# Pappus’s Theorem

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

Find the centroid of a right triangle with legs *a, b*.

### Example 6

Find the centroid of the region enclosed by a half-wave of the sine curve and the *x*-axis.

### Example 7

A curve shown in Figure \(10\) is rotated about the \(y-\)axis. Find the area of the surface of revolution.

### Example 8

A square of side \(a\) is rotated about an axis lying in its plane and passing through one of the vertices. The angle between the square side and positive direction of the axis of rotation is \(\alpha.\) What is the volume of the obtained solid of revolution?

### Example 9

Find the centroid of the region bounded by the parabola \(y = 4 - {x^2}\) and the \(x-\)axis.

### Example 10

A circular arc of radius \(R\) subtending the central angle \(2\alpha\) is rotated about the \(x-\)axis as shown in Figure \(13.\) Determine the centroid of the arc.

### Example 5.

Find the centroid of a right triangle with legs \(a, b.\)

Solution.

To determine the coordinates of the centroid, we will use the \(2\text{nd}\) theorem of Pappus.

Suppose first that the triangle is rotated about the \(y-\)axis. The volume of the obtained cone is given by

The area of the triangle is

Then, by the Pappus's theorem,

Let the triangle rotate now about the \(x-\)axis. Similarly, we find the volume

and the \(\bar y-\)coordinate of the centroid:

Thus, the centroid of the triangle is located at the point

which is the point of intersection of its medians.

### Example 6.

Find the centroid of the region enclosed by a half-wave of the sine curve and the \(x-\)axis.

Solution.

Let the point \(G\left( {\bar x,\bar y} \right)\) denote the centroid of the figure. By symmetry, \(\bar x = \frac{\pi }{2},\) so we need to calculate only the coordinate \(\bar y = m.\)

Using the disk method, we find the volume of the solid of revolution:

The area under the sine curve is

The \(2\text{nd}\) theorem of Pappus states that

Hence,

Thus, the centroid of the region has the coordinates

### Example 7.

A curve shown in Figure \(10\) is rotated about the \(y-\)axis. Find the area of the surface of revolution.

Solution.

We consider separately three sections of the curve and compute their centroids.

- Horizontal line segment \(AB.\)

The length is \({L_{AB}} = 3.\) The centroid is located at the point \({G_{AB}} = \left( {3.5,10} \right);\) - Vertical line segment \(BC.\)

The length is \({L_{BC}} = 2.\) The centroid is located at the point \({G_{BC}} = \left( {2,9} \right);\) - Semicircular arc \(CD.\)

The length is \({L_{CD}} = \pi R = 3\pi.\) The centroid is located at the point \({G_{CD}} = \left( {\bar x_{CD},\bar y_{CD}} \right),\) where\[{{\bar x}_{CD}} = 2 + \frac{{2R}}{\pi } = 2 + \frac{6}{\pi };\;\; {{\bar y}_{CD}} = 5.\]

Calculate the \(\bar x-\)coordinate of the centroid \(G\) of the whole curve:

where \(L = {L_{AB}} + {L_{BC}} + {L_{CD}}\) is the total length of the curve.

By the \(1\text{st}\) theorem of Pappus, the surface area is given by

where \(d\) is the path traversed by the centroid of the curve in one turn and \(m = \bar x\) is the distance from the centroid to the \(y-\)axis.

Hence,

### Example 8.

A square of side \(a\) is rotated about an axis lying in its plane and passing through one of the vertices. The angle between the square side and positive direction of the axis of rotation is \(\alpha.\) What is the volume of the obtained solid of revolution?

Solution.

The half of the diagonal of the square \(AG\) has the length

The angle \(\beta = \angle KGA\) is expressed in terms of \(\alpha\) as follows:

Hence, the distance \(m\) from the centroid \(G\) to the axis of revolution is given by

Using the cosine subtraction identity

we write the distance \(m\) in the form

so the path \(d\) traversed by the centroid \(G\) of the square is given by

Applying the \(2\text{nd}\) theorem of Pappus, we find the volume of the solid of revolution:

Considering the volume as a function of angle \(\alpha\) we can determine its largest value:

By the second derivative test,

and, consequently,

So the volume has the maximum at \(\alpha = \frac{\pi }{4}:\)

### Example 9.

Find the centroid of the region bounded by the parabola \(y = 4 - {x^2}\) and the \(x-\)axis.

Solution.

Using the Pappus's theorem for volume, we have

where \(A\) is the area of the region and \(m\) is the \(\bar y-\)coordinate of the centroid \(G\left( {\bar x,\bar y} \right).\)

We compute the volume of the solid of revolution using the disk method:

The area of the region is

Then the \(\bar y-\)coordinate of the centroid is given by

Due to symmetry of the region, the \(\bar x-\)coordinate is equal to \(0,\) so the final answer is

### Example 10.

A circular arc of radius \(R\) subtending the central angle \(2\alpha\) is rotated about the \(x-\)axis as shown in Figure \(13.\) Determine the centroid of the arc.

Solution.

By symmetry, the centroid \(G\) is located on the \(y-\)axis, so its coordinates are

where \(m = \bar y\) is the distance from the centroid to the axis of rotation that we're going to find.

When the arc is rotated it forms a spherical segment. The surface area \(A\) of the spherical segment is given by

where \(h\) is the distance between the parallel planes cutting the sphere.

Since \(h = 2R\sin\alpha,\) we can write

From the other side, by the \(1\text{st}\) theorem of Pappus, we have

where \(d = 2\pi m\) is the path traversed by the centroid in one turn and \(L = 2\alpha R\) is the length of the arc.

From here it follows that

We can highlight some particular cases:

- If \(\alpha = 0,\) then
\[m\left( {\alpha = 0} \right) = \lim\limits_{\alpha \to 0} \frac{{R\sin \alpha }}{\alpha } = R\,\underbrace {\lim\limits_{\alpha \to 0} \frac{{\sin \alpha }}{\alpha }}_1 = R;\]
- If \(\alpha = \pi,\) then
\[m\left( {\alpha = \pi } \right) = \frac{{R\sin \pi }}{\pi } = 0;\]
- If \({\alpha = \frac{\pi }{2}},\) then
\[m\left( {\alpha = \frac{\pi }{2}} \right) = \frac{{2R\sin \frac{\pi }{2}}}{\pi } = \frac{{2R}}{\pi }.\]