Pappus’s theorem (also known as Pappus’s centroid theorem, Pappus-Guldinus theorem or the Guldinus theorem) deals with the areas of surfaces of revolution and with the volumes of solids of revolution.

The Pappus’s theorem is actually two theorems that allow us to find surface areas and volumes without using integration.

### Pappus’s Theorem for Surface Area

The first theorem of Pappus states that the surface area \(A\) of a surface of revolution obtained by rotating a plane curve \(C\) about a non-intersecting axis which lies in the same plane is equal to the product of the curve length \(L\) and the distance \(d\) traveled by the centroid of \(C:\)

\[A = Ld.\]

### Pappus’s Theorem for Volume

The second theorem of Pappus states that the volume of a solid of revolution obtained by rotating a lamina \(F\) about a non-intersecting axis lying in the same plane is equal to the product of the area \(A\) of the lamina \(F\) and the distance \(d\) traveled by the centroid of \(F:\)

\[V = Ad.\]

### Surface Area and Volume of a Torus

A torus is the solid of revolution obtained by rotating a circle about an external coplanar axis.

We can easily find the surface area of a torus using the \(1\text{st}\) Theorem of Pappus. If the radius of the circle is \(r\) and the distance from the center of circle to the axis of revolution is \(R,\) then the surface area of the torus is

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

The volume inside the torus is given by the \(2\text{nd}\) Theorem of Pappus:

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

The Pappus’s theorem can also be used in reverse to find the centroid of a curve or figure.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A regular hexagon of side length \(a\) is rotated about one of the sides. Find the volume of the solid of revolution.### Example 2

Find the centroid of a uniform semicircle of radius \(R.\)### Example 3

An ellipse with the semimajor axis \(a\) and semiminor axis \(b\) is rotated about a straight line parallel to the axis \(a\) and spaced from it at a distance \(m \gt b.\) Find the volume of the solid of revolution.### Example 4

A triangle with the vertices \(A\left( {1,2} \right),\) \(B\left( {2,6} \right),\) \(C\left( {6,2} \right)\) is rotated about the \(x-\)axis. Find the volume of the solid of revolution thus obtained.### 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 1.

A regular hexagon of side length \(a\) is rotated about one of the sides. Find the volume of the solid of revolution.Solution.

Given the side of the hexagon \(a,\) we can easily find the the apothem length \(m:\)

\[m = \frac{a}{2}\cot 30^{\circ} = \frac{{a\sqrt 3 }}{2}.\]

Hence, the distance \(d\) traveled by the centroid \(C\) when rotating the hexagon is written in the form

\[{d = 2\pi m = 2\pi \cdot \frac{{a\sqrt 3 }}{2} }={ \pi a\sqrt 3 .}\]

The area \(A\) of the hexagon is equal to

\[A = {a^2}\frac{{3\sqrt 3 }}{2}.\]

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

\[{V = Ad = {a^2}\frac{{3\sqrt 3 }}{2} \cdot \pi a\sqrt 3 }={ \frac{{9\pi {a^3}}}{2}.}\]

### Example 2.

Find the centroid of a uniform semicircle of radius \(R.\)Solution.

Let \(m\) be the distance between the centroid \(G\) and the axis of rotation. When the semicircle makes the full turn, the path \(d\) traversed by the centroid is equal to

\[d = 2\pi m.\]

The solid of rotation is a ball of volume

\[V = \frac{{4\pi {R^3}}}{3}.\]

By the \(2\text{nd}\) theorem of Pappus, we have the relationship

\[V = Ad,\]

where \(A = \large{\frac{{\pi {R^2}}}{2}}\normalsize\) is the area of the semicircle.

Hence,

\[{m = \frac{V}{{2\pi A}} }={ \frac{{\frac{{4\pi {R^3}}}{3}}}{{2\pi \cdot \frac{{\pi {R^2}}}{2}}} }={ \frac{{4R}}{{3\pi }} }\approx{ 0.42R}\]

### Example 3.

An ellipse with the semimajor axis \(a\) and semiminor axis \(b\) is rotated about a straight line parallel to the axis \(a\) and spaced from it at a distance \(m \gt b.\) Find the volume of the solid of revolution.Solution.

The volume of the solid of revolution can be determined using the \(2\text{nd}\) theorem of Pappus:

\[V = Ad.\]

The path \(d\) traversed in one turn by the centroid of the ellipse is equal to

\[d = 2\pi m.\]

The area of the ellipse is given by the formula

\[A = \pi ab.\]

Hence, the volume of the solid is

\[{V = Ad = \pi ab \cdot 2\pi m }={ 2{\pi ^2}mab.}\]

In particular, when \(m = 2b,\) the volume is equal to \(V = 4{\pi^2}a{b^2}.\)

### Example 4.

A triangle with the vertices \(A\left( {1,2} \right),\) \(B\left( {2,6} \right),\) \(C\left( {6,2} \right)\) is rotated about the \(x-\)axis. Find the volume of the solid of revolution thus obtained.Solution.

