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

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

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

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

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

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.

1. Horizontal line segment $$AB.$$
The length is $${L_{AB}} = 3.$$ The centroid is located at the point $${G_{AB}} = \left( {3.5,10} \right);$$
2. Vertical line segment $$BC.$$
The length is $${L_{BC}} = 2.$$ The centroid is located at the point $${G_{BC}} = \left( {2,9} \right);$$
3. 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:

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

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,

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

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 $$\alpha = \frac{\pi }{4}:$$

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

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:

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