Calculus

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Pappus’s Theorem

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 surface area
Figure 1.

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 \(F:\)

\[V = Ad.\]
Pappus's theorem for volume
Figure 2.

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

\[A = Ld = 2\pi r \cdot 2\pi R = 4{\pi ^2}rR.\]
Torus as a solid of revolution
Figure 3.

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 1.

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

Solution.

A regular hexagon rotating about one of the sides.
Figure 4.

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.

Centroid of a semicircle of radius R.
Figure 5.

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 \(A = \frac{{\pi {R^2}}}{2}\) 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.

Solid of revolution obtained by rotating an ellipse around an external axis.
Figure 6.

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.

Solid of revolution obtained by rotating a triangle with vertices A(1,2), B(2,6), C(6,2).
Figure 7.

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:

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

See more problems on Page 2.

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