form sqrt(1 + 4 x^2) * `dx.

Can we use these ideas to obtain an accurate expression for the arc length of the y = x^2 curve between x = 0 and x = 3?

If we consider a typical small interval `dx containing coordinate x, we find from the preceding analysis that the arc distance is very close to `dL = sqrt(1 + 4x^2) * `dx.

Summing for a set of intervals which covers the interval from x = 0 to x = 3 we obtain sum ( sqrt(1+4x^2) `dx), from x = 0 to 3.

In the limit as `dx -> 0 on all intervals, the approximations   `dL = sqrt(1 + 4x^2) * `dx become more and more accurate, and the sum of all the approximations approaches the accurate arc length.

The limiting value of the sum is the integral indicated below.

It is possible to integrate sqrt(1 + 4x^2) with respect to x, from x = 0 to x = 3, using first a standard trigonometric substitution.

• We first substitute x = 1/2 tan(theta), as indicated below.
• We end up having to integrate 1 / cos^3(theta).
• This integral is not at all simple.  We get antiderivative LN(TAN((2 theta + pi)/4))/2 + SIN(theta)/(2 COS^2(theta)); substituting theta = arctan(2 x) we get LN(sqrt(4x^2 + 1) + 2x)/4 + x sqrt(4·x^2 + 1)/2
• Evaluating the antiderivative at the endpoints and subtracting we find that the arc length is 9.747.

Find the volume of the solid of revolution formed by revolving y = x^2 about the x axis.

The solid is depicted in the figure below.

A side view of the solid shows a typical interval near coordinate x with width `dx.

• This interval will correspond to a slice of the solid.  A typical slice is shown in the second figure below.
• If `dx is small then the average cross-sectional area of the slice must be nearly equal to the cross-sectional area of a cut made at x.
• The radius of a cross-section at x is the distance from the x axis to the curve y = x^2; the radius is therefore x^2.
• The c.s. area of a cut made at x is therefore pi r^2 = pi (x^2)^2 = pi x^4, as shown in the second figure below.
• The thickness of the slice is `dx so the volume of the slice is

`dV = c.s. area * thickness = pi x^4 `dx.

• Summing such volumes from x = 0 to x = 4 and taking the limit as `dx -> 0 results, as shown below, in the integral

volume = int( pi x^4, x, 0, 3).

This integral is easily calculated.  The antiderivative will be pi x^5 / 3.  Between x = 0 and x = 3 this antiderivative changes by pi * 3^5 / 3 - pi * 0^5 / 3 = 81 pi.

The volume of this solid is therefore 81 pi.