Since the coordinates of the vertices are known, we can easily find the area of the triangle. First we calculate the determinant:

\[{\Delta = \left| {\begin{array}{*{20}{c}} {{x_B} – {x_A}}&{{y_B} – {y_A}}\\ {{x_C} – {x_A}}&{{y_C} – {y_A}} \end{array}} \right| }={ \left| {\begin{array}{*{20}{c}} 1&4\\ 5&0 \end{array}} \right| }={ – 20.}\]

Then the area of the triangle is

\[A = \frac{1}{2}\left| \Delta \right| = 10.\]

Now we determine the centroid of the triangle:

\[{\bar x = \frac{{{x_A} + {x_B} + {x_C}}}{3} }={ \frac{{1 + 2 + 6}}{3} }={ 3;}\]

\[{\bar y = \frac{{{y_A} + {y_B} + {y_C}}}{3} }={ \frac{{2 + 6 + 2}}{3} }={ \frac{{10}}{3}.}\]

Thus,

\[G\left( {\bar x,\bar y} \right) = G\left( {3,\frac{{10}}{3}} \right).\]

By the \(2\text{nd}\) theorem of Pappus, the volume of the solid of revolution is given by

\[V = Ad = 2\pi mA,\]

where \(m = \bar y\) is the distance from the centroid \(G\) to the axis of rotation.

This yields:

\[{V = 2\pi \cdot \frac{{10}}{3} \cdot 10 }={ \frac{{200\pi }}{3}}\]

### 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

\[{V_y} = \frac{{\pi {a^2}b}}{3}.\]

The area of the triangle is

\[A = \frac{{ab}}{2}.\]

Then, by the Pappus’s theorem,

\[{{V_y} = 2\pi \bar xA,}\;\; \Rightarrow {\bar x = \frac{{V_y}}{{2\pi A}} = \frac{{\frac{{\pi {a^2}b}}{3}}}{{2\pi \cdot \frac{{ab}}{2}}} }={ \frac{a}{3}}\]

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

\[{V_x} = \frac{{\pi a{b^2}}}{3}\]

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

\[{{V_x} = 2\pi \bar yA,}\;\; \Rightarrow {\bar y = \frac{{{V_x}}}{{2\pi A}} = \frac{{\frac{{\pi a{b^2}}}{3}}}{{2\pi \cdot \frac{{ab}}{2}}} }={ \frac{b}{3}}\]

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

\[G\left( {\bar x,\bar y} \right) = G\left( {\frac{a}{3},\frac{b}{3}} \right),\]

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, \(\bar x = \large{\frac{\pi }{2}}\normalsize,\) so we need to calculate only the coordinate \(\bar y = m.\)

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

\[{V = \pi \int\limits_a^b {{f^2}\left( x \right)dx} }={ \pi \int\limits_0^\pi {{{\sin }^2}xdx} }={ \frac{\pi }{2}\int\limits_0^\pi {\left( {1 – \cos 2x} \right)dx} }={ \frac{\pi }{2}\left. {\left( {x – \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi }={ \frac{{{\pi ^2}}}{2}.}\]

The area under the sine curve is

\[{A = \int\limits_a^b {f\left( x \right)dx} }={ \int\limits_0^\pi {\sin xdx} }={ \left. {\left( { – \cos x} \right)} \right|_0^\pi }={ – \cos \pi + \cos 0 }={ 2.}\]

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

\[V = Ad = 2\pi mA.\]

Hence,

\[{m = \frac{V}{{2\pi A}} = \frac{{\frac{{{\pi ^2}}}{2}}}{{4\pi }} }={ \frac{\pi }{8} }\approx{ 0.39}\]

Thus, the centroid of the region has the coordinates

\[G\left( {\bar x,\bar y} \right) = G\left( {\frac{\pi }{2},\frac{\pi }{8}} \right).\]

### 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 \[{{{\bar x}_{CD}} = 2 + \frac{{2R}}{\pi } = 2 + \frac{6}{\pi };\;\;}\kern0pt{{{\bar y}_{CD}} = 5.}\]

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

\[\bar x = \frac{{{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}}}{L},\]

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

\[A = Ld = 2\pi mL,\]

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,

\[\require{cancel}{A \text{ = }}\kern0pt{2\pi \cdot \frac{{{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}}}{\cancel{L}} \cdot \cancel{L} }={ 2\pi \left( {{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}} \right) }={ 2\pi \left[ {3.5 \cdot 3 + 2 \cdot 2 + \left( {2 + \frac{6}{\pi }} \right) \cdot 3\pi } \right] }={ 61\pi + 12{\pi ^2} }\approx{ 310}\]

### 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

\[AG = \frac{{a\sqrt 2 }}{2}.\]

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

\[\beta = 45 ^{\circ}- \alpha .\]

