We find the volume of the 'lower third' of a hemisphere of radius R--i.e., of the indicated portion of the sphere lying between its lowest point and a horizontal plane at y = 1/3 R.

• As usual we begin by looking at a slice of thickness `dyi and vertical position yi.
• We will be able to determine the volume of this slice once we know its radius ri.

The volume of the slice is therefore Vi = c.s. area * thickness = Ai `dyi = `pi ri^2 `dyi, as indicated below.

• If we partition the y axis in the usual manner, the total volume will therefore be approximated by the sum in the third line below.
• The sum is rearranged and taken to the limit in lines 4-7.
• The total volume will therefore be the integral in the last line.

Video Clip #01

Suppose now that we wish to find the total mass of a sphere if the density changes linearly with altitude from `rho = 2 to `rho = 3, as indicated below.

• We choose to measure yi from the bottom of the sphere, and we consider a slice of thickness `dyi near vertical position yi.
• Using the geometry of a circle we obtain the indicated expression for the radius ri. 
• The approximate mass mi contained in this slice will be the product of the density function `rho evaluated position yi and the area Ai `dyi, as indicated in the last two lines of the figure below.

The density function can be represented by a straight line from the point (0,3) to (2R, 2) on a graph of `rho vs. y.

• Using any of several methods we find that the density function is `rho(y) = 3 - y / (2 R).

Assuming the usual type of partition of the y interval from 0 to 2R, we obtain the mass approximation in the first line by substituting 3 - yi / (2 R) for the density function.

• The sum approaches the indicated integral as its limit.

The torque about the left-hand side of the sphere associated with a slice at position yi will be the first moment `taui = mi yi, as indicated below.

• Setting up the Riemann sum and taking it to the limit we obtain the indicated integral for the total torque.
• The position of the center of mass will be the quotient yCM = `tau / m, where `tau is the total torque and m the total mass.

Video Clip #02

The torque of an object about a point is, as we have seen, defined as the sum of the m x contributions, where x is the horizontal distance from an assumed center of rotation to the mass m.

• The torque, as we have stated, is a first moment.

The moment of inertia of an object, which determines among other things its energy as it rotates about a given axis, is the sum of the m x^2 contributions, where x is the distance from the axis of rotation to the mass.

• The moment of inertia, which involves the mass and the second power of x, is called a second moment.

In the figures below we will calculate the moment of inertia I = sum( m x^2 ) about the center of mass yCM, for the sphere of the present example.

• We begin by finding the distance xi = | yi - yCM | of the slice at position yi from the center of mass yCM.

We will obtain the sum of terms Ii = (yi - yCM) * mi.

• Using the expression for mi used in the previous analysis, we obtain a sum which approaches the integral indicated in the figure below.

Recall the tangent-line approximation f(x) = f(a) + f'(a) (x - a).

The linear approximation starts at the point (a, f(a)) and follows the slope of the graph of f(x) at the point f(a).

If there is a curvature to the graph, then the graph of f(x) will typically curve away from that of the linear approximation slowly at first, then more and more rapidly until the linear approximation no longer provides a good approximation to the function.

The quadratic approximation allows us to follow the curvature at point a of the graph of f(x).

If the second derivative of f(x) is not constant, then this quadratic approximation will at first gradually, then more and more rapidly, diverge from the actual graph.

The linear approximation is said to be a first-order approximation; quadratic approximation is said to be a second-order approximation.

• We can continue this process to higher and higher orders of approximation, using cubic (degree-3) approxinations and beyond.
• Higher-order approximations typically continue to provide better and better approximations over greater in greater distances.

• We say that the function f(x) is equal to its nth-order approximation, expressed in the first to lines below, plus an error term En(x).
• We will study the error term later, and concentrate now on approximation process.

The nth-order approximation is referred to as the Taylor polynomial of degree n, or as the nth-order Taylor polynomial.

Video Clip #04

As an example, we calculate the third-order approximation of the function f(x) = sin(x), about the point x = a.

We calculate the first five derivatives of sin(x):

• f'(x) = ( sin(x) ) ' = cos(x)
• f''(x) = ( sin(x) ) '' = -sin(x)
• f'''(x) = ( sin(x) ) ''' = -cos(x)
• f''''(x) = ( sin(x) ) '''' = sin(x)
• f'''''(x) = ( sin(x) ) ''''' = cos(x).

We then evaluate f(x) and its derivatives that a = 0:

• f(0) = sin(0) = 0
• f'(0) = cos(0) = 1
• f'(0) = -sin(0) = 0
• f'''(0) = -cos(0) = -1
• f'''(0) = sin(0) = 0
• f'''''(0) = cos(0) = 1.

We now plug the values into the expression for the fifth-order Taylor polynomial:

• f(a) + f'(a) (x-a) + f''(a) (x-a)^2 / 2! + f'''(a) (x-a)^3 / 3! + f''''(a) (x-a)^4 / 4! + f'''''(a) (x-a)^5 / 5!
• = 0 + 1(x-0) + 0(x-0)^2 / 2! + -1(x-0)^3 / 3! + 0(x-0)^4 / 4! + 1(x-0)^5 / 5!
• = x - x^3/3! + x^5 / 5!
• = x - x^3/6 + x^5 / 120.

From the pattern established here, you should be able to write down as many terms as desired of the undending Taylor series

• sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + .....
• This series, which by the ... at the end is indicated to go on forever, is equal to the function sin(x) in the sense that for any x, the series converges the value sin(x).
• Note that the factorials in the denominators become extremely large very quickly.