Pappus Theorem

Pappus theorem (also known as Pappus - Guldinus theorem, pappus theorem) is  named after the Greek Mathematician Pappus of Alexandria (4th century A.D.) It is a theorem that deals with the area of surface of revolution and volume of surfaces and solids of revolution. 

The Pappus Guldinus theorem helps to find the surface area and volume of surfaces and solids of revolution.

The statement of the Pappus theorem was introduced in the print for the first time in 1659, but it was known before by Kepler in 1659, and Guldin in 1640.

The First Pappus Guldinus Theorem

Let C be any curve in the plane. The area of the surface (A) obtained when C is revolved around an external axis is equivalent to the product of the arc length L of C and distance (d) travelled by the Centroid of C.

\[A = Ld\]

For example, the surface area of torus with minor and major radius r and R respectively is:  V = 2\[\pi r\]+ 2\[\pi R\]=  4\[\pi^{2}r R\]

The Second Pappus Guldinus Theorem

Let R be any region in the plane. The volume of the solid obtained when R is revolved around a central axis and is equivalent to the product of the area A of R and the distance covered (d) by the centroid of R.

\[V = Ad\]

For example, the volume of torus with minor and major radius r and R respectively is:

\[V = (\pi r^{2})(2\pi R) = 2\pi^{2}Rr^{2}\]

This special case was introduced by Johannes Kepler using infinitesimals.

Examples

Let us consider the cylinder obtained by revolving a rectangle with vertical side h and horizontal side r around one of its vertical sides (say its left side). The surface area of the cylinder, excluding the top and bottom, is calculated using the Pappus theorem as the surface is obtained by revolving its right side around its left side. The length of the arc of its right side is h and the distance travelled by its centroid is simply 2r. Therefore, the area of the circular cylinder will be 2\[\pi rh\].

The volume of the cylinder is the distance travelled by its centroid multiplied by the area rh of the rectangle. The centroid of the rectangle is considered as its centre, which is a distance \[\frac{r}{2}\] from the axis of revolution. Hence, it covers a distance of  2\[\pi\] \[\frac{r}{2}\]  as it revolves. Therefore, the volume of the cylinder is \[(\pi r)\] \[(rh)\]=\[\pi r^{2}h\].

Pappus And Guldinus Theorem Examples With Solutions

Example 1:

The surface area and volume of torus can easily be calculated using the Pappus theorem. Considering the radius of circle r, and distance from the centre of the circle to the axis of its revolution is R, then the surface area of torus can be calculated using the first Pappus theorem as:

\[A = Ld\] = \[2 \pi r\].\[2\pi R\]= \[4\pi^{2}rR\] 

(Image will be uploaded soon)

The volume inside the torus can be calculated using the second Pappus theorem as:

\[V = Ad\] = \[\pi r^{2}\].\[2\pi R\]= \[2\pi^{2}r^{2}R\]

Example 2: 

Find the volume of the right circular cone with radius r and height (h) = 10 

The Pappus theorem for volume is defined as\[V = Ad\]. Before calculating the volume, it is important to find the area of the triangle to which we are revolving. The right circular cone shape can be considered as a triangle rotated around an axis. The formula for the area of triangle is given as:

\[\frac{1}{2}\]\[base \times height\]

The base of the triangle will be considered as the radius (r) = 6 and the height (h) of the triangle = 10.

A = \[\frac{1}{2} \times 6 \times 10\] 

A = 30

The next step is to solve for distance (d). Distance will include the relationship of the centroid of the triangle and its rotation. 

In other words, the distance (d) = \[2\pi \bar{x}\], where \[\bar{x}\] is the coordinate of the centroid and 2refers to the fact that the object is being rotated around an axis. Therefore, the equation for \[\bar{x}\] is given as:

\[\bar{x}\]= \[\frac{1}{A}\] \[\int_{a}^{b}\] \[x f(x) dx\]

Here, we are still missing the f(x), which is the third side of the triangle. There are different ways to calculate the third side of the triangle. But, we will use the pythagorean theorem which states that

H² = h² + b²

Here, H is the hypotenuse, b is the base, and h is the height

Now, we will solve for H and substitute the values as shown below:

H = \[\sqrt{h^{2} + b^{2}}\]

H = \[\sqrt{(10)^{2} + (6)^{2}}\]

H = \[\sqrt{136}\]

Note : H =\[f(x)\]

Hence, \[f (x)\] = 0

Let us now solve for \[\bar{x}\]

\[\bar{x}\] = \[\frac{1}{30}\] \[\int_{0}^{6}\] \[x \sqrt{136} dx\]

\[\bar{x}\] = \[\frac{\sqrt{136}}{30}\] \[\int_{0}^{6}\] \[x\] \[dx\]

\[\bar{x}\] =\[\frac{\sqrt{136}}{30}\] \[(\frac{x^{2}}{x})\] \[\int_{0}^{6}\] 

\[\bar{x}\] =\[\frac{\sqrt{136}}{30}\] \[[\frac{(6)^{2}}{2} - \frac{(0)^{2}}{2}]\]

\[\bar{x}\] =\[\frac{6\sqrt{34}}{5}\]

Now, let us solve for distance d = 2\[\lambda\] \[\bar{x}\] 

\[d = 2 \pi \frac{6\sqrt{34}}{5}\]

\[d = \frac{12\pi\sqrt{34}}{5}\]

At last, we will solve for volume using V = Ad

V = (20) \[(\frac{12\pi\sqrt{34}}{5})\]

V = 72\[\pi \sqrt{34}\]

Therefore, the volume of right circular cone with base radius 6 and height 10 is V = 72\[\pi \sqrt{34}\]