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

\[{m = KG = AG\cos \beta }={ \frac{{a\sqrt 2 }}{2}\cos \left( {45^{\circ} – \alpha } \right).}\]

Using the cosine subtraction identity

\[{\cos \left( {A – B} \right) }={ \cos A\cos B + \sin A\sin B,}\]

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

\[{m = \frac{{a\sqrt 2 }}{2}\cos \left( {45^{\circ} – \alpha } \right) }={ \frac{{a\sqrt 2 }}{2}\left( {\cos 45^{\circ}\cos \alpha + \sin 45^{\circ}\sin \alpha } \right) }={ \frac{{a\sqrt 2 }}{2}\left( {\frac{{\sqrt 2 }}{2}\cos \alpha + \frac{{\sqrt 2 }}{2}\sin \alpha } \right) }={ \frac{a}{2}\left( {\cos \alpha + \sin \alpha } \right),}\]

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

\[{d = 2\pi m }={ \pi a\left( {\cos \alpha + \sin \alpha } \right).}\]

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

\[{V = Ad }={ {a^2} \cdot \pi a\left( {\cos \alpha + \sin \alpha } \right) }={ \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right).}\]

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

\[{V\left( \alpha \right) }={ \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right),}\]

\[{V^\prime\left( \alpha \right) }={ \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right)^\prime }={ \pi {a^3}\left( { – \sin \alpha + \cos \alpha } \right).}\]

\[{V^\prime\left( \alpha \right) = 0,}\;\; \Rightarrow {\pi {a^3}\left( {\cos \alpha – \sin \alpha } \right) = 0,}\;\; \Rightarrow {\tan \alpha = 1,}\;\; \Rightarrow {\alpha = \frac{\pi }{4}.}\]

By the second derivative test,

\[{V^{\prime\prime}\left( \alpha \right) = \pi {a^3}\left( { – \sin \alpha + \cos \alpha } \right)^{\prime} }={ – \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right),}\]

and, consequently,

\[{V^{\prime\prime}\left( {\frac{\pi }{4}} \right) }={ – \pi {a^3}\left( {\cos \frac{\pi }{4} + \sin \frac{\pi }{4}} \right) }={ – \pi {a^3}\sqrt 2 \lt 0.}\]

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

\[{{V_{\max }} = V\left( {\frac{\pi }{4}} \right) }={ \pi {a^3}\left( {\cos \frac{\pi }{4} + \sin \frac{\pi }{4}} \right) }={ \pi {a^3}\sqrt 2 .}\]

### 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

\[V = Ad = 2\pi mA,\]

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:

\[{V = \pi \int\limits_a^b {{f^2}\left( x \right)dx} }={ \pi \int\limits_{ – 2}^2 {{{\left( {4 – {x^2}} \right)}^2}dx} }={ 2\pi \int\limits_0^2 {{{\left( {4 – {x^2}} \right)}^2}dx} }={ 2\pi \int\limits_0^2 {\left( {16 – 8{x^2} + {x^4}} \right)dx} }={ 2\pi \left. {\left( {16x – \frac{{8{x^3}}}{3} + \frac{{{x^5}}}{5}} \right)} \right|_0^2 }={ 2\pi \left( {32 – \frac{{64}}{3} + \frac{{32}}{5}} \right) }={ \frac{{512\pi }}{{15}}.}\]

The area of the region is

\[{A = \int\limits_a^b {f\left( x \right)dx} }={ \int\limits_{ – 2}^2 {\left( {4 – {x^2}} \right)dx} }={ 2\int\limits_0^2 {\left( {4 – {x^2}} \right)dx} }={ 2\left. {\left( {4x – \frac{{{x^3}}}{3}} \right)} \right|_0^2 }={ 2\left( {8 – \frac{8}{3}} \right) }={ \frac{{32}}{3}.}\]

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

\[{m = \frac{V}{{2\pi A}} }={ \frac{{\frac{{512\pi }}{{15}}}}{{2\pi \cdot \frac{{32}}{3}}} }={ \frac{8}{5}.}\]

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

\[G\left( {\bar x,\bar y} \right) = G\left( {0,\frac{8}{5}} \right).\]

### 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

\[G\left( {\bar x,\bar y} \right) = G\left( {0,m} \right),\]

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

\[A = 2\pi Rh,\]

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

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

\[A = 2\pi R \cdot 2R\sin \alpha = 4\pi {R^2}\sin \alpha .\]

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

\[A = dL = 2\pi mL,\]

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

\[{m = \bar y = \frac{A}{{2\pi L}} }={ \frac{{4\pi {R^2}\sin \alpha }}{{2\pi \cdot 2\alpha R}} }={ \frac{{R\sin \alpha }}{\alpha }.}\]

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 \({\alpha = \large{\frac{\pi }{2}}\normalsize},\) then \[m\left( {\alpha = \frac{\pi }{2}} \right) = \frac{{2R\sin \frac{\pi }{2}}}{\pi } = \frac{{2R}}{\pi }.\]